Performance of iterating over Array vs List - performance

Inspired by this question, I wanted to see if there were any performance differences between iterating over an Array vs a List.
Since we would iterating over the entire collection, my initial thought was that there shouldn't really be a performance difference between the two. Furthermore, I thought that using a tail recursive function to do a count should be as fast as just using a mutable variable. However, when I wrote a simple script to test the difference, I found the following (run in Release Mode with VS2015):
add_k_list, elapsed 15804 ms, result 0L
add_k_list_mutable, elapsed 12800 ms, result 0L
add_k_array, elapsed 15719 ms, result 0L
I'm wonder why the list addition implementation that uses a mutable variable is decently faster than both tail recursive version and the one using a mutable variable and an array.
Here's my code:
open System.Diagnostics
let d = 100000
let n = 100000
let stopWatch =
let sw = Stopwatch ()
sw.Start ()
sw
let testList = [1..d]
let testArray = [|1..d|]
let timeIt (name : string) (a : int -> int list -> 'T) : unit =
let t = stopWatch.ElapsedMilliseconds
let v = a 0 (testList)
for i = 1 to (n) do
a i testList |> ignore
let d = stopWatch.ElapsedMilliseconds - t
printfn "%s, elapsed %d ms, result %A" name d v
let timeItArr (name : string) (a : int -> int [] -> 'T) : unit =
let t = stopWatch.ElapsedMilliseconds
let v = a 0 (testArray)
for i = 1 to (n) do
a i testArray |> ignore
let d = stopWatch.ElapsedMilliseconds - t
printfn "%s, elapsed %d ms, result %A" name d v
let add_k_list x (k_range: int list) =
let rec add k_range x acc =
match k_range with
| [] -> acc
| k::ks -> let y = x ^^^ k
if (y < k || y > d) then
add ks x (acc + 1L)
else
add ks x acc
add k_range x 0L
let add_k_list_mutable x (k_range: int list) =
let mutable count = 0L
for k in k_range do
let y = x ^^^ k
if (y < k || y > d) then
count <- count + 1L
count
let add_k_array x (k_range: int []) =
let mutable count = 0L
for k in k_range do
let y = x ^^^ k
if (y < k || y > d) then
count <- count + 1L
count
[<EntryPoint>]
let main argv =
let x = 5
timeItArr "add_k_array" add_k_array
timeIt "add_k_list" add_k_list
timeIt "add_k_list_mutable" add_k_list_mutable
printfn "%A" argv
0 // return an integer exit code
EDIT: The above test was run on 32bit Release mode in VS2015. At the suggestion of s952163, I ran it at 64 bit and found the results differ quite a bit:
add_k_list, elapsed 17918 ms, result 0L
add_k_list_mutable, elapsed 17898 ms, result 0L
add_k_array, elapsed 8261 ms, result 0L
I'm especially surprised that the difference between using tail recursion with an accumulator vs a mutable variable seems to have disappeared.

When running a slightly modified program (posted below) these are the numbers I received:
x64 Release .NET 4.6.1
TestRun: Total: 1000000000, Outer: 100, Inner: 10000000
add_k_array, elapsed 1296 ms, accumulated result 495000099L
add_k_list, elapsed 2675 ms, accumulated result 495000099L
add_k_list_mutable, elapsed 2678 ms, accumulated result 495000099L
TestRun: Total: 1000000000, Outer: 1000, Inner: 1000000
add_k_array, elapsed 869 ms, accumulated result 499624318L
add_k_list, elapsed 2486 ms, accumulated result 499624318L
add_k_list_mutable, elapsed 2483 ms, accumulated result 499624318L
TestRun: Total: 1000000000, Outer: 10000, Inner: 100000
add_k_array, elapsed 750 ms, accumulated result 507000943L
add_k_list, elapsed 1602 ms, accumulated result 507000943L
add_k_list_mutable, elapsed 1603 ms, accumulated result 507000943L
x86 Release .NET 4.6.1
TestRun: Total: 1000000000, Outer: 100, Inner: 10000000
add_k_array, elapsed 1601 ms, accumulated result 495000099L
add_k_list, elapsed 2014 ms, accumulated result 495000099L
add_k_list_mutable, elapsed 1835 ms, accumulated result 495000099L
TestRun: Total: 1000000000, Outer: 1000, Inner: 1000000
add_k_array, elapsed 1495 ms, accumulated result 499624318L
add_k_list, elapsed 1714 ms, accumulated result 499624318L
add_k_list_mutable, elapsed 1595 ms, accumulated result 499624318L
TestRun: Total: 1000000000, Outer: 10000, Inner: 100000
add_k_array, elapsed 1363 ms, accumulated result 507000943L
add_k_list, elapsed 1406 ms, accumulated result 507000943L
add_k_list_mutable, elapsed 1221 ms, accumulated result 507000943L
(As usual it's important to not run with the debugger attached as that changes how the JIT:er works. With debugger attached the JIT:er produces code that is easier for the debugger but also slower.)
The way this works is that the total number of iterations is kept constant but it varies the count of the outer loop and the size of the list/array.
For me the only measurement that is odd is that the array loop is worse in some cases than the list loop.
If the total amount of work is the same why do we see different results when outer/inner is varied?
The answer is most likely related to the CPU cache. When we iterate over an array of size 10,000,000 the actual size of it in memory is 40,000,000 bytes. My machine have "just" 6,000,000 bytes of L3 cache. When the array size is 1,000,000 the size of the array is 4,000,000 bytes which can fit in L3.
The list type in F# is essentially a single-linked list and a rough estimate of the list element is 4(data)+8(64bit pointer)+8(vtable pointer)+4(heap overhead) = 24 bytes. With this estimate the size of a list with 10,000,000 elements is 240,000,000 bytes and with size 1,000,000 elements the size is 24,000,000. Neither fit in the L3 cache on my machine.
When the number of element is 100,000 the size of the array is 400,000 bytes and the list size is 2,400,000. Both fit snugly into the L3 cache.
This reasoning can explain the difference in performance between smaller array/lists compared to bigger ones.
If the elements for the list is not allocated sequentially (ie the heap is fragmented or the GC moved them around) the performance of the list is expected to be much worse if it doesn't fit into the cache because the CPU prefetch strategy no longer works then. The elements in an array is guaranteed to always be sequential thus prefetch will work fine if you iterate sequentially.
Why is tail-recursion slower than the mutable for loop?
This actually isn't true in F# 3 where the for loop is expected to be much slower than the tail-recursion.
For an hint of the answer I used ILSpy to look at the generated IL code.
I found that FSharpList<>::get_TailOrNull() is called twice per loop when using tail-recursion. Once to check if we reached the end and once to get the next element (redundant call).
The for loop version only calls FSharpList<>::get_TailOrNull() once.
This extra call likely explains why tail-recursion is slower but as you noted in x64 bit mode both list versions were about as fast. What's going on?
I checked the JIT:ed assembly code and noted that the x64 JIT:er eliminated the extra call to FSharpList<>::get_TailOrNull(). The x86 JIT:er failed to eliminate the call.
Lastly, why is array version slower than the list version on x86?
In general I expect that arrays to have the least overhead of all collection in .NET. The reason is that it's compact, sequential and there are special instructions in ILAsm to access the elements.
So it's suprising to me that lists performs better in some cases.
Checking the assembly code again what it seems to come from that the array version requires an extra variable to perform its work and the x86 CPU has few registers available leading to an extra read from the stack per iteration. x64 has significantly more registers leading to that the array version only has to read once from memory per iteration where the list version reads twice (head and tail).
Any conclusions?
When it comes to CPU performance x64 is the way to go (this hasn't always been the case)
Expects arrays to perform better than any data structure in .NET for operations where array operations are O(1) (inserts are slow obviously)
The devil is in the details meaning in order to gain true insight we might need to check the assembly code.
Cache locality is very important for large collections. Since arrays are compact and guaranteed to be sequential they are often a good choice.
It's very difficult to predict performance, always measure
Iterate towards zero when possible if you are really hungry for performance. This can save one read from memory.
EDIT: OP wondered why it seemed x86 lists performed better x64 lists
I reran the perf tests with list/array size set to 1,000. This will make sure the entire data structure fit into my L1 Cache (256kB)
x64 Release .NET 4.6.1
TestRun: Total: 1000000000, Outer: 1000000, Inner: 1000
add_k_array, elapsed 1062 ms, accumulated result 999499999L
add_k_list, elapsed 1134 ms, accumulated result 999499999L
add_k_list_mutable, elapsed 1110 ms, accumulated result 999499999L
x86 Release .NET 4.6.1
TestRun: Total: 1000000000, Outer: 1000000, Inner: 1000
add_k_array, elapsed 1617 ms, accumulated result 999499999L
add_k_list, elapsed 1359 ms, accumulated result 999499999L
add_k_list_mutable, elapsed 1100 ms, accumulated result 999499999L
We see that for this size it seems x64 is performing about as well or better than x86. Why do we see the opposite in the other measurements? I speculate that this is because the size of the list elements are larger in the x64 versions meaning we use more bandwidth moving data from L3 to L1.
So if you can try to make sure your data fits into L1 cache.
Final musings
When working with these sort of questions I sometimes wonder if the whole Von Neumann architecture is a big mistake. Instead we should have a data flow architecture as data is slow and instructions are fast.
AFAIK under the hood CPU:s have a data flow architecture. The assembly language though looks like one would expect from a Von Neumann architecture so in some sense it's a high-level abstraction over the data flow architecture. But in order to provide reasonable performant code the CPU die is mostly occupied by cache (~95%). With a pure data flow architecture one would expect a higher percentage of the CPU die would do actual work.
Hope this was interesting, my modified program follows:
open System.Diagnostics
let stopWatch =
let sw = Stopwatch ()
sw.Start ()
sw
let timeIt (name : string) (outer : int) (a : int -> int64) : unit =
let t = stopWatch.ElapsedMilliseconds
let mutable acc = a 0
for i = 2 to outer do
acc <- acc + a i
let d = stopWatch.ElapsedMilliseconds - t
printfn "%s, elapsed %d ms, accumulated result %A" name d acc
let add_k_list x l (k_range: int list) =
let rec add k_range x acc =
match k_range with
| [] -> acc
| k::ks -> let y = x ^^^ k
if (y < k || y > l) then
add ks x (acc + 1L)
else
add ks x acc
add k_range x 0L
let add_k_list_mutable x l (k_range: int list) =
let mutable count = 0L
for k in k_range do
let y = x ^^^ k
if (y < k || y > l) then
count <- count + 1L
count
let add_k_array x l (k_range: int []) =
let mutable count = 0L
for k in k_range do
let y = x ^^^ k
if (y < k || y > l) then
count <- count + 1L
count
[<EntryPoint>]
let main argv =
let total = 1000000000
let outers = [|100; 1000; 10000|]
for outer in outers do
let inner = total / outer
printfn "TestRun: Total: %d, Outer: %d, Inner: %d" total outer inner
ignore <| System.GC.WaitForFullGCComplete ()
let testList = [1..inner]
let testArray = [|1..inner|]
timeIt "add_k_array" outer <| fun x -> add_k_array x inner testArray
timeIt "add_k_list" outer <| fun x -> add_k_list x inner testList
timeIt "add_k_list_mutable" outer <| fun x -> add_k_list_mutable x inner testList
0

Related

Matlab allocating A=zeros(N) vs. A(N,N)=0 yields strange performance results [duplicate]

I've stumbled upon the weird way (in my view) that Matlab is dealing with empty matrices. For example, if two empty matrices are multiplied the result is:
zeros(3,0)*zeros(0,3)
ans =
0 0 0
0 0 0
0 0 0
Now, this already took me by surprise, however, a quick search got me to the link above, and I got an explanation of the somewhat twisted logic of why this is happening.
However, nothing prepared me for the following observation. I asked myself, how efficient is this type of multiplication vs just using zeros(n) function, say for the purpose of initialization? I've used timeit to answer this:
f=#() zeros(1000)
timeit(f)
ans =
0.0033
vs:
g=#() zeros(1000,0)*zeros(0,1000)
timeit(g)
ans =
9.2048e-06
Both have the same outcome of 1000x1000 matrix of zeros of class double, but the empty matrix multiplication one is ~350 times faster! (a similar result happens using tic and toc and a loop)
How can this be? are timeit or tic,toc bluffing or have I found a faster way to initialize matrices?
(this was done with matlab 2012a, on a win7-64 machine, intel-i5 650 3.2Ghz...)
EDIT:
After reading your feedback, I have looked more carefully into this peculiarity, and tested on 2 different computers (same matlab ver though 2012a) a code that examine the run time vs the size of matrix n. This is what I get:
The code to generate this used timeit as before, but a loop with tic and toc will look the same. So, for small sizes, zeros(n) is comparable. However, around n=400 there is a jump in performance for the empty matrix multiplication. The code I've used to generate that plot was:
n=unique(round(logspace(0,4,200)));
for k=1:length(n)
f=#() zeros(n(k));
t1(k)=timeit(f);
g=#() zeros(n(k),0)*zeros(0,n(k));
t2(k)=timeit(g);
end
loglog(n,t1,'b',n,t2,'r');
legend('zeros(n)','zeros(n,0)*zeros(0,n)',2);
xlabel('matrix size (n)'); ylabel('time [sec]');
Are any of you experience this too?
EDIT #2:
Incidentally, empty matrix multiplication is not needed to get this effect. One can simply do:
z(n,n)=0;
where n> some threshold matrix size seen in the previous graph, and get the exact efficiency profile as with empty matrix multiplication (again using timeit).
Here's an example where it improves efficiency of a code:
n = 1e4;
clear z1
tic
z1 = zeros( n );
for cc = 1 : n
z1(:,cc)=cc;
end
toc % Elapsed time is 0.445780 seconds.
%%
clear z0
tic
z0 = zeros(n,0)*zeros(0,n);
for cc = 1 : n
z0(:,cc)=cc;
end
toc % Elapsed time is 0.297953 seconds.
However, using z(n,n)=0; instead yields similar results to the zeros(n) case.
This is strange, I am seeing f being faster while g being slower than what you are seeing. But both of them are identical for me. Perhaps a different version of MATLAB ?
>> g = #() zeros(1000, 0) * zeros(0, 1000);
>> f = #() zeros(1000)
f =
#()zeros(1000)
>> timeit(f)
ans =
8.5019e-04
>> timeit(f)
ans =
8.4627e-04
>> timeit(g)
ans =
8.4627e-04
EDIT can you add + 1 for the end of f and g, and see what times you are getting.
EDIT Jan 6, 2013 7:42 EST
I am using a machine remotely, so sorry about the low quality graphs (had to generate them blind).
Machine config:
i7 920. 2.653 GHz. Linux. 12 GB RAM. 8MB cache.
It looks like even the machine I have access to shows this behavior, except at a larger size (somewhere between 1979 and 2073). There is no reason I can think of right now for the empty matrix multiplication to be faster at larger sizes.
I will be investigating a little bit more before coming back.
EDIT Jan 11, 2013
After #EitanT's post, I wanted to do a little bit more of digging. I wrote some C code to see how matlab may be creating a zeros matrix. Here is the c++ code that I used.
int main(int argc, char **argv)
{
for (int i = 1975; i <= 2100; i+=25) {
timer::start();
double *foo = (double *)malloc(i * i * sizeof(double));
for (int k = 0; k < i * i; k++) foo[k] = 0;
double mftime = timer::stop();
free(foo);
timer::start();
double *bar = (double *)malloc(i * i * sizeof(double));
memset(bar, 0, i * i * sizeof(double));
double mmtime = timer::stop();
free(bar);
timer::start();
double *baz = (double *)calloc(i * i, sizeof(double));
double catime = timer::stop();
free(baz);
printf("%d, %lf, %lf, %lf\n", i, mftime, mmtime, catime);
}
}
Here are the results.
$ ./test
1975, 0.013812, 0.013578, 0.003321
2000, 0.014144, 0.013879, 0.003408
2025, 0.014396, 0.014219, 0.003490
2050, 0.014732, 0.013784, 0.000043
2075, 0.015022, 0.014122, 0.000045
2100, 0.014606, 0.014480, 0.000045
As you can see calloc (4th column) seems to be the fastest method. It is also getting significantly faster between 2025 and 2050 (I'd assume it would at around 2048 ?).
Now I went back to matlab to check for the same. Here are the results.
>> test
1975, 0.003296, 0.003297
2000, 0.003377, 0.003385
2025, 0.003465, 0.003464
2050, 0.015987, 0.000019
2075, 0.016373, 0.000019
2100, 0.016762, 0.000020
It looks like both f() and g() are using calloc at smaller sizes (<2048 ?). But at larger sizes f() (zeros(m, n)) starts to use malloc + memset, while g() (zeros(m, 0) * zeros(0, n)) keeps using calloc.
So the divergence is explained by the following
zeros(..) begins to use a different (slower ?) scheme at larger sizes.
calloc also behaves somewhat unexpectedly, leading to an improvement in performance.
This is the behavior on Linux. Can someone do the same experiment on a different machine (and perhaps a different OS) and see if the experiment holds ?
The results might be a bit misleading. When you multiply two empty matrices, the resulting matrix is not immediately "allocated" and "initialized", rather this is postponed until you first use it (sort of like a lazy evaluation).
The same applies when indexing out of bounds to grow a variable, which in the case of numeric arrays fills out any missing entries with zeros (I discuss afterwards the non-numeric case). Of course growing the matrix this way does not overwrite existing elements.
So while it may seem faster, you are just delaying the allocation time until you actually first use the matrix. In the end you'll have similar timings as if you did the allocation from the start.
Example to show this behavior, compared to a few other alternatives:
N = 1000;
clear z
tic, z = zeros(N,N); toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
tic, z = zeros(N,0)*zeros(0,N); toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
tic, z(N,N) = 0; toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
tic, z = full(spalloc(N,N,0)); toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
tic, z(1:N,1:N) = 0; toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
val = 0;
tic, z = val(ones(N)); toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
tic, z = repmat(0, [N N]); toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
The result shows that if you sum the elapsed time for both instructions in each case, you end up with similar total timings:
// zeros(N,N)
Elapsed time is 0.004525 seconds.
Elapsed time is 0.000792 seconds.
// zeros(N,0)*zeros(0,N)
Elapsed time is 0.000052 seconds.
Elapsed time is 0.004365 seconds.
// z(N,N) = 0
Elapsed time is 0.000053 seconds.
Elapsed time is 0.004119 seconds.
The other timings were:
// full(spalloc(N,N,0))
Elapsed time is 0.001463 seconds.
Elapsed time is 0.003751 seconds.
// z(1:N,1:N) = 0
Elapsed time is 0.006820 seconds.
Elapsed time is 0.000647 seconds.
// val(ones(N))
Elapsed time is 0.034880 seconds.
Elapsed time is 0.000911 seconds.
// repmat(0, [N N])
Elapsed time is 0.001320 seconds.
Elapsed time is 0.003749 seconds.
These measurements are too small in the milliseconds and might not be very accurate, so you might wanna run these commands in a loop a few thousand times and take the average. Also sometimes running saved M-functions is faster than running scripts or on the command prompt, as certain optimizations only happen that way...
Either way allocation is usually done once, so who cares if it takes an extra 30ms :)
A similar behavior can be seen with cell arrays or arrays of structures. Consider the following example:
N = 1000;
tic, a = cell(N,N); toc
tic, b = repmat({[]}, [N,N]); toc
tic, c{N,N} = []; toc
which gives:
Elapsed time is 0.001245 seconds.
Elapsed time is 0.040698 seconds.
Elapsed time is 0.004846 seconds.
Note that even if they are all equal, they occupy different amount of memory:
>> assert(isequal(a,b,c))
>> whos a b c
Name Size Bytes Class Attributes
a 1000x1000 8000000 cell
b 1000x1000 112000000 cell
c 1000x1000 8000104 cell
In fact the situation is a bit more complicated here, since MATLAB is probably sharing the same empty matrix for all the cells, rather than creating multiple copies.
The cell array a is in fact an array of uninitialized cells (an array of NULL pointers), while b is a cell array where each cell is an empty array [] (internally and because of data sharing, only the first cell b{1} points to [] while all the rest have a reference to the first cell). The final array c is similar to a (uninitialized cells), but with the last one containing an empty numeric matrix [].
I looked around the list of exported C functions from the libmx.dll (using Dependency Walker tool), and I found a few interesting things.
there are undocumented functions for creating uninitialized arrays: mxCreateUninitDoubleMatrix, mxCreateUninitNumericArray, and mxCreateUninitNumericMatrix. In fact there is a submission on the File Exchange makes use of these functions to provide a faster alternative to zeros function.
there exist an undocumented function called mxFastZeros. Googling online, I can see you cross-posted this question on MATLAB Answers as well, with some excellent answers over there. James Tursa (same author of UNINIT from before) gave an example on how to use this undocumented function.
libmx.dll is linked against tbbmalloc.dll shared library. This is Intel TBB scalable memory allocator. This library provides equivalent memory allocation functions (malloc, calloc, free) optimized for parallel applications. Remember that many MATLAB functions are automatically multithreaded, so I wouldn't be surprised if zeros(..) is multithreaded and is using Intel's memory allocator once the matrix size is large enough (here is recent comment by Loren Shure that confirms this fact).
Regarding the last point about the memory allocator, you could write a similar benchmark in C/C++ similar to what #PavanYalamanchili did, and compare the various allocators available. Something like this. Remember that MEX-files have a slightly higher memory management overhead, since MATLAB automatically frees any memory that was allocated in MEX-files using the mxCalloc, mxMalloc, or mxRealloc functions. For what it's worth, it used to be possible to change the internal memory manager in older versions.
EDIT:
Here is a more thorough benchmark to compare the discussed alternatives. It specifically shows that once you stress the use of the entire allocated matrix, all three methods are on equal footing, and the difference is negligible.
function compare_zeros_init()
iter = 100;
for N = 512.*(1:8)
% ZEROS(N,N)
t = zeros(iter,3);
for i=1:iter
clear z
tic, z = zeros(N,N); t(i,1) = toc;
tic, z(:) = 9; t(i,2) = toc;
tic, z = z + 1; t(i,3) = toc;
end
fprintf('N = %4d, ZEROS = %.9f\n', N, mean(sum(t,2)))
% z(N,N)=0
t = zeros(iter,3);
for i=1:iter
clear z
tic, z(N,N) = 0; t(i,1) = toc;
tic, z(:) = 9; t(i,2) = toc;
tic, z = z + 1; t(i,3) = toc;
end
fprintf('N = %4d, GROW = %.9f\n', N, mean(sum(t,2)))
% ZEROS(N,0)*ZEROS(0,N)
t = zeros(iter,3);
for i=1:iter
clear z
tic, z = zeros(N,0)*zeros(0,N); t(i,1) = toc;
tic, z(:) = 9; t(i,2) = toc;
tic, z = z + 1; t(i,3) = toc;
end
fprintf('N = %4d, MULT = %.9f\n\n', N, mean(sum(t,2)))
end
end
Below are the timings averaged over 100 iterations in terms of increasing matrix size. I performed the tests in R2013a.
>> compare_zeros_init
N = 512, ZEROS = 0.001560168
N = 512, GROW = 0.001479991
N = 512, MULT = 0.001457031
N = 1024, ZEROS = 0.005744873
N = 1024, GROW = 0.005352638
N = 1024, MULT = 0.005359236
N = 1536, ZEROS = 0.011950846
N = 1536, GROW = 0.009051589
N = 1536, MULT = 0.008418878
N = 2048, ZEROS = 0.012154002
N = 2048, GROW = 0.010996315
N = 2048, MULT = 0.011002169
N = 2560, ZEROS = 0.017940950
N = 2560, GROW = 0.017641046
N = 2560, MULT = 0.017640323
N = 3072, ZEROS = 0.025657999
N = 3072, GROW = 0.025836506
N = 3072, MULT = 0.051533432
N = 3584, ZEROS = 0.074739924
N = 3584, GROW = 0.070486857
N = 3584, MULT = 0.072822335
N = 4096, ZEROS = 0.098791732
N = 4096, GROW = 0.095849788
N = 4096, MULT = 0.102148452
After doing some research, I've found this article in "Undocumented Matlab", in which Mr. Yair Altman had already come to the conclusion that MathWork's way of preallocating matrices using zeros(M, N) is indeed not the most efficient way.
He timed x = zeros(M,N) vs. clear x, x(M,N) = 0 and found that the latter is ~500 times faster. According to his explanation, the second method simply creates an M-by-N matrix, the elements of which being automatically initialized to 0. The first method however, creates x (with x having automatic zero elements) and then assigns a zero to every element in x again, and that is a redundant operation that takes more time.
In the case of empty matrix multiplication, such as what you've shown in your question, MATLAB expects the product to be an M×N matrix, and therefore it allocates an M×N matrix. Consequently, the output matrix is automatically initialized to zeroes. Since the original matrices are empty, no further calculations are performed, and hence the elements in the output matrix remain unchanged and equal to zero.
Interesting question, apparently there are several ways to 'beat' the built-in zeros function. My only guess as to why this is happening would be that it could be more memory efficient (after all, zeros(LargeNumer) will sooner cause Matlab to hit the memory limit than form a devestating speed bottleneck in most code), or more robust somehow.
Here is another fast allocation method using a sparse matrix, i have added the regular zeros function as a benchmark:
tic; x=zeros(1000,1000); toc
Elapsed time is 0.002863 seconds.
tic; clear x; x(1000,1000)=0; toc
Elapsed time is 0.000282 seconds.
tic; x=full(spalloc(1000,1000,0)); toc
Elapsed time is 0.000273 seconds.
tic; x=spalloc(1000,1000,1000000); toc %Is this the same for practical purposes?
Elapsed time is 0.000281 seconds.

Julia allocates huge amount of memory for unknown reason

I'm trying to parallelize a little scientific code I wrote. But when I add #parallelize, similar code on just one processor suddenly takes 10 times as long to execute. It should take roughly the same amount of time. The first code makes one memory allocation, while the second makes 20. But zeros(Float64, num_bins) should not be a bottleneck. num_bins is 1800. So each call to zeros() should be allocating 8*1800 bytes. 20 calls to allocate 14,400 bytes should not be taking this long.
I can't figure out what I'm doing wrong, and the Julia documentation is vague and non-specific about how variables are accessed within #parallel. Both versions of the code below compute the correct value for the rdf vector. Can anyone tell by looking at it what is making it allocate so much memory and take so long?
atoms = readAtoms(file)
rdf = zeros(Float64, num_bins)
#time for k = 1:20
for i = 1:num_atoms
for j = 1:num_atoms
r = distance(k, atoms, i, atoms, j)
bin_number = floor(r / dr) + 1
rdf[bin_number] += 1
end
end
end
elapsed time: 8.1 seconds (0 bytes allocated)
atoms = readAtoms(file)
#time rdf = #parallel (+) for k = 1:20
rdf_part = zeros(Float64, num_bins)
for i = 1:num_atoms
for j = 1:num_atoms
r = distance(k, atoms, i, atoms, j)
bin_number = floor(r / dr) + 1
rdf_part[bin_number] += 1
end
end
rdf_part
end
elapsed time: 81.2 seconds (33472513332 bytes allocated, 17.40% gc time)

What's the ideal implementation for the Sieve of Eratosthenes between Lists, Arrays, and Mutable Arrays?

In Haskell, I've found three simple implementations of the Sieve of Eratosthenes on the Rosetta Code page.
Now my question is, which one should be used in which situations?
Correcting my initial reasoning would be helpful too:
I'm assuming the List one is the most idiomatic and easy to read for a Haskeller. Is it correct, though? I'm wondering if it suffers from the same problems as another list-based sieve that I then learned was not actually implementing the algorithm:
(edit: shown here is the list-based sieve I know has problems, not the one from RosettaCode, which I pasted at the bottom)
primes = sieve [2..] where
sieve (p:x) = p : sieve [ n | n <- x, n `mod` p > 0 ]
In terms of performance, the immutable Array seems to be the winner. With an upper bound m of 2000000, the times were about:
1.3s for List
0.6s for Array
1.8s for Mutable Array
So I'd pick Array for performance.
And of course, the Mutable Array one is also easy to reason about since I have a more imperative language background. I'm not sure why I would pick this one if I'm coding in Haskell, though, since it's both slower than the others and non-idiomatic.
Code copied here for reference:
List:
primesTo m = 2 : eratos [3,5..m] where
eratos (p : xs) | p*p>m = p : xs
| True = p : eratos (xs `minus` [p*p, p*p+2*p..])
minus a#(x:xs) b#(y:ys) = case compare x y of
LT -> x : minus xs b
EQ -> minus xs ys
GT -> minus a ys
minus a b = a
Immutable Array:
import Data.Array.Unboxed
primesToA m = sieve 3 (array (3,m) [(i,odd i) | i<-[3..m]] :: UArray Int Bool)
where
sieve p a
| p*p > m = 2 : [i | (i,True) <- assocs a]
| a!p = sieve (p+2) $ a//[(i,False) | i <- [p*p, p*p+2*p..m]]
| otherwise = sieve (p+2) a
Mutable Array:
import Control.Monad (forM_, when)
import Control.Monad.ST
import Data.Array.ST
import Data.Array.Unboxed
primeSieve :: Integer -> UArray Integer Bool
primeSieve top = runSTUArray $ do
a <- newArray (2,top) True -- :: ST s (STUArray s Integer Bool)
let r = ceiling . sqrt $ fromInteger top
forM_ [2..r] $ \i -> do
ai <- readArray a i
when ai $ do
forM_ [i*i,i*i+i..top] $ \j -> do
writeArray a j False
return a
-- Return primes from sieve as list:
primesTo :: Integer -> [Integer]
primesTo top = [p | (p,True) <- assocs $ primeSieve top]
EDIT
I showed Turner's Sieve at the top but that's not the list-based example I'm comparing with the other two. I just wanted to know if the list-based example suffers from the same "not the correct Sieve of Eratosthenes" problems as Turner's.
It appears the performance comparison is unfair because the Mutable Array example goes through evens as well and uses Integer rather than Int...
This
primes = sieve [2..] where
sieve (p:x) = p : sieve [ n | n <- x, n `mod` p > 0 ]
is not a sieve. It's very inefficient trial division. Don't use that!
I'm curious about how you got your times, there is no way that the Turner "sieve" could produce the primes not exceeding 2,000,000 in mere seconds. Letting it find the primes to 200,000 took
MUT time 6.38s ( 6.39s elapsed)
GC time 9.19s ( 9.20s elapsed)
EXIT time 0.00s ( 0.00s elapsed)
Total time 15.57s ( 15.59s elapsed)
on my box (64-bit Linux, ghc-7.6.1, compiled with -O2). The complexity of that algorithm is O(N² / log² N), almost quadratic. Letting it proceed to 2,000,000 would take about twenty minutes.
Your times for the array versions are suspicious too, though in the other direction. Did you measure interpreted code?
Sieving to 2,000,000, compiled with optimisations, the mutable array code took 0.35 seconds to run, and the immutable array code 0.12 seconds.
Now, that still has the mutable array about three times slower than the immutable array.
But, it's an unfair comparison. For the immutable array, you used Ints, and for the mutable array Integers. Changing the mutable array code to use Ints - as it should, since under the hood, arrays are Int-indexed, so using Integer is an unnecessary performance sacrifice that buys nothing - made the mutable array code run in 0.15 seconds. Close to the mutable array code, but not quite there. However, you let the mutable array do more work, since in the immutable array code you only eliminate odd multiples of the odd primes, but in the mutable array code, you mark all multiples of all primes. Changing the mutable array code to treat 2 specially, and only eliminate odd multiples of odd primes brings that down to 0.12 seconds too.
But, you're using range-checked array indexing, which is slow, and, since the validity of the indices is checked in the code itself, unnecessary. Changing that to using unsafeRead and unsafeWrite brings down the time for the immutable array to 0.09 seconds.
Then you have the problem that using
forM_ [x, y .. z]
uses boxed Ints (fortunately, GHC eliminates the list). Writing a loop yourself, so that only unboxed Int#s are used, the time goes down to 0.02 seconds.
{-# LANGUAGE MonoLocalBinds #-}
import Control.Monad (forM_, when)
import Control.Monad.ST
import Data.Array.ST
import Data.Array.Unboxed
import Data.Array.Base
primeSieve :: Int -> UArray Int Bool
primeSieve top = runSTUArray $ do
a <- newArray (0,top) True
unsafeWrite a 0 False
unsafeWrite a 1 False
let r = ceiling . sqrt $ fromIntegral top
mark step idx
| top < idx = return ()
| otherwise = do
unsafeWrite a idx False
mark step (idx+step)
sift p
| r < p = return a
| otherwise = do
prim <- unsafeRead a p
when prim $ mark (2*p) (p*p)
sift (p+2)
mark 2 4
sift 3
-- Return primes from sieve as list:
primesTo :: Int -> [Int]
primesTo top = [p | (p,True) <- assocs $ primeSieve top]
main :: IO ()
main = print .last $ primesTo 2000000
So, wrapping up, for a Sieve of Eratosthenes, you should use an array - not surprising, its efficiency depends on being able to step from one multiple to the next in short constant time.
You get very simple and straightforward code with immutable arrays, and that code performs decently for not too high limits (it gets relatively worse for higher limits, since the arrays are still copied and garbage-collected, but that's not too bad).
When you need better performance, you need mutable arrays. Writing efficient mutable array code is not entirely trivial, one has to know how the compiler translates the various constructs to choose the right one, and some would consider such code unidiomatic. But you can also use a library (disclaimer: I wrote it) that provides a fairly efficient implementation rather than writing it yourself.
Mutable array will always be the winner in terms of performance (and you really should've copied the version that works on odds only as a minimum; it should be the fastest of the three - also because it uses Int and not Integer).
For lists, tree-shaped merging incremental sieve should perform better than the one you show. You can always use it with takeWhile (< limit) if needed. I contend that it conveys the true nature of the sieve most clearly:
import Data.List (unfoldr)
primes :: [Int]
primes = 2 : _Y ((3 :) . gaps 5 . _U . map (\p -> [p*p, p*p+2*p..]))
_Y g = g (_Y g) -- recursion
_U ((x:xs):t) = (x :) . union xs . _U -- ~= nub . sort . concat
. unfoldr (\(a:b:c) -> Just (union a b, c)) $ t
gaps k s#(x:xs) | k < x = k : gaps (k+2) s -- ~= [k,k+2..]\\s, when
| otherwise = gaps (k+2) xs -- k<=x && null(s\\[k,k+2..])
union a#(x:xs) b#(y:ys) = case compare x y of -- ~= nub . sort .: (++)
LT -> x : union xs b
EQ -> x : union xs ys
GT -> y : union a ys
_U reimplements Data.List.Ordered.unionAll, and gaps 5 is (minus [5,7..]), fused for efficiency, with minus and union from the same package.
Of course nothing beats the brevity of Data.List.nubBy (((>1).).gcd) [2..] (but it's very slow).
To your 1st new question: not. It does find the multiples by counting up, as any true sieve should (although "minus" on lists is of course under-performant; the above improves on that by re-arranging a linear subtraction chain ((((xs-a)-b)-c)- ... ) into a subtraction of tree-folded additions, xs-(a+((b+c)+...))).
As has been said, using mutable arrays has the best performance. The following code is derived from this "TemplateHaskell" version converted back to something more in line with the straight mutable Array solution as the "TemplateHaskell" doesn't seem to make any difference, with some further optimizations. I believe it is faster than the usual mutable unboxed array versions due to the further optimizations and especially due to the use of the "unsafeRead" and "unsafeWrite" functions that avoid range checking of the arrays, probably also internally using pointers for array access:
{-# OPTIONS -O2 -optc-O3 #-}
import Control.Monad
import Control.Monad.ST
import Data.Array.ST
import Data.Array.Unboxed
import Data.Array.Base
primesToUA :: Word32-> [Word32]
primesToUA top = do
let sieveUA top = runSTUArray $ do
let m = ((fromIntegral top) - 3) `div` 2 :: Int
buf <- newArray (0,m) True -- :: ST s (STUArray s Int Bool)
let cullUA i = do
let p = i + i + 3
strt = p * (i + 1) + i
let cull j = do
if j > m then cullUA (i + 1)
else do
unsafeWrite buf j False
cull (j + p)
if strt > m then return ()
else do
e <- unsafeRead buf i
if e then cull strt else cullUA (i + 1)
cullUA 0; return buf
if top > 1 then 2 : [2 * (fromIntegral i) + 3 | (i,True) <- assocs $ sieveUA top]
else []
main = do
x <- read `fmap` getLine -- 1mln 2mln 10mln 100mln
print (last (primesToUA x)) -- 0.01 0.02 0.09 1.26 seconds
EDIT: The above code has been corrected and the times below edited to reflect the correction and as well the comments comparing the paged to the non-paged version have been edited.
The times to run this to the indicated top ranges are as shown in the comment table at the bottom of the above source code as measured by ideone.com and are about exactly five times faster than the answer posted by #WillNess as also measured at ideone.com. This code takes a trivial amount of time to cull the primes to two million and only 1.26 seconds to cull to a hundred million. These times are about 2.86 times faster when run on my i7 (3.5 GHz) at 0.44 seconds to a hundred million, and takes 6.81 seconds to run to one billion. The memory use is just over six megabytes for the former and sixty megabytes for the latter, which is the memory used by the huge (bit packed) array. This array also explains the non-linear performance in that as the array size exceeds the CPU cache sizes the average memory access times get worse per composite number representation cull.
EDIT_ADD: A page segmented sieve is more efficient in that it has better memory access efficiency when the buffer size is kept smaller than the L1 or L2 CPU caches, and as well has the advantage that it is unbounded so that the upper range does not have to be specified in advance and a much smaller memory footprint being just the base primes less than the square root of the range used plus the page buffer memory. The following code has been written as a page segmented implementation and is somewhat faster than the non-paged version; it also offers the advantage that one can change the output range specification at the top of the code to 'Word64' from 'Word32' so it is then not limited to the 32-bit number range, at only a slight cost in processing time (for 32-bit compiled code) for any range that is in common. The code is as follows:
-- from http://www.haskell.org/haskellwiki/Prime_numbers#Using_ST_Array
{-# OPTIONS -O2 -optc-O3 #-}
import Data.Word
import Control.Monad
import Control.Monad.ST
import Data.Array.ST
import Data.Array.Unboxed
import Data.Array.Base
primesUA :: () -> [Word32]
primesUA () = do
let pgSZBTS = 262144 * 8
let sieveUA low bps = runSTUArray $ do
let nxt = (fromIntegral low) + (fromIntegral pgSZBTS)
buf <- newArray (0,pgSZBTS - 1) True -- :: ST s (STUArray s Int Bool)
let cullUAbase i = do
let p = i + i + 3
strt = p * (i + 1) + i
when (strt < pgSZBTS) $ do
e <- unsafeRead buf i
if e then do
let loop j = do
if j < pgSZBTS then do
unsafeWrite buf j False
loop (j + p)
else cullUAbase (i + 1)
loop strt
else cullUAbase (i + 1)
let cullUA ~(p:t) = do
let bp = (fromIntegral p)
i = (bp - 3) `div` 2
s = bp * (i + 1) + i
when (s < nxt) $ do
let strt = do
if s >= low then fromIntegral (s - low)
else do
let b = (low - s) `rem` bp
if b == 0 then 0 else fromIntegral (bp - b)
let loop j = do
if j < pgSZBTS then do
unsafeWrite buf j False
loop (j + (fromIntegral p))
else cullUA t
loop strt
if low <= 0 then cullUAbase 0 else cullUA bps
return buf
let sieveList low bps = do
[2 * ((fromIntegral i) + low) + 3 | (i,True) <- assocs $ sieveUA low bps]
let sieve low bps = do
(sieveList low bps) ++ sieve (low + (fromIntegral pgSZBTS)) bps
let primes' = ((sieveList 0 []) ++ sieve (fromIntegral pgSZBTS) primes') :: [Word32]
2 : sieve 0 primes'
main = do
x <- read `fmap` getLine -- 1mln 2mln 10mln 100mln
-- 0.02 0.03 0.13 1.13 seconds
print (length (takeWhile ((>=) (fromIntegral x)) (primesUA ())))
The above code has quite a few more lines of code than the non-paged case due to the need to cull the composite number representations from the first page array differently than succeeding pages. The code also has the fixes so that there aren't memory leaks due to the base primes list and the output list now not being the same list (thus avoiding holding onto the whole list in memory).
Note that this code takes close to linear time (over the range sieved) as the range gets larger due to the culling buffer being of a constant size less than the L2 CPU cache. Memory use is a fraction of that used by the non-paged version at just under 600 kilobytes for a hundred million and just over 600 kilobytes for one billion, which slight increase is just the extra space required for the base primes less than the square root of the range list.
On ideone.com this code produces the number of primes to a hundred million in about 1.13 seconds and about 12 seconds to one billion (32-bit setting). Probably wheel factorization and definitely multi-core processing would make it even faster on a multi-core CPU. On my i7 (3.5 GHz), it takes 0.44 seconds to sieve to a hundred million and 4.7 seconds to one billion, with the roughly linear performance with increasing range as expected. It seems that there is some sort of non-linear overhead in the version of GHC run on ideone.com that has some performance penalty for larger ranges that is not present for the i7 and that is perhaps related to different garbage collection, as the page buffers are being created new for each new page. END_EDIT_ADD
EDIT_ADD2: It seems that much of the processing time for the above page segmented code is used in (lazy) list processing, so the code is accordingly reformulated with several improvements as follows:
Implemented a prime counting function that does not use list processing and uses "popCount" table look ups to count the number of 'one' bits in a 32 bit word at a time. In this way, the time to find the results is insignificant compared to the actual sieve culling time.
Stored the base primes as a list of bit packed page segments, which is much more space efficient than storing a list of primes, and the the time to convert the page segments to primes as required is not much of a computational overhead.
Tuned the prime segment make function so that for the initial zero page segment, it uses its own bit pattern as a source page, thus making the composite number cull code shorter and simpler.
The code then becomes as follows:
{-# OPTIONS -O3 -rtsopts #-} -- -fllvm ide.com doesn't support LLVM
import Data.Word
import Data.Bits
import Control.Monad
import Control.Monad.ST
import Data.Array.ST (runSTUArray)
import Data.Array.Unboxed
import Data.Array.Base
pgSZBTS = (2^18) * 8 :: Int -- size of L2 data cache
type PrimeType = Word32
type Chunk = UArray PrimeType Bool
-- makes a new page chunk and culls it
-- if the base primes list provided is empty then
-- it uses the current array as source (for zero page base primes)
mkChnk :: Word32 -> [Chunk] -> Chunk
mkChnk low bschnks = runSTUArray $ do
let nxt = (fromIntegral low) + (fromIntegral pgSZBTS)
buf <- nxt `seq` newArray (fromIntegral low, fromIntegral nxt - 1) True
let cull ~(p:ps) =
let bp = (fromIntegral p)
i = (bp - 3) `shiftR` 1
s = bp * (i + 1) + i in
let cullp j = do
if j >= pgSZBTS then cull ps
else do
unsafeWrite buf j False
cullp (j + (fromIntegral p)) in
when (s < nxt) $ do
let strt = do
if s >= low then fromIntegral (s - low)
else do
let b = (low - s) `rem` bp
if b == 0 then 0 else fromIntegral (bp - b)
cullp strt
case bschnks of
[] -> do bsbf <- unsafeFreezeSTUArray buf
cull (listChnkPrms [bsbf])
_ -> cull $ listChnkPrms bschnks
return buf
-- creates a page chunk list starting at the lw value
chnksList :: Word32 -> [Chunk]
chnksList lw =
mkChnk lw basePrmChnks : chnksList (lw + fromIntegral pgSZBTS)
-- converts a page chunk list to a list of primes
listChnkPrms :: [Chunk] -> [PrimeType]
listChnkPrms [] = []
listChnkPrms ~(hdchnk#(UArray lw _ rng _):tlchnks) =
let nxtp i =
if i >= rng then [] else
if unsafeAt hdchnk i then
(case ((lw + fromIntegral i) `shiftL` 1) + 3 of
np -> np) : nxtp (i + 1)
else nxtp (i + 1) in
(hdchnk `seq` lw `seq` nxtp 0) ++ listChnkPrms tlchnks
-- the base page chunk list used to cull the higher order pages,
-- note that it has special treatment for the zero page.
-- It is more space efficient to store this as chunks rather than
-- as a list of primes or even a list of deltas (gaps), with the
-- extra processing to convert as needed not too much.
basePrmChnks :: [Chunk]
basePrmChnks = mkChnk 0 [] : chnksList (fromIntegral pgSZBTS)
-- the full list of primes could be accessed with the following function.
primes :: () -> [PrimeType]
primes () = 2 : (listChnkPrms $ chnksList 0)
-- a quite fast prime counting up to the given limit using
-- chunk processing to avoid innermost list processing loops.
countPrimesTo :: PrimeType -> Int
countPrimesTo limit =
let lmtb = (limit - 3) `div` 2 in
let sumChnks acc chnks#(chnk#(UArray lo hi rng _):chnks') =
let cnt :: UArray PrimeType Word32 -> Int
cnt bfw =
case if lmtb < hi then fromIntegral (lmtb - lo) else rng of
crng -> case crng `shiftR` 5 of
rngw ->
let cnt' i ac =
ac `seq` if i >= rngw then
if (i `shiftL` 5) >= rng then ac else
case (-2) `shiftL` fromIntegral (lmtb .&. 31) of
msk -> msk `seq`
case (unsafeAt bfw rngw) .&.
(complement msk) of
bts -> bts `seq` case popCount bts of
c -> c `seq` case ac + c of nac -> nac
else case ac + (popCount $ unsafeAt bfw i) of
nacc -> nacc `seq` cnt' (i + 1) (nacc)
in cnt' 0 0 in
acc `seq` case runST $ do -- make UArray _ Bool into a UArray _ Word32
stbuf <- unsafeThawSTUArray chnk
stbufw <- castSTUArray stbuf
bufw <- unsafeFreezeSTUArray stbufw
return $ cnt bufw of
c -> c `seq` case acc + c of
nacc -> nacc `seq` if hi >= lmtb then nacc
else sumChnks nacc chnks' in
if limit < 2 then 0 else if limit < 3 then 1 else
lmtb `seq` sumChnks 1 (chnksList 0)
main = do
x <- read `fmap` getLine -- 1mln 2mln 10mln 100mln 1000mln
-- 0.02 0.03 0.06 0.45 4.60 seconds
-- 7328 7328 8352 8352 9424 Kilobytes
-- this takes 14.34 seconds and 9424 Kilobytes to 3 billion on ideone.com,
-- and 9.12 seconds for 3 billion on an i7-2700K (3.5 GHz).
-- The above ratio of about 1.6 is much better than usual due to
-- the extremely low memory use of the page segmented algorithm.
-- It seems thaat the Windows Native Code Generator (NCG) from GHC
-- is particularly poor, as the Linux 32-bit version takes
-- less than two thirds of the time for exactly the same program...
print $ countPrimesTo x
-- print $ length $ takeWhile ((>=) x) $ primes () -- the slow way to do this
The times and memory requirements given in the code are as observed when run on ideone.com, with 0.02, 0.03, 0.05, 0.30, 3.0, and 9.1 seconds required to run on my i7-2700K (3.5 GHz) for one, two, ten, a hundred, a thousand (one billion), and three thousand (three billion) million range, respectively, with a pretty much constant memory footprint slowly increasing with the number of base primes less than the square root of the range as required. When compiled with the LLVM compiler back end, these times become 0.01, 0.02, 0.02, 0.12, 1.35, and 4.15 seconds, respectively, due to its more efficient use of registers and machine instructions; this last is quite close to the same speed as if compiled with a 64-bit compiler rather than the 32-bit compiler used as the efficient use of registers means that availability of extra registers doesn't make much difference.
As commented in the code, the ratio between performance on my real machine and the ideone.com servers becomes much less than for much more memory wasteful algorithms due to not being throttled by memory access bottlenecks so the limit to speed is mostly just the ratio of CPU clock speeds as well as CPU processing efficiency per clock cycle. However, as commented there, there is some strange inefficiency with the GHC Native Code Generator (NCG) when run under Windows (32-bit compiler) in that the run times are over 50% slower than if run under Linux (as the ideone.com server uses). AFAIK they both have a common code base for the same Haskell GHC version and the only divergence is in the linker used (which is also used with the LLVM backend, which is not affected) as GHC NCG does not use GCC but only the mingw32 assembler, which should also be the same.
Note that this code when compiled with the LLVM compiler back end is about the same speed as the same algorithm written for highly optimized 'C/C++' implementations indicating that Haskell really has the ability to develop very tight loop coding. It might be said that the Haskell code is quite a bit more readable and secure than equivalent 'C/C++' code once one gets used to the Haskell paradigms of monadic and non-strict code. The further refinements in execution speed for the Sieve of Eratosthenes are purely a function of the tuning of the implementations used and not the choice of language between Haskell and 'C/C++'.
Summary: Of course, this isn't yet the ultimate in speed for a Haskell version of the Sieve of Eratosthenes in that we still haven't tuned the memory access to more efficiently use the fast CPU L1 cache, nor have we significantly reduced the total number of composite culling operations necessary using extreme wheel factorization other than to eliminate the odds processing. However, this is enough to answer the question in showing that mutable arrays are the most efficient way of addressing such tight loop type of problems, with potential speed gains of about 100 times over the use of lists or immutable arrays. END_EDIT_ADD2

Faster way to initialize arrays via empty matrix multiplication? (Matlab)

I've stumbled upon the weird way (in my view) that Matlab is dealing with empty matrices. For example, if two empty matrices are multiplied the result is:
zeros(3,0)*zeros(0,3)
ans =
0 0 0
0 0 0
0 0 0
Now, this already took me by surprise, however, a quick search got me to the link above, and I got an explanation of the somewhat twisted logic of why this is happening.
However, nothing prepared me for the following observation. I asked myself, how efficient is this type of multiplication vs just using zeros(n) function, say for the purpose of initialization? I've used timeit to answer this:
f=#() zeros(1000)
timeit(f)
ans =
0.0033
vs:
g=#() zeros(1000,0)*zeros(0,1000)
timeit(g)
ans =
9.2048e-06
Both have the same outcome of 1000x1000 matrix of zeros of class double, but the empty matrix multiplication one is ~350 times faster! (a similar result happens using tic and toc and a loop)
How can this be? are timeit or tic,toc bluffing or have I found a faster way to initialize matrices?
(this was done with matlab 2012a, on a win7-64 machine, intel-i5 650 3.2Ghz...)
EDIT:
After reading your feedback, I have looked more carefully into this peculiarity, and tested on 2 different computers (same matlab ver though 2012a) a code that examine the run time vs the size of matrix n. This is what I get:
The code to generate this used timeit as before, but a loop with tic and toc will look the same. So, for small sizes, zeros(n) is comparable. However, around n=400 there is a jump in performance for the empty matrix multiplication. The code I've used to generate that plot was:
n=unique(round(logspace(0,4,200)));
for k=1:length(n)
f=#() zeros(n(k));
t1(k)=timeit(f);
g=#() zeros(n(k),0)*zeros(0,n(k));
t2(k)=timeit(g);
end
loglog(n,t1,'b',n,t2,'r');
legend('zeros(n)','zeros(n,0)*zeros(0,n)',2);
xlabel('matrix size (n)'); ylabel('time [sec]');
Are any of you experience this too?
EDIT #2:
Incidentally, empty matrix multiplication is not needed to get this effect. One can simply do:
z(n,n)=0;
where n> some threshold matrix size seen in the previous graph, and get the exact efficiency profile as with empty matrix multiplication (again using timeit).
Here's an example where it improves efficiency of a code:
n = 1e4;
clear z1
tic
z1 = zeros( n );
for cc = 1 : n
z1(:,cc)=cc;
end
toc % Elapsed time is 0.445780 seconds.
%%
clear z0
tic
z0 = zeros(n,0)*zeros(0,n);
for cc = 1 : n
z0(:,cc)=cc;
end
toc % Elapsed time is 0.297953 seconds.
However, using z(n,n)=0; instead yields similar results to the zeros(n) case.
This is strange, I am seeing f being faster while g being slower than what you are seeing. But both of them are identical for me. Perhaps a different version of MATLAB ?
>> g = #() zeros(1000, 0) * zeros(0, 1000);
>> f = #() zeros(1000)
f =
#()zeros(1000)
>> timeit(f)
ans =
8.5019e-04
>> timeit(f)
ans =
8.4627e-04
>> timeit(g)
ans =
8.4627e-04
EDIT can you add + 1 for the end of f and g, and see what times you are getting.
EDIT Jan 6, 2013 7:42 EST
I am using a machine remotely, so sorry about the low quality graphs (had to generate them blind).
Machine config:
i7 920. 2.653 GHz. Linux. 12 GB RAM. 8MB cache.
It looks like even the machine I have access to shows this behavior, except at a larger size (somewhere between 1979 and 2073). There is no reason I can think of right now for the empty matrix multiplication to be faster at larger sizes.
I will be investigating a little bit more before coming back.
EDIT Jan 11, 2013
After #EitanT's post, I wanted to do a little bit more of digging. I wrote some C code to see how matlab may be creating a zeros matrix. Here is the c++ code that I used.
int main(int argc, char **argv)
{
for (int i = 1975; i <= 2100; i+=25) {
timer::start();
double *foo = (double *)malloc(i * i * sizeof(double));
for (int k = 0; k < i * i; k++) foo[k] = 0;
double mftime = timer::stop();
free(foo);
timer::start();
double *bar = (double *)malloc(i * i * sizeof(double));
memset(bar, 0, i * i * sizeof(double));
double mmtime = timer::stop();
free(bar);
timer::start();
double *baz = (double *)calloc(i * i, sizeof(double));
double catime = timer::stop();
free(baz);
printf("%d, %lf, %lf, %lf\n", i, mftime, mmtime, catime);
}
}
Here are the results.
$ ./test
1975, 0.013812, 0.013578, 0.003321
2000, 0.014144, 0.013879, 0.003408
2025, 0.014396, 0.014219, 0.003490
2050, 0.014732, 0.013784, 0.000043
2075, 0.015022, 0.014122, 0.000045
2100, 0.014606, 0.014480, 0.000045
As you can see calloc (4th column) seems to be the fastest method. It is also getting significantly faster between 2025 and 2050 (I'd assume it would at around 2048 ?).
Now I went back to matlab to check for the same. Here are the results.
>> test
1975, 0.003296, 0.003297
2000, 0.003377, 0.003385
2025, 0.003465, 0.003464
2050, 0.015987, 0.000019
2075, 0.016373, 0.000019
2100, 0.016762, 0.000020
It looks like both f() and g() are using calloc at smaller sizes (<2048 ?). But at larger sizes f() (zeros(m, n)) starts to use malloc + memset, while g() (zeros(m, 0) * zeros(0, n)) keeps using calloc.
So the divergence is explained by the following
zeros(..) begins to use a different (slower ?) scheme at larger sizes.
calloc also behaves somewhat unexpectedly, leading to an improvement in performance.
This is the behavior on Linux. Can someone do the same experiment on a different machine (and perhaps a different OS) and see if the experiment holds ?
The results might be a bit misleading. When you multiply two empty matrices, the resulting matrix is not immediately "allocated" and "initialized", rather this is postponed until you first use it (sort of like a lazy evaluation).
The same applies when indexing out of bounds to grow a variable, which in the case of numeric arrays fills out any missing entries with zeros (I discuss afterwards the non-numeric case). Of course growing the matrix this way does not overwrite existing elements.
So while it may seem faster, you are just delaying the allocation time until you actually first use the matrix. In the end you'll have similar timings as if you did the allocation from the start.
Example to show this behavior, compared to a few other alternatives:
N = 1000;
clear z
tic, z = zeros(N,N); toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
tic, z = zeros(N,0)*zeros(0,N); toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
tic, z(N,N) = 0; toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
tic, z = full(spalloc(N,N,0)); toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
tic, z(1:N,1:N) = 0; toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
val = 0;
tic, z = val(ones(N)); toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
clear z
tic, z = repmat(0, [N N]); toc
tic, z = z + 1; toc
assert(isequal(z,ones(N)))
The result shows that if you sum the elapsed time for both instructions in each case, you end up with similar total timings:
// zeros(N,N)
Elapsed time is 0.004525 seconds.
Elapsed time is 0.000792 seconds.
// zeros(N,0)*zeros(0,N)
Elapsed time is 0.000052 seconds.
Elapsed time is 0.004365 seconds.
// z(N,N) = 0
Elapsed time is 0.000053 seconds.
Elapsed time is 0.004119 seconds.
The other timings were:
// full(spalloc(N,N,0))
Elapsed time is 0.001463 seconds.
Elapsed time is 0.003751 seconds.
// z(1:N,1:N) = 0
Elapsed time is 0.006820 seconds.
Elapsed time is 0.000647 seconds.
// val(ones(N))
Elapsed time is 0.034880 seconds.
Elapsed time is 0.000911 seconds.
// repmat(0, [N N])
Elapsed time is 0.001320 seconds.
Elapsed time is 0.003749 seconds.
These measurements are too small in the milliseconds and might not be very accurate, so you might wanna run these commands in a loop a few thousand times and take the average. Also sometimes running saved M-functions is faster than running scripts or on the command prompt, as certain optimizations only happen that way...
Either way allocation is usually done once, so who cares if it takes an extra 30ms :)
A similar behavior can be seen with cell arrays or arrays of structures. Consider the following example:
N = 1000;
tic, a = cell(N,N); toc
tic, b = repmat({[]}, [N,N]); toc
tic, c{N,N} = []; toc
which gives:
Elapsed time is 0.001245 seconds.
Elapsed time is 0.040698 seconds.
Elapsed time is 0.004846 seconds.
Note that even if they are all equal, they occupy different amount of memory:
>> assert(isequal(a,b,c))
>> whos a b c
Name Size Bytes Class Attributes
a 1000x1000 8000000 cell
b 1000x1000 112000000 cell
c 1000x1000 8000104 cell
In fact the situation is a bit more complicated here, since MATLAB is probably sharing the same empty matrix for all the cells, rather than creating multiple copies.
The cell array a is in fact an array of uninitialized cells (an array of NULL pointers), while b is a cell array where each cell is an empty array [] (internally and because of data sharing, only the first cell b{1} points to [] while all the rest have a reference to the first cell). The final array c is similar to a (uninitialized cells), but with the last one containing an empty numeric matrix [].
I looked around the list of exported C functions from the libmx.dll (using Dependency Walker tool), and I found a few interesting things.
there are undocumented functions for creating uninitialized arrays: mxCreateUninitDoubleMatrix, mxCreateUninitNumericArray, and mxCreateUninitNumericMatrix. In fact there is a submission on the File Exchange makes use of these functions to provide a faster alternative to zeros function.
there exist an undocumented function called mxFastZeros. Googling online, I can see you cross-posted this question on MATLAB Answers as well, with some excellent answers over there. James Tursa (same author of UNINIT from before) gave an example on how to use this undocumented function.
libmx.dll is linked against tbbmalloc.dll shared library. This is Intel TBB scalable memory allocator. This library provides equivalent memory allocation functions (malloc, calloc, free) optimized for parallel applications. Remember that many MATLAB functions are automatically multithreaded, so I wouldn't be surprised if zeros(..) is multithreaded and is using Intel's memory allocator once the matrix size is large enough (here is recent comment by Loren Shure that confirms this fact).
Regarding the last point about the memory allocator, you could write a similar benchmark in C/C++ similar to what #PavanYalamanchili did, and compare the various allocators available. Something like this. Remember that MEX-files have a slightly higher memory management overhead, since MATLAB automatically frees any memory that was allocated in MEX-files using the mxCalloc, mxMalloc, or mxRealloc functions. For what it's worth, it used to be possible to change the internal memory manager in older versions.
EDIT:
Here is a more thorough benchmark to compare the discussed alternatives. It specifically shows that once you stress the use of the entire allocated matrix, all three methods are on equal footing, and the difference is negligible.
function compare_zeros_init()
iter = 100;
for N = 512.*(1:8)
% ZEROS(N,N)
t = zeros(iter,3);
for i=1:iter
clear z
tic, z = zeros(N,N); t(i,1) = toc;
tic, z(:) = 9; t(i,2) = toc;
tic, z = z + 1; t(i,3) = toc;
end
fprintf('N = %4d, ZEROS = %.9f\n', N, mean(sum(t,2)))
% z(N,N)=0
t = zeros(iter,3);
for i=1:iter
clear z
tic, z(N,N) = 0; t(i,1) = toc;
tic, z(:) = 9; t(i,2) = toc;
tic, z = z + 1; t(i,3) = toc;
end
fprintf('N = %4d, GROW = %.9f\n', N, mean(sum(t,2)))
% ZEROS(N,0)*ZEROS(0,N)
t = zeros(iter,3);
for i=1:iter
clear z
tic, z = zeros(N,0)*zeros(0,N); t(i,1) = toc;
tic, z(:) = 9; t(i,2) = toc;
tic, z = z + 1; t(i,3) = toc;
end
fprintf('N = %4d, MULT = %.9f\n\n', N, mean(sum(t,2)))
end
end
Below are the timings averaged over 100 iterations in terms of increasing matrix size. I performed the tests in R2013a.
>> compare_zeros_init
N = 512, ZEROS = 0.001560168
N = 512, GROW = 0.001479991
N = 512, MULT = 0.001457031
N = 1024, ZEROS = 0.005744873
N = 1024, GROW = 0.005352638
N = 1024, MULT = 0.005359236
N = 1536, ZEROS = 0.011950846
N = 1536, GROW = 0.009051589
N = 1536, MULT = 0.008418878
N = 2048, ZEROS = 0.012154002
N = 2048, GROW = 0.010996315
N = 2048, MULT = 0.011002169
N = 2560, ZEROS = 0.017940950
N = 2560, GROW = 0.017641046
N = 2560, MULT = 0.017640323
N = 3072, ZEROS = 0.025657999
N = 3072, GROW = 0.025836506
N = 3072, MULT = 0.051533432
N = 3584, ZEROS = 0.074739924
N = 3584, GROW = 0.070486857
N = 3584, MULT = 0.072822335
N = 4096, ZEROS = 0.098791732
N = 4096, GROW = 0.095849788
N = 4096, MULT = 0.102148452
After doing some research, I've found this article in "Undocumented Matlab", in which Mr. Yair Altman had already come to the conclusion that MathWork's way of preallocating matrices using zeros(M, N) is indeed not the most efficient way.
He timed x = zeros(M,N) vs. clear x, x(M,N) = 0 and found that the latter is ~500 times faster. According to his explanation, the second method simply creates an M-by-N matrix, the elements of which being automatically initialized to 0. The first method however, creates x (with x having automatic zero elements) and then assigns a zero to every element in x again, and that is a redundant operation that takes more time.
In the case of empty matrix multiplication, such as what you've shown in your question, MATLAB expects the product to be an M×N matrix, and therefore it allocates an M×N matrix. Consequently, the output matrix is automatically initialized to zeroes. Since the original matrices are empty, no further calculations are performed, and hence the elements in the output matrix remain unchanged and equal to zero.
Interesting question, apparently there are several ways to 'beat' the built-in zeros function. My only guess as to why this is happening would be that it could be more memory efficient (after all, zeros(LargeNumer) will sooner cause Matlab to hit the memory limit than form a devestating speed bottleneck in most code), or more robust somehow.
Here is another fast allocation method using a sparse matrix, i have added the regular zeros function as a benchmark:
tic; x=zeros(1000,1000); toc
Elapsed time is 0.002863 seconds.
tic; clear x; x(1000,1000)=0; toc
Elapsed time is 0.000282 seconds.
tic; x=full(spalloc(1000,1000,0)); toc
Elapsed time is 0.000273 seconds.
tic; x=spalloc(1000,1000,1000000); toc %Is this the same for practical purposes?
Elapsed time is 0.000281 seconds.

Speeding up MATLAB code for FDR estimation

I have 2 input variables:
a vector of p-values (p) with N elements (unsorted)
and N x M matrix with p-values obtained by random permutations (pr) with M iterations. N is quite large, 10K to 100K or more. M let's say 100.
I'm estimating the False Discovery Rate (FDR) for each element of p representing how many p-values from random permutations will pass if the current p-value (from p) will be the threshold.
I wrote the function with ARRAYFUN, but it takes lot of time for large N (2 min for N=20K), comparable to for-loop.
function pfdr = fdr_from_random_permutations(p, pr)
%# ... skipping arguments checks
pfdr = arrayfun( #(x) mean(sum(pr<=x))./sum(p<=x), p);
Any ideas how to make it faster?
Comments about statistical issues here are also welcome.
The test data can be generated as p = rand(N,1); pr = rand(N,M);.
Well, the trick was indeed sorting the vectors. I give credit to #EgonGeerardyn for that. Also, there is no need to use mean. You can just divide everything afterwards by M. When p is sorted, finding the amount of values that are less than current x, is just a running index. pr is a more interesting case - I used a running index called place to discover how many elements are less than x.
Edit(2): Here is the fastest version I come up with:
function Speedup2()
N = 10000/4 ;
M = 100/4 ;
p = rand(N,1); pr = rand(N,M);
tic
pfdr = arrayfun( #(x) mean(sum(pr<=x))./sum(p<=x), p);
toc
tic
out = zeros(numel(p),1);
[p,sortIndex] = sort(p);
pr = sort(pr(:));
pr(end+1) = Inf;
place = 1;
N = numel(pr);
for i=1:numel(p)
x = p(i);
while pr(place)<=x
place = place+1;
end
exp1a = place-1;
exp2 = i;
out(i) = exp1a/exp2;
end
out(sortIndex) = out/ M;
toc
disp(max(abs(pfdr-out)));
end
And the benchmark results for N = 10000/4 ; M = 100/4 :
Elapsed time is 0.898689 seconds.
Elapsed time is 0.007697 seconds.
2.220446049250313e-016
and for N = 10000 ; M = 100 ;
Elapsed time is 39.730695 seconds.
Elapsed time is 0.088870 seconds.
2.220446049250313e-016
First of all, tr to analyze this using the profiler. Profiling should ALWAYS be the first step when trying to improve performance. We can all guess at what is causing your performance drop, but the only way to be sure and focus on the right part is to inspect the profiler report.
I didn't run the profiler on your code, as I don't want to generate test data to do so; but I have some ideas about what work is being carried out in vain. In your function mean(sum(pr<=x))./sum(p<=x), you are repeatedly summing over p<=x. All in all, one call includes N comparisons and N-1 summations. So for both, you have behavior that is quadratic in N when all N values of p are calculated.
If you step through a sorted version of p, you need less calculations and comparisons, as you can keep track of a running sum (i.e. behavior that is linear in N). I guess a similar method could be applied to the other part of the calculation.
edit:
The implementation of my idea as expressed above:
function pfdr = fdr(p,pr)
[N, M] = size(pr);
[p, idxP] = sort(p);
[pr] = sort(pr(:));
pfdr = NaN(N,1);
parfor iP = 1:N
x = p(iP);
m = sum(pr<=x)/M;
pfdr(iP) = m/iP;
end
pfdr(idxP) = pfdr;
If you have access to the parallel computing toolbox, the parfor loop will allow you to gain some performance. I used two basic ideas: mean(sum(pr<=x)) is actually equal to sum(pr(:)<=x)/M. On the other hand, since p is sorted, this allows you to just take the index as the number of elements (in the assumption that every element is unique, otherwise you'll have to work with unique to do the full rigorous analysis).
As you should already know very well by running the profiler yourself, the line m = sum(pr<=x)/M; is the main resource hog. This can be tackled similarly to p by making use of the sorted nature of pr.
I tested my code (both for identical results and for time consumption) against yours. For N=20e3; M=100, I get about 63 seconds to run your code and 43 seconds to run mine on my main computer (MATLAB 2011a on 64 bit Arch Linux, 8 GiB RAM, Core i7 860). For smaller values of M the gain is larger. But this gain is in part due to parallelization.
edit2: Apparently, I came to very similar results as Andrey, my result would have been very similar had I pursued the same approach.
However, I realised that there are some built-in functions that do more or less what you need, i.e. quite similar to determining the empirical cumulative density function. And this can be done by constructing the histogram:
function pfdr = fdr(p,pr)
[N, M] = size(pr);
[p, idxP] = sort(p);
count = histc(pr(:), [0; p]);
count = cumsum(count(1:N));
pfdr = count./(1:N).';
pfdr(idxP) = pfdr/M;
For the same M and N as above, this code takes 228 milliseconds on my computer. It takes 104 milliseconds for Andrey's parameters, so on my computer it turns out a bit slower, but I think this code is far more readable than intricate for loops (as was the case in both our examples).
Following the discussion between me and Andrey in this question, this very late answer is just to prove to Andrey that vectorized solutions are still faster than JIT'ed loops, they sometimes just aren't as easy to find.
I am more than willing to remove this answer if it is deemed inappropriate by the OP.
Now, on to business, here's the original arrayfun, looped version by Andrey, and vectorized version by Egon:
function test
clc
N = 10000/4 ;
M = 100/4 ;
p = rand(N,1);
pr = rand(N,M);
%% first option
tic
pfdr = arrayfun( #(x) mean(sum(pr<=x))./sum(p<=x), p);
toc
%% second option
tic
out = zeros(numel(p),1);
[p2,sortIndex] = sort(p);
pr2 = sort(pr(:));
pr2(end+1) = Inf;
place = 1;
for i=1:numel(p2)
x = p2(i);
while pr2(place)<=x
place = place+1;
end
exp1a = place-1;
exp2 = i;
out(i) = exp1a/exp2;
end
out(sortIndex) = out/ M;
toc
%% third option
tic
[p2,sortIndex] = sort(p);
count = histc(pr2(:), [0; p2]);
count = cumsum(count(1:N));
out = count./(1:N).';
out(sortIndex) = out/M;
toc
end
Results on my laptop:
Elapsed time is 0.916196 seconds.
Elapsed time is 0.011429 seconds.
Elapsed time is 0.007328 seconds.
and for N=1000; M = 100; :
Elapsed time is 38.082718 seconds.
Elapsed time is 0.127052 seconds.
Elapsed time is 0.042686 seconds.
So: vectorized is 2-3 times faster.

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