If i want to find the union of two unordered sets, represented as 1D vectors, eg:
a = [2 4 6 8 1]
b = [1 2 5 7 9]
i can use the union function:
c = union(a,b)
which gives the answer:
c = [1 2 4 5 6 7 8 9]
however, this appears to be quite slow (relatively speaking). If i run a tic-toc test on it i get:
>> for test = 1
tic
c = union(a,b);
toc
end
Elapsed time is 0.000906 seconds.
whereas, if i use this much more convoluted method, i get a much quicker result:
>> for test = 1
tic
a_1 = zeros(1,9);
b_1 = zeros(1,9);
a_1(a) = 1;
b_1(b) = 1;
c_1 = or(a_1,b_1);
c = find(c_1);
toc
end
Elapsed time is 0.000100 seconds.
this still gives the same answer for c, but is about 9 times as fast (with this small example, im not sure about how well it scales)
what is the advantage of ever using union? and can anyone suggest a more compact way of expressing the second method that i used?
thanks
Several comments:
Your code isn't as general as union. You're assuming set members are strictly positive integers.
Your code will blow up if a set member has a large value, eg. 10^20, as it tries to allocate an absurd quantity of memory.
A bunch of Matlab functions that don't use BLAS/LAPACK or aren't built-in are actually quite slow. They're there for convenience but don't be shocked when you can roll something faster yourself, especially if you can specialize for your particular problem.
Representing sets using logical arrays (which is what you do) can be highly efficient for problems where the set of possible set members is quite small.
Related
I am trying to find a solution in which a given resource (eg. budget) will be best distributed to different options which yields different results on the resource provided.
Let's say I have N = 1200 and some functions. (a, b, c, d are some unknown variables)
f1(x) = a * x
f2(x) = b * x^c
f3(x) = a*x + b*x^2 + c*x^3
f4(x) = d^x
f5(x) = log x^d
...
And also, let's say there n number of these functions that yield different results based on its input x, where x = 0 or x >= m, where m is a constant.
Although I am not able to find exact formula for the given functions, I am able to find the output. This means that I can do:
X = f1(N1) + f2(N2) + f3(N3) + ... + fn(Nn) where (N1 + ... Nn) = N as many times as there are ways of distributing N into n numbers, and find a specific case where X is the greatest.
How would I actually go about finding the best distribution of N with the least computation power, using whatever libraries currently available?
If you are happy with allocations constrained to be whole numbers then there is a dynamic programming solution of cost O(Nn) - so you can increase accuracy by scaling if you want, but this will increase cpu time.
For each i=1 to n maintain an array where element j gives the maximum yield using only the first i functions giving them a total allowance of j.
For i=1 this is simply the result of f1().
For i=k+1 consider when working out the result for j consider each possible way of splitting j units between f_{k+1}() and the table that tells you the best return from a distribution among the first k functions - so you can calculate the table for i=k+1 using the table created for k.
At the end you get the best possible return for n functions and N resources. It makes it easier to find out what that best answer is if you maintain of a set of arrays telling the best way to distribute k units among the first i functions, for all possible values of i and k. Then you can look up the best allocation for f100(), subtract off the value this allocated to f100() from N, look up the best allocation for f99() given the resulting resources, and carry on like this until you have worked out the best allocations for all f().
As an example suppose f1(x) = 2x, f2(x) = x^2 and f3(x) = 3 if x>0 and 0 otherwise. Suppose we have 3 units of resource.
The first table is just f1(x) which is 0, 2, 4, 6 for 0,1,2,3 units.
The second table is the best you can do using f1(x) and f2(x) for 0,1,2,3 units and is 0, 2, 4, 9, switching from f1 to f2 at x=2.
The third table is 0, 3, 5, 9. I can get 3 and 5 by using 1 unit for f3() and the rest for the best solution in the second table. 9 is simply the best solution in the second table - there is no better solution using 3 resources that gives any of them to f(3)
So 9 is the best answer here. One way to work out how to get there is to keep the tables around and recalculate that answer. 9 comes from f3(0) + 9 from the second table so all 3 units are available to f2() + f1(). The second table 9 comes from f2(3) so there are no units left for f(1) and we get f1(0) + f2(3) + f3(0).
When you are working the resources to use at stage i=k+1 you have a table form i=k that tells you exactly the result to expect from the resources you have left over after you have decided to use some at stage i=k+1. The best distribution does not become incorrect because that stage i=k you have worked out the result for the best distribution given every possible number of remaining resources.
It is really stupid all I am trying to do is having a 7 column matrix consisiting all mod 7 numbers and it takes a huge time to generate such a matrix utilizing the following code
to = 7^k;
msgValue = zeros(to,k);
for l=0:to
for kCounter=0:(k-1)
msgValue(l+1,kCounter+1)=mod((l/7^kCounter),7);
end
end
msgValue = floor(msgValue);
How can I do this faster?
Or another vectorized approach (direct matrix multiplication):
msgValue = floor( mod( (0:7^k).' * (1./(7.^(0:k-1))),7 ) ) ;
a wee bit faster than the famous bsxfun ;-)
%// For 10000 iterations, k=3
Elapsed time is 2.280774 seconds. %// double loop
Elapsed time is 1.329179 seconds. %// bsxfun
Elapsed time is 0.958945 seconds. %// matrix multiplication
You can use a vectorized approach with bsxfun -
msgValue = floor(mod(bsxfun(#rdivide,[0:to]',7.^(0:(k-1))),7));
Quick runtime tests for k = 7:
-------------------- With Original Approach
Elapsed time is 1.519023 seconds.
-------------------- With Proposed Approach
Elapsed time is 0.279547 seconds.
I used a submssion from matlab central called rude, which I tend to use from time to time and was able to eliminate one for loop and vectorize the code to some extent.
tic
k=7;
modval = 7;
to=modval^k;
mods = mod(0:(modval-1),modval);
msgValue=zeros(to,k);
for kCounter=1:k
aux = rude(modval^(kCounter-1)*ones(1,modval),mods)';
msgValue(:,kCounter) = repmat(aux,to/(7^kCounter),1);
end
toc
The idea behind the code is to make at the beginning of each iteration the building block of the column vector using the rude function. Rude, in turn, uses mods = [0 1 2 3 4 5 6] as a starting point for the manipulation. The real work is done through vectorization.
You did not mention how long your code takes to run. So I timed it just once to give you a rough idea. It ran in 0.43 seconds in my machine, a Windows 7 Ultimate, 2.4 GHz, 4GB Ram, Dual CPU.
Also, the way you defined your loop adds a repetition in your msgValue matrix. The first row consists of zero values throughout all columns, and so the last row, which I also fixed. For a toy example with k=3, your code returns a 344x1 matrix, while you explicitly initialize it as a 7³x1 (343x1) matrix.
I am trying to evaluate the following integral:
I can find the area for the following polynomial as follows:
pn =
-0.0250 0.0667 0.2500 -0.6000 0
First using the integration by Simpson's rule
fn=#(x) exp(polyval(pn,x));
area=quad(fn,-10,10);
fprintf('area evaluated by Simpsons rule : %f \n',area)
and the result is area evaluated by Simpsons rule : 11.483072
Then with the following code that evaluates the summation in the above formula with gamma function
a=pn(1);b=pn(2);c=pn(3);d=pn(4);f=pn(5);
area=0;
result=0;
for n=0:40;
for m=0:40;
for p=0:40;
if(rem(n+p,2)==0)
result=result+ (b^n * c^m * d^p) / ( factorial(n)*factorial(m)*factorial(p) ) *...
gamma( (3*n+2*m+p+1)/4 ) / (-a)^( (3*n+2*m+p+1)/4 );
end
end
end
end
result=result*1/2*exp(f)
and this returns 11.4831. More or less the same result with the quad function. Now my question is whether or not it is possible for me to get rid of this nested loop as I will construct the cumulative distribution function so that I can get samples from this distribution using the inverse CDF transform. (for constructing the cdf I will use gammainc i.e. the incomplete gamma function instead of gamma)
I will need to sample from such densities that may have different polynomial coefficients and speed is of concern to me. I can already sample from such densities using Monte Carlo methods but I would like to see whether or not it is possible for me to use exact sampling from the density in order to speed up.
Thank you very much in advance.
There are several things one might do. The simplest is to avoid calling factorial. Instead one can use the relation that
factorial(n) = gamma(n+1)
Since gamma seems to be actually faster than a call to factorial, you can save a bit there. Even better, you can
>> timeit(#() factorial(40))
ans =
4.28681157826087e-05
>> timeit(#() gamma(41))
ans =
2.06671024634146e-05
>> timeit(#() gammaln(41))
ans =
2.17632543333333e-05
Even better, one can do all 4 calls in a single call to gammaln. For example, think about what this does:
gammaln([(3*n+2*m+p+1)/4,n+1,m+1,p+1])*[1 -1 -1 -1]'
Note that this call has no problem with overflows either in case your numbers get large enough. And since gammln is vectorized, that one call is fast. It costs little more time to compute 4 values than it does to compute one.
>> timeit(#() gammaln([15 20 40 30]))
ans =
2.73937416896552e-05
>> timeit(#() gammaln(40))
ans =
2.46521943333333e-05
Admittedly, if you use gammaln, you will need a call to exp at the end to recover the final result. You could do it with a single call to gamma however too. Perhaps like this:
g = gamma([(3*n+2*m+p+1)/4,n+1,m+1,p+1]);
g = g(1)/(g(2)*g(3)*g(4));
Next, you can be more creative in the inner loop on p. Rather than a full loop, coupled with a test to ignore the combinations you don't need, why not just do this?
for p=mod(n,2):2:40
That statement will select only those values of p that would have been used anyway, so now you can drop the if statement completely.
All of the above will give you what I'll guess is about a 5x speed increase in your loops. But it still has a set of nested loops. With some effort, you might be able to improve that too.
For example, rather than computing all of those factorials (or gamma functions) many times, do it ONCE. This should work:
a=pn(1);b=pn(2);c=pn(3);d=pn(4);f=pn(5);
area=0;
result=0;
nlim = 40;
facts = factorial(0:nlim);
gammas = gamma((0:(6*nlim+1))/4);
for n=0:nlim
for m=0:nlim
for p=mod(n,2):2:nlim
result = result + (b.^n * c.^m * d.^p) ...
.*gammas(3*n+2*m+p+1 + 1) ...
./ (facts(n+1).*facts(m+1).*facts(p+1)) ...
./ (-a)^( (3*n+2*m+p+1)/4 );
end
end
end
result=result*1/2*exp(f)
In my test on my machine, I find that your triply nested loops required 4.3 seconds to run. My version above produces the same result, yet required only 0.028418 seconds, a speedup of roughly 150 to 1, despite the triply nested loops.
Well, without even making changes to your code you could install an excellent package from Tom Minka at Microsoft called lightspeed which replaces some built-in matlab functions with much faster versions. I know there's a replacement for gammaln().
You'll get nontrivial speed improvements, though I'm not sure how much, and it's straight-forward to install.
Lets have matrix A say A = magic(100);. I have seen 2 ways of computing sum of all elements of matrix A.
sumOfA = sum(sum(A));
Or
sumOfA = sum(A(:));
Is one of them faster (or better practise) then other? If so which one is it? Or are they both equally fast?
It seems that you can't make up your mind about whether performance or floating point accuracy is more important.
If floating point accuracy were of paramount accuracy, then you would segregate the positive and negative elements, sorting each segment. Then sum in order of increasing absolute value. Yeah, I know, its more work than anyone would do, and it probably will be a waste of time.
Instead, use adequate precision such that any errors made will be irrelevant. Use good numerical practices about tests, etc, such that there are no problems generated.
As far as the time goes, for an NxM array,
sum(A(:)) will require N*M-1 additions.
sum(sum(A)) will require (N-1)*M + M-1 = N*M-1 additions.
Either method requires the same number of adds, so for a large array, even if the interpreter is not smart enough to recognize that they are both the same op, who cares?
It is simply not an issue. Don't make a mountain out of a mole hill to worry about this.
Edit: in response to Amro's comment about the errors for one method over the other, there is little you can control. The additions will be done in a different order, but there is no assurance about which sequence will be better.
A = randn(1000);
format long g
The two solutions are quite close. In fact, compared to eps, the difference is barely significant.
sum(A(:))
ans =
945.760668102446
sum(sum(A))
ans =
945.760668102449
sum(sum(A)) - sum(A(:))
ans =
2.72848410531878e-12
eps(sum(A(:)))
ans =
1.13686837721616e-13
Suppose you choose the segregate and sort trick I mentioned. See that the negative and positive parts will be large enough that there will be a loss of precision.
sum(sort(A(A<0),'descend'))
ans =
-398276.24754782
sum(sort(A(A<0),'descend')) + sum(sort(A(A>=0),'ascend'))
ans =
945.7606681037
So you really would need to accumulate the pieces in a higher precision array anyway. We might try this:
[~,tags] = sort(abs(A(:)));
sum(A(tags))
ans =
945.760668102446
An interesting problem arises even in these tests. Will there be an issue because the tests are done on a random (normal) array? Essentially, we can view sum(A(:)) as a random walk, a drunkard's walk. But consider sum(sum(A)). Each element of sum(A) (i.e., the internal sum) is itself a sum of 1000 normal deviates. Look at a few of them:
sum(A)
ans =
Columns 1 through 6
-32.6319600960983 36.8984589766173 38.2749084367497 27.3297721091922 30.5600109446534 -59.039228262402
Columns 7 through 12
3.82231962760523 4.11017616179294 -68.1497901792032 35.4196443983385 7.05786623564426 -27.1215387236418
Columns 13 through 18
When we add them up, there will be a loss of precision. So potentially, the operation as sum(A(:)) might be slightly more accurate. Is it so? What if we use a higher precision for the accumulation? So first, I'll form the sum down the columns using doubles, then convert to 25 digits of decimal precision, and sum the rows. (I've displayed only 20 digits here, leaving 5 digits hidden as guard digits.)
sum(hpf(sum(A)))
ans =
945.76066810244807408
Or, instead, convert immediately to 25 digits of precision, then summing the result.
sum(hpf(A(:))
945.76066810244749807
So both forms in double precision were equally wrong here, in opposite directions. In the end, this is all moot, since any of the alternatives I've shown are far more time consuming compared to the simple variations sum(A(:)) or sum(sum(A)). Just pick one of them and don't worry.
Performance-wise, I'd say both are very similar (assuming a recent MATLAB version). Here is quick test using the TIMEIT function:
function sumTest()
M = randn(5000);
timeit( #() func1(M) )
timeit( #() func2(M) )
end
function v = func1(A)
v = sum(A(:));
end
function v = func2(A)
v = sum(sum(A));
end
the results were:
>> sumTest
ans =
0.0020917
ans =
0.0017159
What I would worry about is floating-point issues. Example:
>> M = randn(1000);
>> abs( sum(M(:)) - sum(sum(M)) )
ans =
3.9108e-11
Error magnitude increases for larger matrices
i think a simple way to understand is apply " tic_ toc "function in first and last of your code.
tic
A = randn(5000);
format long g
sum(A(:));
toc
but when you used randn function ,elements of it are random and time of calculation can
different in each cycle CPU calculation .
This better you used a unique matrix whit so large elements to compare time of calculation.
I am just learning MATLAB and I find it hard to understand the performance factors of loops vs vectorized functions.
In my previous question: Nested for loops extremely slow in MATLAB (preallocated) I realized that using a vectorized function vs. 4 nested loops made a 7x times difference in running time.
In that example instead of looping through all dimensions of a 4 dimensional array and calculating median for each vector, it was much cleaner and faster to just call median(stack, n) where n meant the working dimension of the median function.
But median is just a very easy example and I was just lucky that it had this dimension parameter implemented.
My question is that how do you write a function yourself which works as efficiently as one which has this dimension range implemented?
For example you have a function my_median_1D which only works on a 1-D vector and returns a number.
How do you write a function my_median_nD which acts like MATLAB's median, by taking an n-dimensional array and a "working dimension" parameter?
Update
I found the code for calculating median in higher dimensions
% In all other cases, use linear indexing to determine exact location
% of medians. Use linear indices to extract medians, then reshape at
% end to appropriate size.
cumSize = cumprod(s);
total = cumSize(end); % Equivalent to NUMEL(x)
numMedians = total / nCompare;
numConseq = cumSize(dim - 1); % Number of consecutive indices
increment = cumSize(dim); % Gap between runs of indices
ixMedians = 1;
y = repmat(x(1),numMedians,1); % Preallocate appropriate type
% Nested FOR loop tracks down medians by their indices.
for seqIndex = 1:increment:total
for consIndex = half*numConseq:(half+1)*numConseq-1
absIndex = seqIndex + consIndex;
y(ixMedians) = x(absIndex);
ixMedians = ixMedians + 1;
end
end
% Average in second value if n is even
if 2*half == nCompare
ixMedians = 1;
for seqIndex = 1:increment:total
for consIndex = (half-1)*numConseq:half*numConseq-1
absIndex = seqIndex + consIndex;
y(ixMedians) = meanof(x(absIndex),y(ixMedians));
ixMedians = ixMedians + 1;
end
end
end
% Check last indices for NaN
ixMedians = 1;
for seqIndex = 1:increment:total
for consIndex = (nCompare-1)*numConseq:nCompare*numConseq-1
absIndex = seqIndex + consIndex;
if isnan(x(absIndex))
y(ixMedians) = NaN;
end
ixMedians = ixMedians + 1;
end
end
Could you explain to me that why is this code so effective compared to the simple nested loops? It has nested loops just like the other function.
I don't understand how could it be 7x times faster and also, that why is it so complicated.
Update 2
I realized that using median was not a good example as it is a complicated function itself requiring sorting of the array or other neat tricks. I re-did the tests with mean instead and the results are even more crazy:
19 seconds vs 0.12 seconds.
It means that the built in way for sum is 160 times faster than the nested loops.
It is really hard for me to understand how can an industry leading language have such an extreme performance difference based on the programming style, but I see the points mentioned in the answers below.
Update 2 (to address your updated question)
MATLAB is optimized to work well with arrays. Once you get used to it, it is actually really nice to just have to type one line and have MATLAB do the full 4D looping stuff itself without having to worry about it. MATLAB is often used for prototyping / one-off calculations, so it makes sense to save time for the person coding, and giving up some of C[++|#]'s flexibility.
This is why MATLAB internally does some loops really well - often by coding them as a compiled function.
The code snippet you give doesn't really contain the relevant line of code which does the main work, namely
% Sort along given dimension
x = sort(x,dim);
In other words, the code you show only needs to access the median values by their correct index in the now-sorted multi-dimensional array x (which doesn't take much time). The actual work accessing all array elements was done by sort, which is a built-in (i.e. compiled and highly optimized) function.
Original answer (about how to built your own fast functions working on arrays)
There are actually quite a few built-ins that take a dimension parameter: min(stack, [], n), max(stack, [], n), mean(stack, n), std(stack, [], n), median(stack,n), sum(stack, n)... together with the fact that other built-in functions like exp(), sin() automatically work on each element of your whole array (i.e. sin(stack) automatically does four nested loops for you if stack is 4D), you can built up a lot of functions that you might need just be relying on the existing built-ins.
If this is not enough for a particular case you should have a look at repmat, bsxfun, arrayfun and accumarray which are very powerful functions for doing things "the MATLAB way". Just search on SO for questions (or rather answers) using one of these, I learned a lot about MATLABs strong points that way.
As an example, say you wanted to implement the p-norm of stack along dimension n, you could write
function result=pnorm(stack, p, n)
result=sum(stack.^p,n)^(1/p);
... where you effectively reuse the "which-dimension-capability" of sum.
Update
As Max points out in the comments, also have a look at the colon operator (:) which is a very powerful tool for selecting elements from an array (or even changing it shape, which is more generally done with reshape).
In general, have a look at the section Array Operations in the help - it contains repmat et al. mentioned above, but also cumsum and some more obscure helper functions which you should use as building blocks.
Vectorization
In addition to whats already been said, you should also understand that vectorization involves parallelization, i.e. performing concurrent operations on data as opposed to sequential execution (think SIMD instructions), and even taking advantage of threads and multiprocessors in some cases...
MEX-files
Now although the "interpreted vs. compiled" point has already been argued, no one mentioned that you can extend MATLAB by writing MEX-files, which are compiled executables written in C, that can be called directly as normal function from inside MATLAB. This allows you to implement performance-critical parts using a lower-level language like C.
Column-major order
Finally, when trying to optimize some code, always remember that MATLAB stores matrices in column-major order. Accessing elements in that order can yield significant improvements compared to other arbitrary orders.
For example, in your previous linked question, you were computing the median of set of stacked images along some dimension. Now the order in which those dimensions are ordered greatly affect the performance. Illustration:
%# sequence of 10 images
fPath = fullfile(matlabroot,'toolbox','images','imdemos');
files = dir( fullfile(fPath,'AT3_1m4_*.tif') );
files = strcat(fPath,{filesep},{files.name}'); %'
I = imread( files{1} );
%# stacked images along the 1st dimension: [numImages H W RGB]
stack1 = zeros([numel(files) size(I) 3], class(I));
for i=1:numel(files)
I = imread( files{i} );
stack1(i,:,:,:) = repmat(I, [1 1 3]); %# grayscale to RGB
end
%# stacked images along the 4th dimension: [H W RGB numImages]
stack4 = permute(stack1, [2 3 4 1]);
%# compute median image from each of these two stacks
tic, m1 = squeeze( median(stack1,1) ); toc
tic, m4 = median(stack4,4); toc
isequal(m1,m4)
The timing difference was huge:
Elapsed time is 0.257551 seconds. %# stack1
Elapsed time is 17.405075 seconds. %# stack4
Could you explain to me that why is this code so effective compared to the simple nested loops? It has nested loops just like the other function.
The problem with nested loops is not the nested loops themselves. It's the operations you perform inside.
Each function call (especially to a non-built-in function) generates a little bit of overhead; more so if the function performs e.g. error checking that takes the same amount of time regardless of input size. Thus, if a function has only a 1 ms overhead, if you call it 1000 times, you will have wasted a second. If you can call it once to perform a vectorized calculation, you pay overhead only once.
Furthermore, the JIT compiler (pdf) can help vectorize simple for-loops, where you, for example, only perform basic arithmetic operations. Thus, the loops with simple calculations in your post are sped up by a lot, while the loops calling median are not.
In this case
M = median(A,dim) returns the median values for elements along the dimension of A specified by scalar dim
But with a general function you can try splitting your array with mat2cell (which can work with n-D arrays and not just matrices) and applying your my_median_1D function through cellfun. Below I will use median as an example to show that you get equivalent results, but instead you can pass it any function defined in an m-file, or an anonymous function defined with the #(args) notation.
>> testarr = [[1 2 3]' [4 5 6]']
testarr =
1 4
2 5
3 6
>> median(testarr,2)
ans =
2.5000
3.5000
4.5000
>> shape = size(testarr)
shape =
3 2
>> cellfun(#median,mat2cell(testarr,repmat(1,1,shape(1)),[shape(2)]))
ans =
2.5000
3.5000
4.5000