I wonder why operating on Float64 values is faster than operating on Float16:
julia> rnd64 = rand(Float64, 1000);
julia> rnd16 = rand(Float16, 1000);
julia> #benchmark rnd64.^2
BenchmarkTools.Trial: 10000 samples with 10 evaluations.
Range (min … max): 1.800 μs … 662.140 μs ┊ GC (min … max): 0.00% … 99.37%
Time (median): 2.180 μs ┊ GC (median): 0.00%
Time (mean ± σ): 3.457 μs ± 13.176 μs ┊ GC (mean ± σ): 12.34% ± 3.89%
▁██▄▂▂▆▆▄▂▁ ▂▆▄▁ ▂▂▂▁ ▂
████████████████▇▇▆▆▇▆▅▇██▆▆▅▅▆▄▄▁▁▃▃▁▁▄▁▃▄▁▃▁▄▃▁▁▆▇██████▇ █
1.8 μs Histogram: log(frequency) by time 10.6 μs <
Memory estimate: 8.02 KiB, allocs estimate: 5.
julia> #benchmark rnd16.^2
BenchmarkTools.Trial: 10000 samples with 6 evaluations.
Range (min … max): 5.117 μs … 587.133 μs ┊ GC (min … max): 0.00% … 98.61%
Time (median): 5.383 μs ┊ GC (median): 0.00%
Time (mean ± σ): 5.716 μs ± 9.987 μs ┊ GC (mean ± σ): 3.01% ± 1.71%
▃▅█▇▅▄▄▆▇▅▄▁ ▁ ▂
▄██████████████▇▆▇▆▆▇▆▇▅█▇████▇█▇▇▆▅▆▄▇▇▆█▇██▇█▇▇▇▆▇▇▆▆▆▆▄▄ █
5.12 μs Histogram: log(frequency) by time 7.48 μs <
Memory estimate: 2.14 KiB, allocs estimate: 5.
Maybe you ask why I expect the opposite: Because Float16 values have less floating point precision:
julia> rnd16[1]
Float16(0.627)
julia> rnd64[1]
0.4375452455597999
Shouldn't calculations with fewer precisions take place faster? Then I wonder why someone should use Float16? They can do it even with Float128!
As you can see, the effect you are expecting is present for Float32:
julia> rnd64 = rand(Float64, 1000);
julia> rnd32 = rand(Float32, 1000);
julia> rnd16 = rand(Float16, 1000);
julia> #btime $rnd64.^2;
616.495 ns (1 allocation: 7.94 KiB)
julia> #btime $rnd32.^2;
330.769 ns (1 allocation: 4.06 KiB) # faster!!
julia> #btime $rnd16.^2;
2.067 μs (1 allocation: 2.06 KiB) # slower!!
Float64 and Float32 have hardware support on most platforms, but Float16 does not, and must therefore be implemented in software.
Note also that you should use variable interpolation ($) when micro-benchmarking. The difference is significant here, not least in terms of allocations:
julia> #btime $rnd32.^2;
336.187 ns (1 allocation: 4.06 KiB)
julia> #btime rnd32.^2;
930.000 ns (5 allocations: 4.14 KiB)
The short answer is that you probably shouldn't use Float16 unless you are using a GPU or an Apple CPU because (as of 2022) other processors don't have hardware support for Float16.
Related
In the Julia package BenchmarkTools, there are macros like #btime, #belapse that seem redundant to me since Julia has built-in #time, #elapse macros. And it seems to me that these macros serve the same purpose. So what's the difference between #time and #btime, and #elapse and #belapsed?
TLDR ;)
#time and #elapsed just run the code once and measure the time. This measurement may or may not include the compile time (depending whether #time is run for the first or second time) and includes time to resolve global variables.
On the the other hand #btime and #belapsed perform warm up so you know that compile time and global variable resolve time (if $ is used) do not affect the time measurement.
Details
For further understand how this works lets used the #macroexpand (I am also stripping comment lines for readability):
julia> using MacroTools, BenchmarkTools
julia> MacroTools.striplines(#macroexpand1 #elapsed sin(x))
quote
Experimental.#force_compile
local var"#28#t0" = Base.time_ns()
sin(x)
(Base.time_ns() - var"#28#t0") / 1.0e9
end
Compilation if sin is not forced and you get different results when running for the first time and subsequent times. For an example:
julia> #time cos(x);
0.110512 seconds (261.97 k allocations: 12.991 MiB, 99.95% compilation time)
julia> #time cos(x);
0.000008 seconds (1 allocation: 16 bytes)
julia> #time cos(x);
0.000006 seconds (1 allocation: : 16 bytes)
The situation is different with #belapsed:
julia> MacroTools.striplines(#macroexpand #belapsed sin($x))
quote
(BenchmarkTools).time((BenchmarkTools).minimum(begin
local var"##314" = begin
BenchmarkTools.generate_benchmark_definition(Main, Symbol[], Any[], [Symbol("##x#315")], (x,), $(Expr(:copyast, :($(QuoteNode(:(sin(var"##x#315"))))))), $(Expr(:copyast, :($(QuoteNode(nothing))))), $(Expr(:copyast, :($(QuoteNode(nothing))))), BenchmarkTools.Parameters())
end
(BenchmarkTools).warmup(var"##314")
(BenchmarkTools).tune!(var"##314")
(BenchmarkTools).run(var"##314")
end)) / 1.0e9
end
You can see that a minimum value is taken (the code is run several times).
Basically most time you should use BenchmarkTools for measuring times when designing your application.
Last but not least try #benchamrk:
julia> #benchmark sin($x)
BenchmarkTools.Trial: 10000 samples with 999 evaluations.
Range (min … max): 13.714 ns … 51.151 ns ┊ GC (min … max): 0.00% … 0.00%
Time (median): 13.814 ns ┊ GC (median): 0.00%
Time (mean ± σ): 14.089 ns ± 1.121 ns ┊ GC (mean ± σ): 0.00% ± 0.00%
█▇ ▂▄ ▁▂ ▃ ▁ ▂
██▆▅██▇▅▄██▃▁▃█▄▃▁▅█▆▁▄▃▅█▅▃▁▄▇▆▁▁▁▁▁▆▄▄▁▁▃▄▇▃▁▃▁▁▁▆▅▁▁▁▆▅▅ █
13.7 ns Histogram: log(frequency) by time 20 ns <
Memory estimate: 0 bytes, allocs estimate: 0.
In Julia I can use argmax(X) to find max element. If I want to find all element satisfying condition C I can use findall(C,X). But how can I combine the two? What's the most efficient/idiomatic/concise way to find maximum element index satisfying some condition in Julia?
If you'd like to avoid allocations, filtering the array lazily would work:
idx_filtered = (i for (i, el) in pairs(X) if C(el))
argmax(i -> X[i], idx_filtered)
Unfortunately, this is about twice as slow as a hand-written version. (edit: in my benchmarks, it's 2x slower on Intel Xeon Platinum but nearly equal on Apple M1)
function byhand(C, X)
start = findfirst(C, X)
isnothing(start) && return nothing
imax, max = start, X[start]
for i = start:lastindex(X)
if C(X[i]) && X[i] > max
imax, max = i, X[i]
end
end
imax, max
end
You can store the index returned by findall and subset it with the result of argmax of the vector fulfilling the condition.
X = [5, 4, -3, -5]
C = <(0)
i = findall(C, X);
i[argmax(X[i])]
#3
Or combine both:
argmax(i -> X[i], findall(C, X))
#3
Assuming that findall is not empty. Otherwise it need to be tested e.g. with isempty.
Benchmark
#Functions
function August(C, X)
idx_filtered = (i for (i, el) in pairs(X) if C(el))
argmax(i -> X[i], idx_filtered)
end
function byhand(C, X)
start = findfirst(C, X)
isnothing(start) && return nothing
imax, max = start, X[start]
for i = start:lastindex(X)
if C(X[i]) && X[i] > max
imax, max = i, X[i]
end
end
imax, max
end
function GKi1(C, X)
i = findall(C, X);
i[argmax(X[i])]
end
GKi2(C, X) = argmax(i -> X[i], findall(C, X))
#Data
using Random
Random.seed!(42)
n = 100000
X = randn(n)
C = <(0)
#Benchmark
using BenchmarkTools
suite = BenchmarkGroup()
suite["August"] = #benchmarkable August(C, $X)
suite["byhand"] = #benchmarkable byhand(C, $X)
suite["GKi1"] = #benchmarkable GKi1(C, $X)
suite["GKi2"] = #benchmarkable GKi2(C, $X)
tune!(suite);
results = run(suite)
#Results
results
#4-element BenchmarkTools.BenchmarkGroup:
# tags: []
# "August" => Trial(641.061 μs)
# "byhand" => Trial(261.135 μs)
# "GKi2" => Trial(259.260 μs)
# "GKi1" => Trial(339.570 μs)
results.data["August"]
#BenchmarkTools.Trial: 7622 samples with 1 evaluation.
# Range (min … max): 641.061 μs … 861.379 μs ┊ GC (min … max): 0.00% … 0.00%
# Time (median): 643.640 μs ┊ GC (median): 0.00%
# Time (mean ± σ): 653.027 μs ± 18.123 μs ┊ GC (mean ± σ): 0.00% ± 0.00%
#
# ▄█▅▄▃ ▂▂▃▁ ▁▃▃▂▂ ▁▃ ▁▁ ▁
# ██████▇████████████▇▆▆▇████▇▆██▇▇▇▆▆▆▅▇▆▅▅▅▅▆██▅▆▆▆▇▆▇▇▆▇▆▆▆▅ █
# 641 μs Histogram: log(frequency) by time 718 μs <
#
# Memory estimate: 16 bytes, allocs estimate: 1.
results.data["byhand"]
#BenchmarkTools.Trial: 10000 samples with 1 evaluation.
# Range (min … max): 261.135 μs … 621.141 μs ┊ GC (min … max): 0.00% … 0.00%
# Time (median): 261.356 μs ┊ GC (median): 0.00%
# Time (mean ± σ): 264.382 μs ± 11.638 μs ┊ GC (mean ± σ): 0.00% ± 0.00%
#
# █ ▁▁▁▁ ▂ ▁▁ ▂ ▁ ▁ ▁
# █▅▂▂▅████▅▄▃▄▆█▇▇▆▄▅███▇▄▄▅▆▆█▄▇█▅▄▅▅▆▇▇▅▄▅▄▄▄▃▄▃▃▃▄▅▆▅▄▇█▆▅▄ █
# 261 μs Histogram: log(frequency) by time 292 μs <
#
# Memory estimate: 32 bytes, allocs estimate: 1.
results.data["GKi1"]
#BenchmarkTools.Trial: 10000 samples with 1 evaluation.
# Range (min … max): 339.570 μs … 1.447 ms ┊ GC (min … max): 0.00% … 0.00%
# Time (median): 342.579 μs ┊ GC (median): 0.00%
# Time (mean ± σ): 355.167 μs ± 52.935 μs ┊ GC (mean ± σ): 1.90% ± 6.85%
#
# █▆▄▅▃▂▁▁ ▁ ▁
# ████████▇▆▆▅▅▅▆▄▄▄▄▁▃▁▁▃▄▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁█ █
# 340 μs Histogram: log(frequency) by time 722 μs <
#
# Memory estimate: 800.39 KiB, allocs estimate: 11.
results.data["GKi2"]
#BenchmarkTools.Trial: 10000 samples with 1 evaluation.
# Range (min … max): 259.260 μs … 752.773 μs ┊ GC (min … max): 0.00% … 54.40%
# Time (median): 260.692 μs ┊ GC (median): 0.00%
# Time (mean ± σ): 270.300 μs ± 40.094 μs ┊ GC (mean ± σ): 1.31% ± 5.60%
#
# █▁▁▅▄▂▂▄▃▂▁▁▁ ▁ ▁
# █████████████████▇██▆▆▇▆▅▄▆▆▆▄▅▄▆▅▇▇▆▆▅▅▄▅▃▃▅▃▄▁▁▁▃▁▃▃▃▄▃▃▁▃▃ █
# 259 μs Histogram: log(frequency) by time 390 μs <
#
# Memory estimate: 408.53 KiB, allocs estimate: 9.
versioninfo()
#Julia Version 1.8.0
#Commit 5544a0fab7 (2022-08-17 13:38 UTC)
#Platform Info:
# OS: Linux (x86_64-linux-gnu)
# CPU: 8 × Intel(R) Core(TM) i7-2600K CPU # 3.40GHz
# WORD_SIZE: 64
# LIBM: libopenlibm
# LLVM: libLLVM-13.0.1 (ORCJIT, sandybridge)
# Threads: 1 on 8 virtual cores
In this example argmax(i -> X[i], findall(C, X)) is close to the performance of the hand written function of #August but uses more memory, but can show better performance in case the data is sorted:
sort!(X)
results = run(suite)
#4-element BenchmarkTools.BenchmarkGroup:
# tags: []
# "August" => Trial(297.519 μs)
# "byhand" => Trial(270.486 μs)
# "GKi2" => Trial(242.320 μs)
# "GKi1" => Trial(319.732 μs)
From what I understand from your question you can use findmax() (requires Julia >= v1.7) to find the maximum index on the result of findall():
julia> v = [10, 20, 30, 40, 50]
5-element Vector{Int64}:
10
20
30
40
50
julia> findmax(findall(x -> x > 30, v))[1]
5
Performance of the above function:
julia> v = collect(10:1:10_000_000);
julia> #btime findmax(findall(x -> x > 30, v))[1]
33.471 ms (10 allocations: 77.49 MiB)
9999991
Update: solution suggested by #dan-getz of using last() and findlast() perform better than findmax() but findlast() is the winner:
julia> #btime last(findall(x -> x > 30, v))
19.961 ms (9 allocations: 77.49 MiB)
9999991
julia> #btime findlast(x -> x > 30, v)
81.422 ns (2 allocations: 32 bytes)
Update 2: Looks like the OP wanted to find the max element and not only the index. In that case, the solution would be:
julia> v[findmax(findall(x -> x > 30, v))[1]]
50
I have simulation program written in Julia that does something equivalent to this as a part of its main loop:
# Some fake data
M = [randn(100,100) for m=1:100, n=1:100]
W = randn(100,100)
work = zip(W,M)
result = mapreduce(x -> x[1]*x[2], +,work)
In other words, a simple sum of weighted matrices. Timing the above code yields
0.691084 seconds (79.03 k allocations: 1.493 GiB, 70.59% gc time, 2.79% compilation time)
I am surprised about the large number of memory allocations, as this problem should be possible to do in-place. To see if it was my use of mapreduce that was wrong I also tested the following equivalent implementation:
#time begin
res = zeros(100,100)
for m=1:100
for n=1:100
res += W[m,n] * M[m,n]
end
end
end
which gave
0.442521 seconds (50.00 k allocations: 1.491 GiB, 70.81% gc time)
So, if I wrote this in C++ or Fortran it would be simple to do all of this in-place. Is this impossible in Julia? Or am I missing something here...?
It is possible to do it in place like this:
function ws(W, M)
res = zeros(100,100)
for m=1:100
for n=1:100
#. res += W[m,n] * M[m, n]
end
end
return res
end
and the timing is:
julia> #time ws(W, M);
0.100328 seconds (2 allocations: 78.172 KiB)
Note that in order to perform this operation in-place I used broadcasting (I could also use loops, but it would be the same).
The problem with your code is that in line:
res += W[m,n] * M[m,n]
You get two allocations:
When you do multiplication W[m,n] * M[m,n] a new matrix is allocated.
When you do addition res += ... again a matrix is allocated
By using broadcasting with #. you perform an in-place operation, see https://docs.julialang.org/en/v1/manual/mathematical-operations/#man-dot-operators for more explanations.
Additionally note that I have wrapped the code inside a function. If you do not do it then access both W and M is type unstable which also causes allocations, see https://docs.julialang.org/en/v1/manual/performance-tips/#Avoid-global-variables.
I'd like to add something to Bogumił's answer. The missing broadcast is the main problem, but in addition, the loop and the mapreduce variant differ in a fundamental semantic way.
The purpose of mapreduce is to reduce by an associative operation with identity element init in an unspecified order. This in particular also includes the (theoretical) option of running parts in parallel and doesn't really play well with mutation. From the docs:
The associativity of the reduction is implementation-dependent. Additionally, some implementations may reuse the return value of f for elements that appear multiple times in itr. Use mapfoldl or
mapfoldr instead for guaranteed left or right associativity and invocation of f for every value.
and
It is unspecified whether init is used for non-empty collections.
What the loop variant really corresponds to is a fold, which has a well-defined order and initial (not necessarily identity) element and can thus use an in-place reduction operator:
Like reduce, but with guaranteed left associativity. If provided, the keyword argument init will be used exactly once.
julia> #benchmark foldl((acc, (m, w)) -> (#. acc += m * w), $work; init=$(zero(W)))
BenchmarkTools.Trial: 45 samples with 1 evaluation.
Range (min … max): 109.967 ms … 118.251 ms ┊ GC (min … max): 0.00% … 0.00%
Time (median): 112.639 ms ┊ GC (median): 0.00%
Time (mean ± σ): 112.862 ms ± 1.154 ms ┊ GC (mean ± σ): 0.00% ± 0.00%
▄▃█ ▁▄▃
▄▁▁▁▁▁▁▁▁▁▁▁▁▁▁▄███▆███▄▁▄▁▁▄▁▁▄▁▁▁▁▁▄▁▁▄▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▄ ▁
110 ms Histogram: frequency by time 118 ms <
Memory estimate: 0 bytes, allocs estimate: 0.
julia> #benchmark mapreduce(Base.splat(*), +, $work)
BenchmarkTools.Trial: 12 samples with 1 evaluation.
Range (min … max): 403.100 ms … 458.882 ms ┊ GC (min … max): 4.53% … 3.89%
Time (median): 445.058 ms ┊ GC (median): 4.04%
Time (mean ± σ): 440.042 ms ± 16.792 ms ┊ GC (mean ± σ): 4.21% ± 0.92%
▁ ▁ ▁ ▁ ▁ ▁ ▁▁▁ █ ▁
█▁▁▁▁▁▁▁▁▁▁▁█▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁█▁▁▁█▁▁▁▁▁▁█▁█▁▁▁▁███▁▁▁▁▁█▁▁▁█ ▁
403 ms Histogram: frequency by time 459 ms <
Memory estimate: 1.49 GiB, allocs estimate: 39998.
Think of it that way: if you would write the function as a parallel for loop with (+) reduction, iteration also would have an unspecified order, and you'd have memory overhead for the necessary copying of the individual results to the accumulating thread.
Thus, there is a trade-off. In your example, allocation/copying dominates. In other cases, the the mapped operation might dominate, and parallel reduction (with unspecified order, but copying overhead) be worth it.
In Julia, what would be an efficient way of turning the diagonal of a matrix to zero?
Assuming m is your matrix of size N x N this could be done as:
setindex!.(Ref(m), 0.0, 1:N, 1:N)
Another option:
using LinearAlgebra
m[diagind(m)] .= 0.0
And some performance tests:
julia> using LinearAlgebra, BenchmarkTools
julia> m=rand(20,20);
julia> #btime setindex!.(Ref($m), 0.0, 1:20, 1:20);
55.533 ns (1 allocation: 240 bytes)
julia> #btime $m[diagind($m)] .= 0.0;
75.386 ns (2 allocations: 80 bytes)
Performance wise, simple loop is faster (and more explicit, but it is taste dependent)
julia> #btime foreach(i -> $m[i, i] = 0, 1:20)
11.881 ns (0 allocations: 0 bytes)
julia> #btime setindex!.(Ref($m), 0.0, 1:20, 1:20);
50.923 ns (1 allocation: 240 bytes)
And it is faster then diagind version, but not by much
julia> m = rand(1000, 1000);
julia> #btime foreach(i -> $m[i, i] = 0.0, 1:1000)
1.456 μs (0 allocations: 0 bytes)
julia> #btime foreach(i -> #inbounds($m[i, i] = 0.0), 1:1000)
1.338 μs (0 allocations: 0 bytes)
julia> #btime $m[diagind($m)] .= 0.0;
1.495 μs (2 allocations: 80 bytes)
Przemyslaw Szufel's solutions bench-marked for 1_000 x 1_000 matrix size show that diagind performs best:
julia> #btime setindex!.(Ref($m), 0.0, 1:1_000, 1:1_000);
2.222 μs (1 allocation: 7.94 KiB)
julia> #btime $m[diagind($m)] .= 0.0;
1.280 μs (2 allocations: 80 bytes)
Here is a more general way how to performantly use setindex! on an Array by accessing custom indices:
Using an array for in array indexing
Linear indexing is best performing, that's why diagind runs better than the Cartesian Indices.
Indexing large matrixes seems to be taking FAR longer in 0.5 and 0.6 than 0.4.7.
For instance:
x = rand(10,10,100,4,4,1000) #Dummy array
tic()
r = squeeze(mean(x[:,:,1:80,:,:,56:800],(1,2,3,4,5)),(1,2,3,4,5))
toc()
Julia 0.5.0 -> elapsed time: 176.357068283 seconds
Julia 0.4.7 -> elapsed time: 1.19991952 seconds
Edit: as per requested, I've updated the benchmark to use BenchmarkTools.jl and wrap the code in a function:
using BenchmarkTools
function testf(x)
r = squeeze(mean(x[:,:,1:80,:,:,56:800],(1,2,3,4,5)),(1,2,3,4,5));
end
x = rand(10,10,100,4,4,1000) #Dummy array
#benchmark testf(x)
In 0.5.0 I get the following (with huge memory usage):
BenchmarkTools.Trial:
samples: 1
evals/sample: 1
time tolerance: 5.00%
memory tolerance: 1.00%
memory estimate: 23.36 gb
allocs estimate: 1043200022
minimum time: 177.94 s (1.34% GC)
median time: 177.94 s (1.34% GC)
mean time: 177.94 s (1.34% GC)
maximum time: 177.94 s (1.34% GC)
In 0.4.7 I get:
BenchmarkTools.Trial:
samples: 11
evals/sample: 1
time tolerance: 5.00%
memory tolerance: 1.00%
memory estimate: 727.55 mb
allocs estimate: 79
minimum time: 425.82 ms (0.06% GC)
median time: 485.95 ms (11.31% GC)
mean time: 482.67 ms (10.37% GC)
maximum time: 503.27 ms (11.22% GC)
Edit: Updated to use sub in 0.4.7 and view in 0.5.0
using BenchmarkTools
function testf(x)
r = mean(sub(x, :, :, 1:80, :, :, 56:800));
end
x = rand(10,10,100,4,4,1000) #Dummy array
#benchmark testf(x)
In 0.5.0 it ran for >20 mins and gave:
BenchmarkTools.Trial:
samples: 1
evals/sample: 1
time tolerance: 5.00%
memory tolerance: 1.00%
memory estimate: 53.75 gb
allocs estimate: 2271872022
minimum time: 407.64 s (1.32% GC)
median time: 407.64 s (1.32% GC)
mean time: 407.64 s (1.32% GC)
maximum time: 407.64 s (1.32% GC)
In 0.4.7 I get:
BenchmarkTools.Trial:
samples: 5
evals/sample: 1
time tolerance: 5.00%
memory tolerance: 1.00%
memory estimate: 1.28 kb
allocs estimate: 34
minimum time: 1.15 s (0.00% GC)
median time: 1.16 s (0.00% GC)
mean time: 1.16 s (0.00% GC)
maximum time: 1.18 s (0.00% GC)
This seems repeatable on other machines, so an issue has been opened: https://github.com/JuliaLang/julia/issues/19174
EDIT 17 March 2017 This regression is fixed in Julia v0.6.0. The discussion still applies if using older versions of Julia.
Try running this crude script in both Julia v0.4.7 and v0.5.0 (change sub to view):
using BenchmarkTools
function testf()
# set seed
srand(2016)
# test array
x = rand(10,10,100,4,4,1000)
# extract array view
y = sub(x, :, :, 1:80, :, :, 56:800) # julia v0.4
#y = view(x, :, :, 1:80, :, :, 56:800) # julia v0.5
# wrap mean(y) into a function
z() = mean(y)
# benchmark array mean
#time z()
#time z()
end
testf()
My machine:
julia> versioninfo()
Julia Version 0.4.7
Commit ae26b25 (2016-09-18 16:17 UTC)
Platform Info:
System: Darwin (x86_64-apple-darwin13.4.0)
CPU: Intel(R) Core(TM) i7-4870HQ CPU # 2.50GHz
WORD_SIZE: 64
BLAS: libopenblas (USE64BITINT DYNAMIC_ARCH NO_AFFINITY Haswell)
LAPACK: libopenblas64_
LIBM: libopenlibm
LLVM: libLLVM-3.3
My output, Julia v0.4.7:
1.314966 seconds (246.43 k allocations: 11.589 MB)
1.017073 seconds (1 allocation: 16 bytes)
My output, Julia v0.5.0:
417.608056 seconds (2.27 G allocations: 53.749 GB, 0.75% gc time)
410.918933 seconds (2.27 G allocations: 53.747 GB, 0.72% gc time)
It would seem that you may have discovered a performance regression. Consider filing an issue.