It's normal that manually inlined code of this kind is faster than wrapped
code unless the compiler manages to see the full inlining potential. In
this case the huge memory allocations for the wrapped solutions indicates
that it's nowhere near doing that at all. I doubt it will take you all the
Reading through the threads and issues on gc rooting for embedded code, as
well as the code comments above the JL_GC_PUSH* macros in julia.h, I'm
still uncertain about how defensive it's necessary to be and best
practices. I'll structure this into a couple of cases with questions.
1. One of the
Thanks. That makes things clearer.
Den fredag 14 oktober 2016 kl. 14:16:54 UTC+2 skrev Yichao Yu:
>
>
>
> On Fri, Oct 14, 2016 at 7:03 AM, Bart Janssens > wrote:
>
>> Hi,
>>
>> Replies below, to the best of my understanding of the Julia C interface:
>>
Is this what you're looking for?
julia> using SIMD
julia> v = rand(2, 5)
2×5 Array{Float64,2}:
0.391832 0.785861 0.352291 0.874891 0.874593
0.697768 0.51637 0.711525 0.400077 0.266559
julia> w = reinterpret(Vec{2, Float64}, vec(v))
5-element Array{SIMD.Vec{2,Float64},1}:
Float64⟨0.39
A double for loop requires a comma between the variables
julia> for i=1:10, j=1:10
println(i," , ",j)
end
To get you're desired result a single loop is sufficient.
julia> for i=1:10
println(i," , ",i)
end
1 , 1
2 , 2
3 , 3
4 , 4
5 , 5
6 , 6
7 , 7
8 , 8
9 , 9
10
Syntactically you're still missing a comma between the loop variables.
Semantically that's not how the double for loop works.
Try either
julia> for i = 1:length(x)
println(x[i], y[i])
end
or
julia> for (i, j) in zip(x, y)
println(i, j)
end
Den tisdag 27 ja
If speed is what's important you almost certainly want some kind of loop.
If it's more important not to have a loop you can satisfy your
specification with
b = reshape(repmat(mean(reshape(a,10,10),1),10),100)
although that's not what I would call a moving average.
Den tisdag 10 februari 2015 k
Den fredag 20 mars 2015 kl. 05:36:15 UTC+1 skrev Kiran Pamnany:
>
> Hey Stefan,
>
>
>> Depending on how well Julia-generated code performs, some important
>>> concurrent data structures might be better implemented in C and wrapped.
>>
>>
>> I would argue that until we can implement efficient con
It doesn't behave the same with respect to duplicates in A but I would do
N = 100
A = rand(N)
n = iceil(0.7 * N)
testindex = sample(1:size(A,1), replace=false, n)
testA = A[testindex]
trainindex = setdiff(1:N, testindex)
trainA = A[trainindex]
Den torsdag 26 mars 2015 kl. 06:21:42 UTC+1 skrev ver
This is reasonably clean:
[y[i] for y in x, i in 1:3]
If performance matters you are probably better off with the other
suggestions.
Den måndag 6 juli 2015 kl. 22:10:01 UTC+2 skrev Júlio Hoffimann:
>
> Hi,
>
> How to convert:
>
> 1000-element Array{Tuple{Integer,Integer,Integer},1}:
> (10,2,1)
julia> [foo[r] for r in some_ranges]
2-element Array{Any,1}:
[6,2]
[4,9]
julia> vcat([foo[r] for r in some_ranges]...)
4-element Array{Int64,1}:
6
2
4
9
Den onsdag 19 augusti 2015 kl. 04:53:55 UTC+2 skrev John Brock:
>
> Is there an easy way to slice an array using an array of ranges? I'm
Here's an example of a language that does a bit more with string operations.
% pike
Pike v7.8 release 700 running Hilfe v3.5 (Incremental Pike Frontend)
> "ab" + "c";
(1) Result: "abc"
> "abc" * 2;
(2) Result: "abcabc"
> "abcabc" - "ca";
(3) Result: "abbc"
> "abcabc" / "b";
(4) Result: ({ /* 3 ele
In medical imaging, and probably most other parts of image processing with
any kind of real-time expectations, single precision is used because the
factor of 2 for SIMD operations is an enormously big deal, regardless
whether the computations are done on a GPU or a CPU. There are pitfalls
relat
BoundsError means you're trying to index out of bounds. I see nothing in
the backtrace indicating the splice function being involved and I have no
way of knowing which line in your function is numbered 204 but there's a
good chance Varraycut[i-1] will try to index at 0 in the first turn of your
This is not the most efficient way, neither the clearest, but it's compact.
In a language like Matlab it would be the preferred approach.
sum((outputarray[1:end-1] .< 0) & (outputarray[2:end] .>= 0))
Den onsdagen den 30:e juli 2014 kl. 22:03:32 UTC+2 skrev yaois...@gmail.com:
>
> Hi guys,
>
> I
Is this easy enough?
julia> x = reshape(1:16, 4, 4)
4x4 Array{Int64,2}:
1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16
julia> a = x[1,:]
1x4 Array{Int64,2}:
1 5 9 13
julia> [a == x[k,:] for k = 1:size(x,1)]
4-element Array{Any,1}:
true
false
false
false
julia> any([a == x[k,:]
gt; > To be honest, the custom for loop is probably your best bet in terms
> > of performance:
> >
> > |function isrow(row, x::Matrix)
> > for k=1:size(x,1)
> >if row == x[k,:]
> > return true
> >end
> > end
Den fredagen den 8:e augusti 2014 kl. 22:25:01 UTC+2 skrev Steven G.
Johnson:
>
>
>
> On Friday, August 8, 2014 3:19:00 AM UTC-4, Gunnar Farnebäck wrote:
>>
>> 1. As in Ethan's response (modulo a trivial bug):
>>
>> Matrix[x[k,:] for k=1:size(x,1)]
>
For maximum efficiency Diego's loop solution is likely best but for
succinctness, try hcat(A...).
Den lördagen den 23:e augusti 2014 kl. 04:05:19 UTC+2 skrev Bradley Setzler:
>
> Good evening,
>
> I often have Any types filled with vectors (especially as the return of a
> pmap), and need them as
You get NaN because the binning has left some bin empty and x * log(2, x)
for x = 0.0 results in 0.0 * -Inf which is NaN. You can avoid this
particular problem by replacing x * log(2,x) with something that special
cases x = 0.0 to return 0.0 rather than NaN.
Note that the result you get from th
This function helped me find a couple of type problems in a somewhat
complex code base. The idea is to first run your code, then use this
function to dig up the type signatures that were actually called and
identify type problematic variables from code_typed.
I strongly suspect that the inspect
In the section "The Loop Body Should Be Straight-Line Code", the first and
second code example look identical with ifelse constructions. I assume the
first one should use ? instead. Also the third code example has a stray
x[i]
> I've posted an article on the @simd feature to
> https://software.
as a nested ifelse in the same way as example 2 (although in the opposite
order, so it also differs for a>b).
Den onsdagen den 17:e september 2014 kl. 16:28:45 UTC+2 skrev Arch Robison:
>
> Thanks. Now fixed.
>
> On Wed, Sep 17, 2014 at 4:14 AM, Gunnar Farnebäck > wrote:
>
It's not like Julia is doing anything strange or uncommon here. Most people
would be really surprised if -10² meant positive 100.
Den torsdagen den 18:e september 2014 kl. 15:01:44 UTC+2 skrev Jutho:
>
> because it is not recognized/parsed as literal but as the application of a
> unary minus, wh
It might help if you explain something about what you want to do with the
binary representation.
For example,
julia> f(x::Float64) = reinterpret(Float64, reinterpret(UInt64, x) &
0x7fff)
f (generic function with 1 method)
generates very simple LLVM code
julia> code_llvm(f, (Float6
Den måndag 19 oktober 2015 kl. 10:28:15 UTC+2 skrev Uliano Guerrini:
>
>
>
> Il giorno lunedì 19 ottobre 2015 08:56:32 UTC+2, Gunnar Farnebäck ha
> scritto:
>>
>> It might help if you explain something about what you want to do with the
>> binary representatio
Array{T, N} is a non-concrete type and thus an inference failure when a
concrete type could have been inferred. Most of the concatenation inference
failures existed in 0.3 but there are also some regressions, most
annoyingly for some constructions referred to by deprecation messages.
Den måndag
If you don't have deprecation warnings I would suspect some change in 0.4
has introduced type instabilities. If you are using typed concatenations
you could be hit by https://github.com/JuliaLang/julia/issues/13254.
Den torsdag 22 oktober 2015 kl. 23:03:00 UTC+2 skrev Kris De Meyer:
>
> Are ther
julia> eltype(Ptr{Float64})
Float64
Den tisdag 3 november 2015 kl. 07:51:43 UTC+1 skrev Petr Hlavenka:
>
> Hi,
> is there a way to obtain a type T from a Ptr{T}, e.g. get UInt32 from
> Ptr{UInt32}.
>
> I'm trying to fix NIDAQ for 0.4 (tupplecalypse)
>
> This is needed if I have a signature of a f
Den torsdag 12 november 2015 kl. 06:36:28 UTC+1 skrev Alireza Nejati
>
> Anyway, the problem I'm facing right now is that even though TensorFlow's
> python interface works fine, I can't get TensorFlow's C library to build!
> Has anyone else had any luck with this? I've had to update java AND gcc
In case the accidental matrix happens to be huge you can also get into all
sorts of memory shortage related inconveniences.
Den tisdag 24 maj 2016 kl. 07:27:24 UTC+2 skrev Tony Kelman:
>
> I can see some situations where getting a matrix instead of a vector
> because one of the inputs was unexpe
fill_W1! allocates memory because it makes copies when constructing the
right hand sides. fill_W2 allocates memory in order to construct the
comprehensions (that you then discard). In both cases memory allocation
could plausibly be avoided by a sufficiently smart compiler, but until
Julia becom
I realize the question is about the performance of the zip approach but may
I suggest just running mean(v)?
On Wednesday, August 10, 2016 at 11:34:08 AM UTC+2, Tamas Papp wrote:
>
> I was benchmarking code for calculating means of vectors, eg
>
> v = [rand(100) for i in 1:100]
>
> m1(v) = map(m
I wrote this when I wanted to cache the results of matread. Your problem
sounds like it could be similar.
let matread_cache = (String => Any)[]
global caching_matread
function caching_matread(filename)
if !haskey(matread_cache, filename)
matread_cache[filename] = matre
I leave it to someone with more insights to comment on the efficiency of
nested functions, but this seems at least to be substantially faster:
type State
a::Uint64
b::Uint64
aa::Uint64
bb::Uint64
cc::Uint64
RANDSIZL::Int
RANDSIZ::Int
end
function ind(s::State, mm, x)
It might be possible to get further doing llvmcall with LLVM inline
assembly but my feeble attempts at that have only resulted in crashes
(well, sort of, getting dumped back to shell prompt without ceremony).
Getting good load/store instructions by default would be a lot better of
course but I
Most likely Julia tries to parse it as x > 10 ? return ((0):nothing)
Try x > 10 ? (return 0) : nothing
instead.
Or the more popular construction x <= 10 || return 0
Den tisdagen den 25:e november 2014 kl. 11:21:39 UTC+1 skrev andreas:
>
>
> Dear list,
>
> I observed that this
>
> function foo(x
It does have issues. I tracked down a part of the story to
https://github.com/JuliaLang/julia/pull/9086
Den onsdagen den 26:e november 2014 kl. 20:52:57 UTC+1 skrev Stefan
Karpinski:
>
> I wonder if \ might be type unstable.
>
Most likely your performance is destroyed by the conditional definition of
the Ti, c1, ... coefficients. It's a rather unfortunate limitation of the
type inference, causing all those variables to be handled as Any, and
everything that depends on them as well, i.e. more or less everything in
you
If you're not concerned about performing unneeded calls to u2_0 you can get
away with (x.>Lbox/2) .* (0.5*(u2_0(x)+u2_0(Lbox-x))-u2_
0(Lbox/2))
Den onsdag 10 februari 2016 kl. 11:58:58 UTC+1 skrev Ferran Mazzanti:
>
> Hi folks,
>
> probably a stupid question but can't find the answer, so please
The environment variable JULIA_PKGDIR may be of use.
Den måndag 7 mars 2016 kl. 05:44:29 UTC+1 skrev Vishnu Raj:
>
> In my system I have Juno with 0.4.2 and julia with 0.4.3. Every time I
> start one, all my packages are getting recompiled. They are all under same
> v0.4 folder and I think that'
This is at least shorter.
julia> (v.^2...)
(1,4,81)
Den torsdag 31 mars 2016 kl. 10:28:09 UTC+2 skrev Tamas Papp:
>
> Hi,
>
> Is there a preferred syntax for mapping a vector to a tuple? Eg an
> alternative for (contrived example)
>
> v = [1,2,9]
> ntuple(i -> v[i]^2, length(v))
> tuple([e^2
Your problem is probably that you try to call your function with four
positional arguments, although the last two should have been named. Compare
julia> f(x, y; z=0) = x + y + z
f (generic function with 1 method)
julia> f(2, 3)
5
julia> f(2, 3, 4)
ERROR: MethodError: `f` has no method matching
It's the same kind of technique I used to generate convolution loops for
different number of dimensions in
https://github.com/GunnarFarneback/julia/commit/51d8c86dd6bdb4c02d03e312dede7e5013da5ae1
See the start and end of the imfilterN_loops function.
I'm obviously biased but I think the construc
At some point when I had trouble loading zlib I wrote a pure Julia
implementation of zlib and gzip decompression.
Pros:
* Written in Julia.
* No external dependencies.
Cons:
* Only supports decompression and only from a buffer (i.e. no streaming).
* Substantially slower than zlib (about five tim
magnitude
faster.
Den lördagen den 4:e januari 2014 kl. 19:03:33 UTC+1 skrev Stefan Karpinski:
>
> Possibly relevant is Julia Evan's gzip implementation:
>
> http://jvns.ca/blog/2013/10/24/day-16-gzip-plus-poetry-equals-awesome/
>
>
> On Sat, Jan 4, 2014 at 6:54 AM, Gunnar F
You can index both dimensions at once.
matrix[subset, subset]
On Wednesday, January 22, 2014 10:20:21 AM UTC+1, Robert Feldt wrote:
>
> I need to repeatedly take subsets of large symmetric matrices
> (symmetrically with the same subset indicies for rows and columns) and
> first tried like so:
>
This is a long-standing issue.
See https://github.com/JuliaLang/julia/issues/265
Den fredagen den 7:e mars 2014 kl. 17:17:49 UTC+1 skrev David Moon:
>
> In Julia-0.3.0-prerelease-d2d9bd93ed.app on Mac OS X 10.6:
>
> julia> p() = q()
> p (generic function with 1 method)
> julia> p2() = q()
> p2 (g
[f(i,j)[k] for k in 1:K, i in 1:n, j in i:m]
Den måndagen den 24:e mars 2014 kl. 15:07:49 UTC+1 skrev Linus Mellberg:
>
> Hi!
>
> I'm trying to construct a 3 dimensional array from a number of 1
> dimensional arrays. Essentially what i would like to do is
>
> a = [f(i, j) for i in 1:n, j in 1:m]
Ok, one might be less redundant if reading the thread properly before
posting. Sorry about the noise.
Den måndagen den 24:e mars 2014 kl. 23:12:03 UTC+1 skrev Gunnar Farnebäck:
>
> [f(i,j)[k] for k in 1:K, i in 1:n, j in i:m]
>
> Den måndagen den 24:e mars 2014 kl. 15:07:49 UTC+1
Ugly workaround:
julia> [Array(Any,3,0) [1:3] float([1:3])]
3x2 Array{Any,2}:
1 1.0
2 2.0
3 3.0
Den tisdagen den 1:e april 2014 kl. 10:39:15 UTC+2 skrev harven:
>
> When I build a matrix from columns using vcat, the type of the columns is
> sometimes modified, e.g.
>
> julia> [[1:3] float(
You can also use parentheses to make the first line incomplete.
if (a>b
&& b>c)
Den torsdagen den 17:e april 2014 kl. 00:30:41 UTC+2 skrev K leo:
>
> Thanks for the answer.
>
> I tested previously this but it did not work:
> if a>b && b>c && c>d
> && d>e && e>f
>
> But now this works
I have, in the process of evaluating Julia, ported some proprietary matlab
image processing code, at the order of 3k lines matlab and 1k lines C mex
code, into pure Julia. The matlab code is fairly well optimized with most
computations vectorized (usually comes natural in image processing) and
You can do multiple assignment with anything you can iterate over on the
right hand side, in this case you could do e.g. aa, bb = 12:3:15 with the
same assignment to aa and bb. One difference is in the value of the
expression, as printed by the REPL, which coincides with the RHS but would
usual
Sum uses pairwise summation, see
https://github.com/JuliaLang/julia/pull/4039
To find the definition:
julia> @which sum(int)
sum{T}(a::AbstractArray{T,N}) at reduce.jl:276
Den torsdagen den 24:e april 2014 kl. 11:46:37 UTC+2 skrev Alex:
>
> I couldn't find the definition of sum right now, but I
The padarray function in the Images package can solve your problem or give
you inspiration if you want to do it differently.
Den fredagen den 25:e april 2014 kl. 05:43:40 UTC+2 skrev Tomas Lycken:
>
> Given the following function definition
>
> function foo{T<:FloatingPoint}(A::Array{T})
> B
If nothing else works you can solve it with a manual repmat.
A different approach is to forgo the logical indexing entirely.
species_codes = ["setosa" => [1 0 2],
"versicolor" => [2 1 0],
"virginica" => [0 2 1]]
N = length(iris_data.columns[5])
iris_output = zer
Other options, both faster:
int(x.*x + y.*y .< radius*radius)
[(x[k]*x[k] + y[k]*y[k] < r*r)?1:0 for k=1:length(x)]
Den söndagen den 8:e juni 2014 kl. 02:02:56 UTC+2 skrev Zahirul ALAM:
>
> thanks guys for all of these :)
>
> On Saturday, 7 June 2014 14:25:08 UTC-4, David Einstein wrote:
>>
>> It
You can do B[E[:],:] or better write E as a vector (rather than a 1x10
array) directly:
E = [5, 5, 5, 4, 5, 5, 5, 5, 1, 5]
Then B[E,:] works as expected.
Den tisdagen den 10:e juni 2014 kl. 15:40:50 UTC+2 skrev Bruno Rodrigues:
>
> Hi, I posted a first thread but can't find it anywhere, so I prob
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