actually, the ismaxfun is not doing that I want. I want to have the maximum
value and its index in the given dimension. while the maximum value is
fine, I can't get the index:
julia> function ismaxfun(x::Array{Float64},d::Int)
z = maximum(x,d)
ismax = x .== z
(z,ismax)
end
A = rand(2,2,2)
x = ismaxfun(A,3)
maxval = squeeze(x[1],3)
maxind = find(x[2])
maxind contains the linear indices of the maxima in dimension 3. I want to
map those indices into an array Z of size(maxval)=(2,2), where the value at
Z(i,j) tells me what the maximal index into along A[i,j,:] is. I can get
this
julia> map(x->ind2sub((2,2,2),x),maxind)
4-element Array{(Int64,Int64,Int64),1}:
(2,1,1)
(1,1,2)
(1,2,2)
(2,2,2)
which is close, but really I want that:
ms=mapslices(findmax,A,3)
julia> map(x->x[2],ms)
2x2x1 Array{Int64,3}:
[:, :, 1] =
2 2
1 2
Any suggestion on how to achieve this using ismaxfun above? thanks.
On Wednesday, 30 April 2014 15:58:16 UTC+1, Florian Oswald wrote:
>
> jeez, you are right! sorry i'm doing my first steps here - forget about
> the compile step. it IS much faster than 2x. after changing ismaxfun to
> output something similar to slicefun, it's 10x faster. but maybe my
> reshaping is not the smartest idea either. anyway, that's pretty good!
>
> julia> function ismaxfun(x::Array{Float64,6})
> z = maximum(x,6)
> ismax = x .== z
> id = reshape(find(ismax),size(x)[1:5])
> end
>
> A = rand(10,10,10,10,10,50)
>
> julia> @time k = slicefun(A)
> elapsed time: 0.912256696 seconds (213759264 bytes allocated)
>
> julia> @time k = ismaxfun(A)
> elapsed time: 0.089194917 seconds (3484324 bytes allocated)
>
>
>
>
> On Wednesday, 30 April 2014 15:47:30 UTC+1, Tim Holy wrote:
>>
>> I time it at 50x faster:
>>
>> julia> @time slicefun(A);
>> elapsed time: 1.135524683 seconds (181765472 bytes allocated)
>>
>> julia> @time ismaxfun(A);
>> elapsed time: 0.020300699 seconds (931120 bytes allocated)
>>
>>
>> Did you run it a second time? (You don't want to include the JIT timing.)
>>
>> --Tim
>>
>> On Wednesday, April 30, 2014 07:38:07 AM Florian Oswald wrote:
>> > I see. thanks!
>> > you are (of course) right with the timing:
>> >
>> > julia> function slicefun(x::Array{Float64,6})
>> > z = mapslices(findmax,x,6)
>> > end
>> > julia> function ismaxfun(x::Array{Float64,6})
>> > z = maximum(x,6)
>> > ismax = x .== z
>> > end
>> >
>> > times as follows. more than twice as fast!
>> >
>> > julia> @time k = ismaxfun(A)
>> > elapsed time: 0.53359244 seconds (14418060 bytes allocated)
>> > julia> @time k = slicefun(A)
>> > elapsed time: 1.396534023 seconds (226762376 bytes allocated)
>> >
>> > On Wednesday, 30 April 2014 15:25:25 UTC+1, Tim Holy wrote:
>> > > There is
>> > >
>> > > maximum(A, 3)
>> > >
>> > > but I gather you also want the index? If so, mapslices is your best
>> > > one-liner
>> > > as of now. But if you're performance-sensitive, you might also look
>> into
>> > >
>> > > this:
>> > > M = maximum(A, dims)
>> > > ismax = A .== M
>> > >
>> > > and then find the `true` elements of ismax. I'll bet that's quite a
>> lot
>> > > faster
>> > > than using mapslices, even though it requires two traversals of the
>> array.
>> > > As
>> > > a bonus, you'll also learn about ties.
>> > >
>> > > --Tim
>> > >
>> > > On Wednesday, April 30, 2014 04:04:47 AM Florian Oswald wrote:
>> > > > I looked at the issue open at
>> > > >
>> > > > https://github.com/JuliaLang/julia/issues/3893
>> > > >
>> > > > but couldn't figure out what's the best thing to do. I gather that
>> there
>> > > > will be a
>> > > >
>> > > > findmax(A,dims)
>> > > >
>> > > > in some future version of julia? that's certainly what im looking
>> for
>> > >
>> > > here.
>> > >
>> > > > in the meantime, is this the best I can do?
>> > > >
>> > > > A = randn(3,3,3)
>> > > > mapslices(findmax,A,3)
>> > > >
>> > > > cheers
>>
>
