Hello all, Can someone please explain to me why the following happens. I know that I am accessing the array in the wrong order, it is necessary to reproduce the allocation and slowdown. This came up in an actual application and I could not figure out why: function f1(A) n1=size(A,1) n2=s…

[image] raminammour: for i=1:n1,j=1:n2-1:n2 [image] raminammour: for i=1:n1, j in [1,n2] One is an array, the other isn’t? You have the cost of creating an array in the second. But I don’t even think that’s the issue. [image] raminammour: @time s1=f1(A); @time s2=f2(A); The f…

It should probably be: function f2(A) n1=size(A,1) n2=size(A,2) s=0. for i=1:n1, j in 1:n2-1:n2 # <--- s+=A[i,j] end return s end in and = are identical in their for-loop usage.

The functions were compiled before, I didn’t post it. I think you are right about the array issue if you change the function to: function f1(A) n1=size(A,1) n2=size(A,2) s=0. for i=1:n1,j=1:n2-1:n2 s+=A[i,j] end return s end function f2(A) n1=size(A,1) n2=s…

Another thing to note, arrays are column-major in Julia, so you’ll get better performance doing: for j = 1:(n2-1):n2, i = 1:n1 ; s += A[i, j] ; end

Hello all, Can someone please explain to me why the following happens. I know that I am accessing the array in the wrong order, it is necessary to reproduce the allocation and slowdown. This came up in an actual application and I could not figure out why: Cannot answer your question as such (…

The j ranges are the same. The issue was that the vector [1,n2] was being created and garbage collected inside the loop at every iteration, which is slow. With the step range 1:n2-1:n2 that does not happen.

I am not sure if you have intended to use the code for matrices of dimensions min(m,n) > 1, but in f1, there is a bug: when n2 == 1, you get an error. Plus, why would you iterate if you know ahead of time that you will use the first and last columns of the matrix? Below code excerpt will save you o…

Thanks for the reply. This code is just meant to reproduce the behavior I was seeing in another application, n>>>1 and I am not all that interested in summing matrices :slight_smile: . This typically comes up when you iterate on an array, and then you want to do something different at the ends (th…

For boundary conditions at the ends, the most elegant, flexible, and efficient technique is often to use ghost elements. See my comment about this in another thread on discourse.

Difference between indexing a for loop

New to Julia

mkborregaard June 19, 2017, 4:12pm 11

link Arrays with periodic boundaries - #4 by stevengj

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Difference between indexing a for loop

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