You could also try something like julia> a = rand(100,100000);
julia> @time [norm(sub(a,:,i)) for i in 1:size(a,2)]; 0.135117 seconds (1.80 M allocations: 40.381 MB, 14.14% gc time) ---- As long as your slices are along the last dimension, i.e. columns in a matrix, you can use FunctionalData, which is optimized exactly for this: https://github.com/rened/FunctionalData.jl#efficiency julia> @time map(a, norm); 0.014685 seconds (200.02 k allocations: 3.815 MB) Am 12.08.2015 um 16:26 schrieb jw3126 <jw3...@gmail.com>: > I have a matrix and I want to compute the norm of each column. I tried > > using TimeIt > N = 10^6 > mat = randn(3, N) > > @timeit mapslices(norm, mat, 1) > > 1 loops, best of 3: > 915.13 ms per loop > > > Dog slow, 100times slower then numpy: > > import numpy as np > N = 10**6 > mat = np.random.randn(3, N) > %timeit np.linalg.norm(mat, axis=0) > > 100 loops, best of 3: > 9.71 ms per loop > > If I understand it correctly, the reason is that mapslices can not really > optimize on the norm argument, e.g. inline norm. I had two ideas how to solve > this problem, both of which are problematic: > > 1) There is the NumericExtensions package, which seems to be aimed at this > kind of problem, however this package is deprecated, so I don't want to use > it. > > 2) Write some code by hand. This gives speedup, but is still slower then > numpy: > > N = 10^6 > mat = randn(3, N) > function f(mat) > return sqrt(mat[1,:].^2 + mat[2,:].^2 + mat[3,:].^2) > end > > @timeit f(mat) > > 10 loops, best of 3: > 38.51 ms per loop > > > So how to make this computation fast? And more generally what are ways speed > up mapslices without using NumericExtensions?
