Hi guys, I would like to know if there is any way to make the following operation faster. def test(): shape=(200,200,200,3) refinds = np.ndindex(shape[:3]) reftmp=np.zeros(shape) for ijk_t in refinds: i,j,k = ijk_t reftmp[i,j,k,0]=i reftmp[i,j,k,1]=j reftmp[i,j,k,2]=k %timeit test() 1 loops, best of 3: 19.5 s per loop I am using ndindex and then a for loop. Is there a better/faster way? Thank you, Eleftherios
On Sat, 04 Dec 2010 19:00:43 +0000, Eleftherios Garyfallidis wrote: [clip]
I am using ndindex and then a for loop. Is there a better/faster way?
Yes: import numpy as np from numpy import newaxis x = np.zeros((200, 200, 200, 3)) x[...,0] = np.arange(200)[:,newaxis,newaxis] x[...,1] = np.arange(200)[newaxis,:,newaxis] x[...,2] = np.arange(200)[newaxis,newaxis,:] x[1,3,2] # -> array([ 1., 3., 2.]) Depending on what you use this array for, it's possible that you can avoid constructing it (and use broadcasting etc. instead). -- Pauli Virtanen
This is beautiful! Thank you Pauli. On Sat, Dec 4, 2010 at 7:16 PM, Pauli Virtanen <pav@iki.fi> wrote:
On Sat, 04 Dec 2010 19:00:43 +0000, Eleftherios Garyfallidis wrote: [clip]
I am using ndindex and then a for loop. Is there a better/faster way?
Yes:
import numpy as np from numpy import newaxis
x = np.zeros((200, 200, 200, 3)) x[...,0] = np.arange(200)[:,newaxis,newaxis] x[...,1] = np.arange(200)[newaxis,:,newaxis] x[...,2] = np.arange(200)[newaxis,newaxis,:] x[1,3,2] # -> array([ 1., 3., 2.])
Depending on what you use this array for, it's possible that you can avoid constructing it (and use broadcasting etc. instead).
-- Pauli Virtanen
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