On Fri, 2013-03-01 at 08:30 +0100, Nicolas Rougier wrote:
Hi,
I'm trying to increment an array using indexing and a second array for increment values (since it might be a little tedious to explain, see below for a short example).
Using "direct" indexing, the values in the example are incremented by 1 only while I want to achieve the alternative behavior. My question is whether there is such function in numpy or if there a re better way to achieve the same result ? (I would like to avoid the while statement)
I found and adapted the alternative solution from: http://stackoverflow.com/questions/2004364/increment-numpy-array-with-repeat... but it is only for a fixed increment from what I've understood.
Nicolas
# ------------------------
import numpy as np
n,p = 5,100 nodes = np.zeros( n, [('value', 'f4', 1)] ) links = np.zeros( p, [('source', 'i4', 1), ('target', 'i4', 1)]) links['source'] = np.random.randint(0, n, p) links['target'] = np.random.randint(0, n, p)
targets = links['target'] # Indices can be repeated K = np.ones(len(targets)) # Note K could be anything
# Direct indexing nodes['value'] = 0 nodes['value'][targets] += K print nodes
# "Alternative" indexing nodes['value'] = 0 B = np.bincount(targets)
bincount takes a weights argument which should do exactly what you are looking for. - Sebastian
while B.any(): I = np.argwhere(B>=1) nodes['value'][I] += K[I] B = np.maximum(B-1,0) print nodes
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