Hi,

#This does not work import numpy a=numpy.zeros(10) b=numpy.ones(4, numpy.int) a[b] += 1

The problem here is that you are setting a[1] to a[1]+1. I think you want: import numpy a=numpy.zeros(10) b=numpy.ones(4, numpy.bool) a[b] += 1 Best, Matthew

On Fri, Jan 23, 2009 at 01:11, V. Armando Sole <sole@esrf.fr> wrote:

Hello,

In an effort to suppress for loops, I have arrived to the following situation.

Through vectorial logical operations I generate a set of indices for which the contents of an array have to be incremented. My problem can be reduced to the following:

#This works import numpy a=numpy.zeros(10) b=numpy.ones(4, numpy.int)

for i in b: a[i] += 1 #a[1] contains 4 at the end

#This does not work import numpy a=numpy.zeros(10) b=numpy.ones(4, numpy.int) a[b] += 1

#a[1] contains 1 at the end

Is that a bug or a feature?

It is an inevitable consequence of several features interacting together. Basically, Python expands "a[b] += 1" into this: c = a[b] d = c.__iadd__(1) a[b] = d Basically, the array c doesn't know that it was created by indexing a, so it can't do the accumulation you want.

Is there a way I can achieve the first result without a for loop? In my application the difference is a factor 10 in execution time (1000 secons instead of 100 ...)

In [6]: bincount? Type: builtin_function_or_method Base Class: <type 'builtin_function_or_method'> String Form: <built-in function bincount> Namespace: Interactive Docstring: bincount(x,weights=None) Return the number of occurrences of each value in x. x must be a list of non-negative integers. The output, b[i], represents the number of times that i is found in x. If weights is specified, every occurrence of i at a position p contributes weights[p] instead of 1. See also: histogram, digitize, unique. -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco