At 01:44 23/01/2009 -0600, Robert Kern wrote: >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.
Well inevitable would not be the word I would use. I would have expected a different behaviour between a[b] = a[b] + 1 and a[b] += 1 In the second case python (or numpy) does not need to generate an intermediate array and could be doing in-place operations. > > 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. Indeed what I am doing is very close to histogramming. Unfortunaly what I have to add is not just one but a value. I have a set of scattered points (x,y,z) and values corresponding to those points. My goal is to get a regular grid in which in each voxel I sum the values of the points falling in them. I guess I will have to write a tiny C extension. I had expected the += sintax would trigger the in-place operation. Best regards, Armando _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
