[Numpy-discussion] Numpy unexpected (for me) behaviour

V. Armando Sole sole@esrf...
Fri Jan 23 02:10:41 CST 2009


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 




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