On Thu, May 22, 2008 at 4:59 PM, Kevin Jacobs <[EMAIL PROTECTED]>
<[EMAIL PROTECTED]> wrote:
> After poking around for a bit, I was wondering if there was a faster method
> for the following:
>
> # Array of index values 0..n
> items = numpy.array([0,3,2,1,4,2],dtype=int)
>
> # Count the number of occurrences of each index
> counts = numpy.zeros(5, dtype=int)
> for i in items:
> counts[i] += 1
>
> In my real code, 'items' contain up to a million values and this loop will
> be in a performance critical area of code. If there is no simple solution,
> I can trivially code this using the C-API.
I would use bincount:
count = bincount(items)
should be all you need:
In [192]: items = [0,3,2,1,4,2]
In [193]: bincount(items)
Out[193]: array([1, 1, 2, 1, 1])
In [194]: 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.
Robin
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