On Thu, May 22, 2008 at 9:22 AM, Robin <[EMAIL PROTECTED]> wrote:
> 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
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 o
On Thu, May 22, 2008 at 9:15 AM, Keith Goodman <[EMAIL PROTECTED]> wrote:
> On Thu, May 22, 2008 at 9:08 AM, Keith Goodman <[EMAIL PROTECTED]> wrote:
>> On Thu, May 22, 2008 at 8:59 AM, Kevin Jacobs <[EMAIL PROTECTED]>
>> <[EMAIL PROTECTED]> wrote:
>>> After poking around for a bit, I was wondering
You're just trying to do this...correct?
>>> import numpy
>>> items = numpy.array([0,3,2,1,4,2],dtype=int)
>>> unique = numpy.unique(items)
>>> unique
array([0, 1, 2, 3, 4])
>>> counts=numpy.histogram(items,unique)
>>> counts
(array([1, 1, 2, 1, 1]), array([0, 1, 2, 3, 4]))
>>> counts[0]
array([1,
On Thu, May 22, 2008 at 9:08 AM, Keith Goodman <[EMAIL PROTECTED]> wrote:
> On Thu, May 22, 2008 at 8:59 AM, 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 valu
On Thu, May 22, 2008 at 12:08 PM, Keith Goodman <[EMAIL PROTECTED]> wrote:
> How big is n? If it is much smaller than a million then loop over that
> instead.
>
n is always relatively small, but I'd rather not do:
for i in range(n):
counts[i] = (items==i).sum()
If that was the best alternativ
On Thu, May 22, 2008 at 8:59 AM, 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 o
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
