So with bincount I can exchange a loop over P for a loop over M? I
guess for me that is still really helpful.
Thanks!
- George
On Mon, Dec 21, 2009 at 6:35 AM, Pauli Virtanen wrote:
> Mon, 21 Dec 2009 09:35:08 +, Neil wrote:
> [clip]
>> I'm also interested to see if there are any answers to
On Mon, Dec 21, 2009 at 6:35 AM, Pauli Virtanen wrote:
> Mon, 21 Dec 2009 09:35:08 +, Neil wrote:
> [clip]
>> I'm also interested to see if there are any answers to this; I came
>> across a similar problem recently. It would have been convenient to do
>> something like U[idx] += dU, but this d
Mon, 21 Dec 2009 09:35:08 +, Neil wrote:
[clip]
> I'm also interested to see if there are any answers to this; I came
> across a similar problem recently. It would have been convenient to do
> something like U[idx] += dU, but this didn't work because there were
> repeated indices in idx. Here's
Hi everyone,
I was wondering if anyone had insight on the best way to solve the
following problem.
Suppose I have a numpy array called U.
U has shape (N,M)
Suppose further that I have another array called dU and that
dU has shape (P,M) and that P has no particular relationship to N, it
could be la
