On Sun, 8 Oct 2006, Matt Knox apparently wrote:
> I have a custom Date class which allows creation of
> different frequency Dates (annual, monthly, etc..). Two
> dates of the same frequency can be subtracted, and
> integers can be added to a Date, with the obvious results
> in each case. I st
On 10/8/06, Daniel Mahler <[EMAIL PROTECTED]> wrote:
>>> aarray([0, 0])>>> barray([0, 1, 0, 1, 0])>>> carray([1, 1, 1, 1, 1])Well for this particular example you could doa=array([len(b)-sum(b), sum(b)])
Since you are just counting the ones and zeros.
This next one is a little closer for the case wh
Bill Baxter wrote:
> Yes, that'd be
>a[b] += c
No, I'm afraid that fancy indexing does not do the loop that you are thinking
it
would (and for reasons that we've discussed previously on this list, *can't* do
that loop). That statement reduces to something like the following:
tmp = a[b]
So what's the answer then? Can it be made faster?--bbOn 10/9/06, Robert Kern <[EMAIL PROTECTED]> wrote:
Bill Baxter wrote:> Yes, that'd be>a[b] += cNo, I'm afraid that fancy indexing does not do the loop that you are thinking it
would (and for reasons that we've discussed previously on this li
On 08/10/06, Robert Kern <[EMAIL PROTECTED]> wrote:
> Bill Baxter wrote:
> > Yes, that'd be
> >a[b] += c
>
> No, I'm afraid that fancy indexing does not do the loop that you are thinking
> it
> would (and for reasons that we've discussed previously on this list, *can't*
> do
> that loop). Tha
Hi Nadav,On 10/8/06, Nadav Horesh <[EMAIL PROTECTED]> wrote:
There is a "tensortdot" function in numpy1.0rc1
The tensordot is not the same thing as a tensor product. What I want is the following:
def tensor(a, b) :
"""Tensor product of a and b
"""
a = asarray(a)
b = asarray(b)
On 10/8/06, Greg Willden <[EMAIL PROTECTED]> wrote:
> On 10/8/06, Daniel Mahler <[EMAIL PROTECTED]> wrote:
> >
> > >>> a
> > array([0, 0])
> > >>> b
> > array([0, 1, 0, 1, 0])
> > >>> c
> > array([1, 1, 1, 1, 1])
> >
>
>
> Well for this particular example you could do
> a=array([len(b)-sum(b), sum(
Daniel Mahler wrote:
> On 10/8/06, Greg Willden <[EMAIL PROTECTED]> wrote:
>> This next one is a little closer for the case when c is not just a bunch of
>> 1's but you still have to know how the highest number in b.
>> a=array([sum(c[b==0]), sum(c[b==1]), ... sum(c[b==N]) ] )
>>
>> So it sort of
