On Wed, Dec 24, 2014 at 10:29 AM, wrote:
>
>
> On Wed, Dec 24, 2014 at 10:30 AM, Julian Taylor <
> jtaylor.deb...@googlemail.com> wrote:
>
>> On 24.12.2014 16:25, Neal Becker wrote:
>> > What would be the most efficient way to compute:
>> >
>> > c[j] = \sum_i (a[i] * b[i,j])
>> >
>> > where a[i]
On Wed, Dec 24, 2014 at 10:30 AM, Julian Taylor <
jtaylor.deb...@googlemail.com> wrote:
> On 24.12.2014 16:25, Neal Becker wrote:
> > What would be the most efficient way to compute:
> >
> > c[j] = \sum_i (a[i] * b[i,j])
> >
> > where a[i] is a 1-d vector, b[i,j] is a 2-d array?
> >
> > This seems
Nathaniel Smith wrote:
> On Wed, Dec 24, 2014 at 3:25 PM, Neal Becker wrote:
>> What would be the most efficient way to compute:
>>
>> c[j] = \sum_i (a[i] * b[i,j])
>>
>> where a[i] is a 1-d vector, b[i,j] is a 2-d array?
>
> I think this formula is just np.dot(a, b). Did you mean c = \sum_j
> \
On Wed, Dec 24, 2014 at 3:25 PM, Neal Becker wrote:
> What would be the most efficient way to compute:
>
> c[j] = \sum_i (a[i] * b[i,j])
>
> where a[i] is a 1-d vector, b[i,j] is a 2-d array?
I think this formula is just np.dot(a, b). Did you mean c = \sum_j
\sum_i (a[i] * b[i, j])?
> This seems
On 24.12.2014 16:25, Neal Becker wrote:
> What would be the most efficient way to compute:
>
> c[j] = \sum_i (a[i] * b[i,j])
>
> where a[i] is a 1-d vector, b[i,j] is a 2-d array?
>
> This seems to be one way:
>
> import numpy as np
> a = np.arange (3)
> b = np.arange (12).reshape (3,4)
> c = n
What would be the most efficient way to compute:
c[j] = \sum_i (a[i] * b[i,j])
where a[i] is a 1-d vector, b[i,j] is a 2-d array?
This seems to be one way:
import numpy as np
a = np.arange (3)
b = np.arange (12).reshape (3,4)
c = np.dot (a, b).sum()
but np.dot returns a vector, which then need
