# [Numpy-discussion] simple array multiplication question

David Warde-Farley dwf@cs.toronto....
Mon Oct 12 23:40:18 CDT 2009

```On 12-Oct-09, at 11:53 PM, per freem wrote:

> what i want to do is pick the columns l of p, and multiply each one by
> the numbers in s. the first step, picking columns l of p, is simply:
>
> cols = p[arange(3), l]

This isn't picking columns of p, this is picking the times at (0, 0),
(1, 0), and (2, 1). Is this what you meant?

In [36]: p[arange(3), [0,0,1]]
Out[36]: array([ 0.2,  0.5,  0.7])

In [37]: p[:, [0,0,1]]
Out[37]:
array([[ 0.2,  0.2,  0.8],
[ 0.5,  0.5,  0.5],
[ 0.3,  0.3,  0.7]])

In [38]: p[arange(3), [0,0,1]] * s.reshape(3,1)
Out[38]:
array([[  2.,   5.,   7.],
[  4.,  10.,  14.],
[  6.,  15.,  21.]])

In [41]: p[:, [0,0,1]] * s.reshape(3,1)
Out[41]:
array([[  2.,   2.,   8.],
[ 10.,  10.,  10.],
[  9.,   9.,  21.]])

Notice the difference.

> then i want to multiply each one by the numbers in s, and i do it
> like this:
>
> cols * s.reshape(3,1)
>
> this seems to work, but i am concerned that it might be inefficient.
> is there a cleaner way of doing this? is 'arange' operation necessary
> to reference all the 'l' columns of p?

That's about as efficient as it gets, I think.

> also, is the reshape operation
> expensive?

No. It will return a view, rather than make a copy.

You could also do cols * s[:, np.newaxis], equivalently.

David
```