[Numpy-discussion] Taking a large number of dot products at once

josef.pktd@gmai... josef.pktd@gmai...
Fri Mar 4 05:18:27 CST 2011


On Fri, Mar 4, 2011 at 4:13 AM, Hector <hector1618@gmail.com> wrote:
>
>
> On Fri, Mar 4, 2011 at 12:59 PM, eat <e.antero.tammi@gmail.com> wrote:
>>
>> Hi,
>>
>> On Fri, Mar 4, 2011 at 8:19 AM, Daniel Hyams <dhyams@gmail.com> wrote:
>>>
>>> This is probably so easy, I'm embarrassed to ask it...but I've been
>>> casting around trying things to no avail for the last hour and a half, so
>>> here goes...
>>> I have a lot of dot products to take.  The length-3 vectors that I want
>>> to dot are stacked in a 2D array like this:
>>> U = [u1 u2 u3....]
>>> and
>>>
>>> V = [v1 v2 v3....]
>>> So both of these arrays, are, say, 3x100 each.  I just want to take the
>>> dot product of each of the corresponding vectors, so that the result is
>>> [u1.v1 u2.v2  u3.v3 ....]
>>> which would be a 1x100 array in this case.
>>> Which function do I need to use?  I thought tensordot() was the one, but
>>> I couldn't make it work....pure user error I'm sure.
>
> Hello Daniel Hyams and group,
>
> I guess the following code serves the purpose but I think something needs to
> be highlighted. I think Eat forgot to paste it here.
>
>
>> No function needed for this case, just:
>
>     In [ ] : import numpy.random as rand
>>
>> In []: x= rand(3, 7)
>> In []: y= rand(3, 7)
>> In []: d= (x* y).sum(0)
>> In [490]: d
>> Out[490]:
>> array([ 1.25404683,  0.19113117,  1.37267133,  0.74219888,  1.55296562,
>>         0.15264303,  0.72039922])
>> In [493]: dot(x[:, 0].T, y[:, 0])
>> Out[493]: 1.2540468282421895
>> Regards,
>> eat
>>>
>>>
>
> I guess you can use 'rand' only this after  import.

to clarify:
random is the module, rand is a function in random

>>> import numpy.random as rand
>>> rand(3,7)
Traceback (most recent call last):
  File "<pyshell#50>", line 1, in <module>
    rand(3,7)
TypeError: 'module' object is not callable


>>> from numpy.random import rand
>>> rand(3,7)
array([[  3.424 ...........

Josef



>
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>
>
> --
> -Regards
> Hector
> Whenever you think you can or you can't, in either way you are right.
>
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