Or
np.multiply.outer(a,b)
Nadav
-הודעה מקורית-
מאת: numpy-discussion-boun...@scipy.org בשם Robert Kern
נשלח: ה 17-ספטמבר-09 00:40
אל: Discussion of Numerical Python
נושא: Re: [Numpy-discussion] array multiplication
2009/9/16 Ernest Adrogu? :
> Hi,
>
> I have two 1-d arrays
2009/9/16 Ernest Adrogué :
> Hi,
>
> I have two 1-d arrays (a and b), and I want to create a
> third 2-d array, whose rows are of the form a[i]*b:
>
> c = np.zeros((len(a),b))
>
> c[0] = a[0]*b
> c[1] = a[1]*b
> .
> .
> .
>
> Is there an easy way to do this (e.g, without a loop)?
c = a[:,np.newaxi
Hi,
I have two 1-d arrays (a and b), and I want to create a
third 2-d array, whose rows are of the form a[i]*b:
c = np.zeros((len(a),b))
c[0] = a[0]*b
c[1] = a[1]*b
.
.
.
Is there an easy way to do this (e.g, without a loop)?
Thanks!
--
Ernest
___
N
Jan Strube wrote:
> I'm having a difficult time understanding the following behavior:
>
> import numpy as N
> # create a new array 4 rows, 3 columns
> x = N.random.random((4, 3))
> # elementwise multiplication
> x*x
>
> newtype = N.dtype([('x', N.float64), ('y', N.float64), ('z', N.float64)])
>
> #
I'm having a difficult time understanding the following behavior:
import numpy as N
# create a new array 4 rows, 3 columns
x = N.random.random((4, 3))
# elementwise multiplication
x*x
newtype = N.dtype([('x', N.float64), ('y', N.float64), ('z', N.float64)])
# interpret the array as an array of
