Python \\ Numpy中的数组行乘法循环

Question

I´m trying to multiply each row of array A by each of all the rows in another array (B) in order to get len(A) number of arrays with same number of rows and columns as the first two arrays.

Any help?

pseudo-code
from numpy  import *
import numpy as np

def multipar():
    A = array( [ (0.1,0.5,0.2,0.2), (0.2,0.5,0.1,0.2), (0.7,0.1,0.1,0.1) ] )
    B = array( [ (1,2,3,4), (2,3,4,5), (3,4,5,6) ] )
    for i in len(A):
        average = A[i]*B
    print average

multipar() 

I would like to have each resulting new array

Array C
(0.1,0.5,0.2,0.2) * (1,2,3,4);
(0.1,0.5,0.2,0.2) * (2,3,4,5);
(...)
Array D
(0.2,0.5,0.1,0.2) * (1,2,3,4);
(...)

Answer 1

You could do something interesting with higher dimensions. Extend either A or B into the third dimension, then multiply that with the one that wasn't extended. eg:

A = array( [ (0.1,0.5,0.2,0.2), (0.2,0.5,0.1,0.2), (0.7,0.1,0.1,0.1) ] )
B = array( [ (1,2,3,4), (2,3,4,5), (3,4,5,6) ] )

tiled = tile (B, (3,1,1)).swapaxes (0,1)
all_results = A*tiled

Now you have all of your result arrays in all_results ; you can easily get them with all_results[0] , all_results[1] , etc

EDIT: In response to the latest question edit: If you really need the result arrays separately, then there are two further options:

• C, D, E = all_results

• replace the last two statements in my first suggestion with:

  C = B * A[0]

  D = B * A[1]

  E = B * A[2]

If you really need separate arrays for the results, and with many more rows so that a loop becomes necessary, then you can do something like (thanks @Jaime for the broadcasting notation)

all_results = A[:, None, :] * B[None, :, :]
for i, res in enumerate (all_results):
    locals () ['result%d'%i] = res

Now the result of multiplying by the first row is in the variable called res1 , the second row in res2 , and so forth.