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I am looking for the Python numpy equivalent of the IDL # operator. Here is what the # operator does:

Computes array elements by multiplying the columns of the first array by the rows of the second array. The second array must have the same number of columns as the first array has rows. The resulting array has the same number of columns as the first array and the same number of rows as the second array.

Here are the numpy arrays I am dealing with:

A = [[ 0.9826128   0.          0.18566662]
     [ 0.          1.          0.        ]
     [-0.18566662  0.          0.9826128 ]]

and

B = [[ 1.          0.          0.        ]
     [ 0.62692564  0.77418869  0.08715574]]

Also, numpy.dot(A,B) results in ValueError: matrices are not aligned.

Jim Lewis
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Serge Stroobandt
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    You are thinking of the same row-column ordering as IDL. NumPy follows mathematical standards, the number of columns in the first matrix must equal the number of rows in the second matrix. Hence, the "strange" expression with all the transposes in it below. – mgalloy Apr 18 '14 at 00:52

2 Answers2

2

Reading the notes on IDL's definition of matrix multiplication, it seems they use the opposite notation to everyone else:

IDL’s convention is to consider the first dimension to be the column and the second dimension to be the row

So # can be achieved by the rather strange looking:

numpy.dot(A.T, B.T).T

from their example values:

import numpy as np
A =  np.array([[0, 1, 2], [3, 4, 5]])
B = np.array([[0, 1], [2, 3], [4, 5]])
C = np.dot(A.T, B.T).T
print(C)

gives

[[ 3  4  5]
 [ 9 14 19]
 [15 24 33]]
dabhaid
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0

If I'm correct you want matrix multiplication.

Vladimir
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