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I have a problem during substitution with data in array: say, 

a = [1, 0, 0]
b = [0, 0, 0]
c = [0, 0]
X = numpy.zeros((3, 3, 2))

and I have Matrix Y with shape (2,3,2) and it is non a zero matrix

Now; I want to equal these elements of X by Y directly;

X[tuple(numpy.where(a==0)[0]), 
  tuple(numpy.where(b==0)[0]),
  tuple(numpy.where(c==0)[0])] = Y

I got the error shape mismatch: objects cannot be broadcast to a single shape

chthonicdaemon
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Khaled Khafagy
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1 Answers1

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You could use np.ix_ to construct index arrays appropriate for indexing X:

import numpy as np
np.random.seed(2016)
a=np.array([1, 0, 0])
b=np.array([0, 0, 0])
c=np.array([0, 0])
X = np.zeros((3,3,2))
Y = np.random.randint(1, 10, size=(2,3,2))

idx = np.ix_(a==0, b==0, c==0)
X[idx] = Y
print(X)

yields

array([[[ 0.,  0.],
        [ 0.,  0.],
        [ 0.,  0.]],

       [[ 9.,  8.],
        [ 3.,  7.],
        [ 4.,  5.]],

       [[ 2.,  2.],
        [ 3.,  3.],
        [ 9.,  9.]]])

Alternatively, you could construct a boolean mask

mask = (a==0)[:,None,None] & (b==0)[None,:,None] & (c==0)[None,None,:]
X[mask] = Y

Indexing (a=0) as in (a==0)[:,None,None] adds new axes to the 1D boolean array (a=0). (a==0)[:,None,None] has shape (3,1,1). Similarly, (b==0)[None,:,None] has shape (1,3,1), and (c==0)[None,None,:] has shape (1,1,2).

When combined with & (bitwise-and), the three arrays are broadcasted to one common shape, (3,3,2). Thus, X gets indexed by one boolean array of shape (3,3,2) in

X[mask] = Y
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unutbu
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  • Thanks a million but, is there any way to do in case Y has (2,3,2) size . what I am saying is that if Y has different size of X what can I do to store it into X ? – Khaled Khafagy Aug 25 '16 at 10:23
  • Thanks a million again and again, it works well...... I said X[mask]=Y .... and it works good – Khaled Khafagy Aug 25 '16 at 10:28