Python / numpy: How to replace matrix elements with specified length along axis

Question

I have an array that specifies how many elements along the last axis of a matrix I want to change to 0. How do I do this efficiently?

x = np.ones((4, 4, 10))
change_to_zeros = np.random.randint(10, size=(4, 4))
# change_to_zeros = 
# [[2 1 6 8]
# [4 0 4 8]
# [7 6 6 2]
# [4 0 7 1]]

Now what I want to do is something like x[:, :, :change_to_zeros] = 0 - for example for the first element of change_to_zeros, I have change_to_zeros[0, 0] = 2 so I want to change the first (or last, or whatever) 2 elements along the last axis of x (length 10) to 0.

Clarification: For example at x[0, 0, :] I have ones with length 10. I want to change 2 of these (change_to_zeros[0, 0] = 2) to 0, keeping the rest with ones.

Sorry for not being clear. For example at x[0, 0, :] I have ones with length 10. I want to change 2 of these (change_to_zeros[0, 0] = 2) to 0, keeping the rest with ones. — ru111
– ru111, Commented Jul 24, 2018 at 19:37

akuiper · Accepted Answer · 2018-07-24 19:36:37Z

1

You can create a boolean array (same shape as x) with change_to_zeros[:,:,None] > np.arange(x.shape[-1]) and then assign zero to trues:

x[change_to_zeros[:,:,None] > np.arange(x.shape[-1])] = 0

Check results:

change_to_zeros[0,0]
# 2

x[0,0]
# array([ 0.,  0.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.])

change_to_zeros[0,2]
# 7

x[0,2]
# array([ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  1.,  1.,  1.])

answered Jul 24, 2018 at 19:36

akuiper

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