index Nd array with list of length N in python

Question

This seems like a simple problem but I can't figure it out.

I have a numpy array of an arbitrary dimension (rank) N. I need to set a single element in the array to 0 given by the index values in a 1D array of length N. So for example:

import numpy as np
A=np.ones((2,2,2))
b=[1,1,1]

so at first I thought

A[b]=0

would do the job, but it did not. If I knew A had a rank of 3 it would be a simple case of doing this:

A[b[0],b[1],b[2]]=0

but the rank of A is not known until runtime, any thoughts?

Community · Accepted Answer · 2020-06-20 09:12:55Z

3

Indexing in numpy has somewhat complicated rules. In your particular case this warning applies:

The definition of advanced indexing means that x[(1,2,3),] is fundamentally different than x[(1,2,3)]. The latter is equivalent to x[1,2,3] which will trigger basic selection while the former will trigger advanced indexing. Be sure to understand why this is occurs.

Also recognize that x[[1,2,3]] will trigger advanced indexing, whereas x[[1,2,slice(None)]] will trigger basic slicing.

You want simple indexing (addressing a particular element), so you'll have to cast your list to a tuple:

A[tuple(b)] = 0

Result:

>>> A
array([[[ 1.,  1.],
        [ 1.,  1.]],

       [[ 1.,  1.],
        [ 1.,  0.]]])

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Sep 4, 2015 at 2:46

fjarri

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1 Comment

derchambers Over a year ago

Beautiful! I knew there would be some intuitive way to do it. Thanks again.

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