Replacing value in a numpy array using as an index another value from the same array

Question

I am looking for an elegant way of doing the following:

I have an array like this:

[[0, 0, 0, 1],
 [0, 0, 0, 2],
 [0, 0, 0, 1]]

I want to replace the element on each row, which index is equal to the last element of that row, with 1. So for the first row I need the element with index 1 to become 1, for the second row the element with an index 2 to become 1 and so on. This is just an example, in reality I have bigger matrices and the last column has values from 0 to 9, which I need to use to indicate which element of the row to become 1.

DSM · Accepted Answer · 2014-11-06 22:32:37Z

4

IIUC, you could use advanced indexing and do something like

>>> s
array([[0, 0, 0, 1],
       [0, 0, 0, 2],
       [0, 0, 0, 1]])
>>> s[np.arange(len(s)),s[:,-1]] = 1
>>> s
array([[0, 1, 0, 1],
       [0, 0, 1, 2],
       [0, 1, 0, 1]])

answered Nov 6, 2014 at 22:32

DSM

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1 Comment

Tsubaki Hana Over a year ago

Thank you! I never managed to organize it like this in my attempts.

Celeo · Accepted Answer · 2014-11-06 22:32:39Z

1

array = [[0, 0, 0, 1], [0, 0, 0, 2], [0, 0, 0, 1]]
for row in array:
    row[row[-1]] = 1

Yields

[[0, 1, 0, 1], [0, 0, 1, 2], [0, 1, 0, 1]]

answered Nov 6, 2014 at 22:32

Celeo

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1 Comment

Tsubaki Hana Over a year ago

Thank you, this helps me too for the sake of learning, just I was preferring something without for loop.

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