Replacing the values of a numpy array of zeros using a array of indexes

Question

I'm working with numpy and I got a problem with index, I have a numpy array of zeros, and a 2D array of indexes, what I need is to use this indexes to change the values of the array of zeros by the value of 1, I tried something, but it's not working, here is what I tried.

import numpy as np

idx = np.array([0, 3, 4], 
               [1, 3, 5],
               [0, 4, 5]]) #Array of index

zeros = np.zeros(6) #Array of zeros [0, 0, 0, 0, 0, 0]

repeat = np.tile(zeros, (idx.shape[0], 1)) #This repeats the array of zeros to match the number of rows of the index array

res = []
for i, j in zip(repeat, idx):
        res.append(i[j] = 1) #Here I try to replace the matching index by the value of 1

output = np.array(res)

but I get the syntax error

expression cannot contain assignment, perhaps you meant "=="?

my desired output should be

output = [[1, 0, 0, 1, 1, 0],
          [0, 1, 0, 1, 0, 1],
          [1, 0, 0, 0, 1, 1]]

This is just an example, the idx array can be bigger, I think the problem is the indexing, and I believe there is a much simple way of doing this without repeating the array of zeros and using the zip function, but I can't figure it out, any help would be aprecciated, thank you!

EDIT: When I change the = by == I get a boolean array which I don't need, so I don't know what's happening there either.

Mark · Accepted Answer · 2021-05-25 22:00:34Z

1

You can use np.put_along_axis to assign values into the array repeat based on indices in idx. This is more efficient than a loop (and easier).

import numpy as np

idx = np.array([[0, 3, 4], 
                [1, 3, 5],
                [0, 4, 5]]) #Array of index

zeros = np.zeros(6).astype(int) #Array of zeros [0, 0, 0, 0, 0, 0]
repeat = np.tile(zeros, (idx.shape[0], 1))

np.put_along_axis(repeat, idx, 1, 1)

repeat will then be:

array([[1, 0, 0, 1, 1, 0],
       [0, 1, 0, 1, 0, 1],
       [1, 0, 0, 0, 1, 1]])

FWIW, you can also make the array of zeros directly by passing in the shape:

np.zeros([idx.shape[0], 6])

answered May 25, 2021 at 22:00

Mark

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2 Comments

Carlos Eduardo Corpus Over a year ago

Hi!, thank you for the quick answer, I didn't know about np.put, it does exactly what I need, quick question, in the arguments of the function, the 1's represents the value I need to replace and I'm guessing the other one is the axis of the array, right?, thanks for the additional answer about the shape of the np.zeros!

Mark Over a year ago

@CarlosEduardoCorpus yes that's correct. You could also call it with keywords which would make it clearer here: np.put_along_axis(repeat, idx, values=1, axis=1)

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