Numpy array, how to replace values that satisfy a list of conditions?

Question

Suppose I have a numpy array x = [1, 2, 3, 4, 5, ...] and I want to replace values that are not in the list a = [1, 3, 5, ...] with 0.

I tried x[x not in a] = 0 but I got the error:

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Anyone know the proper way that doesn't require spelling out the conditions?

Buckeye14Guy · Accepted Answer · 2019-10-10 16:00:13Z

3

import numpy as np
x = np.array([1, 2, 3, 4, 5])
a = np.array([1, 3, 5])
x[~np.isin(x,a)] = 0

### Output
>>> array([1, 0, 3, 0, 5])

answered Oct 10, 2019 at 16:00

Buckeye14Guy

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Comments

Florian Bernard · Accepted Answer · 2019-10-10 16:08:27Z

1

You should use numpy.where :

x = np.array([1, 2, 3, 4, 5])
a = np.array([1, 3, 5])
mask = np.isin(x, a)
x[mask] = 0
print(x)
>>> array([0, 2, 0, 4, 0])

answered Oct 10, 2019 at 16:08

Florian Bernard

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