Convert a NumPy array to a binary array with the condition of each element existing in a list

Question

Is there any way to convert an array to a binary array such that any element that exists within a defined list is 1, and any element not in the list is 0?

For example, if I define a NumPy array as so:

a = np.array([[23,43,1],[43,5,0],[5,0,0]])

and a list as so:

l = [5,43]

I want a function that converts the array to/creates this:

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

I have already tried np.where(a in l, 1, 0), and it gives me this error:

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

Learning is a mess · Accepted Answer · 2021-12-01 00:54:34Z

1

Check out https://numpy.org/doc/stable/reference/generated/numpy.isin.html

arr=np.array([[23,43,1],[43,5,0],[5,0,0]])
l = [5,43]
np.isin(arr, l).astype(int)
#array([[0, 1, 0],
#       [1, 1, 0],
#       [1, 0, 0]])

answered Dec 1, 2021 at 0:54

Learning is a mess

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