delete rows of and 2D array with a condition refering to a different array in python numpy

Question

I have two numpy arrays:

a = np.array([12, 13, 10])
b = np.array([[22, 123], [10, 142], [23, 232], [42, 122], [12, 239]])

I want to delete rows in b if the first element is not in a. Something like:

c = np.delete(b, np.where(a not in b[:, 0]))

print(c) gives:

[ 22 123  10 142  23 232  42 122  12 239]

which doesn't delete an element: c should look like

c = [[10, 142],
     [12, 239]]

Answer 1

Your input:

import numpy as np

a=[12,13,10]
b=[[22, 123],[10,142],[23,232],[42,122],[12,239]]
a, b = np.array(a), np.array(b)

c = b[np.isin(b[:, 0], a)]

Output:

array([[ 10, 142],
       [ 12, 239]])

Answer 2

You need to mark the rows in b whose first elements are not in a. np.isin returns an array of that masks the first input. It also provides an argument invert, to speed up the inversion that you want:

mask = np.isin(b[:, 0], a, invert=True)

You can use this mask in np.delete directly:

c = np.delete(b, mask, axis=0)

Alternatively, you can convert it to indices using np.flatnonzero:

index = np.flatnonzero(mask)
result = np.delete(b, index, axis=0)

Furthermore, you don't need to use np.delete at all, since indexing with the non-inverted mask will drop the undesired rows:

result = b[np.isin(b[:, 0], a)]