here is some code:
c = np.delete(a,b)
print(len(a))
print(a)
print(len(b))
print(b)
print(len(c))
print(c)
it gives back:
24
[32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55]
20
[46, 35, 37, 54, 40, 49, 34, 48, 50, 38, 42, 47, 33, 52, 41, 36, 39, 44, 55,
51]
24
[32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55]
as you can see, all elements of b appear in a, but are not being deleted. can not figure out why. any ideas? thank you.
numpy.delete does not remove the elements contained in b, it deletes a[b], in other words, b needs to contain the indices to remove. Since your b contains only values larger than the length of a, no values are removed. Currently out of bounds indices are ignored, but this will not be true in the future:
/usr/local/bin/ipython3:1: DeprecationWarning: in the future out of bounds indices will raise an error instead of being ignored by `numpy.delete`.
#!/usr/bin/python3
A pure Python solution would be to use set:
set_b = set(b)
c = np.array([x for x in a if x not in set_b])
# array([32, 43, 45, 51, 53])
And using numpy broadcasting to create a mask to determine which values to delete:
c = a[~(a[None,:] == b[:, None]).any(axis=0)]
# array([32, 43, 45, 51, 53])
They are about the same speed with the given example, but the numpy approach and takes more memory (because it generates a 2D matrix that contains all combinations of a and b).
