Remove elements from one array if present in another array, keep duplicates - NumPy / Python

Question

I have two arrays A (len of 3.8million) and B (len of 20k). For the minimal example, lets take this case:

A = np.array([1,1,2,3,3,         


        

Answer 1

Adding to Divakar's answer above -

if the original array A has a wider range than B, that will give you an 'index out of bounds' error. See:

A = np.array([1,1,2,3,3,3,4,5,6,7,8,8,10,12,14])
B = np.array([1,2,8])

A[B[np.searchsorted(B,A)] !=  A]
>> IndexError: index 3 is out of bounds for axis 0 with size 3

This will happen because np.searchsorted will assign index 3 (one-past-the-last in B) as the appropriate position for inserting in B the elements 10, 12 and 14 from A, in this example. Thus you get an IndexError in B[np.searchsorted(B,A)].

To circumvent that, a possible approach is:

def subset_sorted_array(A,B):
    Aa = A[np.where(A <= np.max(B))]
    Bb = (B[np.searchsorted(B,Aa)] !=  Aa)
    Bb = np.pad(Bb,(0,A.shape[0]-Aa.shape[0]), method='constant', constant_values=True)
    return A[Bb]

Which works as follows:

# Take only the elements in A that would be inserted in B
Aa = A[np.where(A <= np.max(B))]

# Pad the resulting filter with 'Trues' - I split this in two operations for
# easier reading
Bb = (B[np.searchsorted(B,Aa)] !=  Aa)
Bb = np.pad(Bb,(0,A.shape[0]-Aa.shape[0]),  method='constant', constant_values=True)

# Then you can filter A by Bb
A[Bb]
# For the input arrays above:
>> array([ 3,  3,  3,  4,  5,  6,  7, 10, 12, 14])

Notice this will also work between arrays of strings and other types (for all types for which the comparison <= operator is defined).