I want to find multiple matching rows from 2d array a
a = np.array([[2, 1], [3, 3], [4, 6], [4, 8], [4, 7], [4, 3]]) I have to search following entries
b = np.array([[4,6], [4,7]]) I know that I can loop over b and do following
for i in range(len(b)) : print(np.where(np.all(a==b[i],axis=1))[0]) And I get following
[2] [4] Can I get [[2],[4]] directly without using any loop?
If you want the indices you will generally reach for the arg_x functions like argmax and argwhere. Here np.argwhere will give you the indices if you can figure out how to pass the correct list of booleans. You can do that with np.isin():
a = np.array([[2, 1], [3, 3], [4, 6], [4, 8], [4, 7], [4, 3]]) b = np.array([[4,6], [4,7]]) np.argwhere(np.isin(a, b).all(axis=1)) Which returns:
array([[2], [4]]) 
np.isin() check each elements individually. For example if the row [6,7] is present in a then this row will be returned as a matching row !This should be a fast solution noticing that the two pairs have the same first coordinate:
np.where((a[:, 0] == 4) & ((a[:, 1] == 6) | (a[:, 1] == 7))) # Out: # (array([2, 4]),) The expression
print((a[:, 0] == 4) & ((a[:, 1] == 6) | (a[:, 1] == 7))) gives
[False False True False True False] 