Finding indices of matches of one array in another array

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Question :

Finding indices of matches of one array in another array

I have two numpy arrays, A and B. A conatains unique values and B is a sub-array of A.
Now I am looking for a way to get the index of B’s values within A.

For example:

A = np.array([1,2,3,4,5,6,7,8,9,10])
B = np.array([1,7,10])
# I need a function fun() that:
fun(A,B)
>> 0,6,9
Asked By: farhawa

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Answer #1:

You can use np.in1d with np.nonzero

np.nonzero(np.in1d(A,B))[0]

You can also use np.searchsorted, if you care about maintaining the order –

np.searchsorted(A,B)

For a generic case, when A & B are unsorted arrays, you can bring in the sorter option in np.searchsorted, like so –

sort_idx = A.argsort()
out = sort_idx[np.searchsorted(A,B,sorter = sort_idx)]

I would add in my favorite broadcasting too in the mix to solve a generic case –

np.nonzero(B[:,None] == A)[1]

Sample run –

In [125]: A
Out[125]: array([ 7,  5,  1,  6, 10,  9,  8])

In [126]: B
Out[126]: array([ 1, 10,  7])

In [127]: sort_idx = A.argsort()

In [128]: sort_idx[np.searchsorted(A,B,sorter = sort_idx)]
Out[128]: array([2, 4, 0])

In [129]: np.nonzero(B[:,None] == A)[1]
Out[129]: array([2, 4, 0])
Answered By: Divakar

Answer #2:

Have you tried searchsorted?

A = np.array([1,2,3,4,5,6,7,8,9,10])
B = np.array([1,7,10])

A.searchsorted(B)
# array([0, 6, 9])
Answered By: Bi Rico

Answer #3:

Just for completeness: If the values in A are non negative and reasonably small:

lookup = np.empty((np.max(A) + 1), dtype=int)
lookup[A] = np.arange(len(A))
indices  = lookup[B]
Answered By: Paul Panzer

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