Search elements of one array in another, row-wise - Python / NumPy

Question

For example, I have a matrix of unique elements,

a=[
    [1,2,3,4],
    [7,5,8,6]
]

and another unique matrix filled with elements which has appeard in the first matrix.

b=[
    [4,1],
    [5,6]
]

And I expect the result of

[
    [3,0],
    [1,3]
].

That is to say, I want to find each row elements of b which equals to some elements of a in the same row, return the indices of these elements in a. How can i do that? Thanks.

You just need to consider each row independently and https://stackoverflow.com/questions/432112/is-there-a-numpy-function-to-return-the-first-index-of-something-in-an-array — sshashank124, Dec 28 '19 at 08:20
Welcome to SO! It seems to not to have any logic in your question? Could you explain which operation/process you wnat to do on it? — David García Bodego, Dec 28 '19 at 08:34
@AgaigetSAgaigetS Did either of the posted solutions work for you? — Divakar, Dec 29 '19 at 10:44

Divakar · Answer 1 · 2019-12-28T09:35:55.920

Here's a vectorized approach -

# https://stackoverflow.com/a/40588862/ @Divakar
def searchsorted2d(a,b):
    m,n = a.shape
    max_num = np.maximum(a.max() - a.min(), b.max() - b.min()) + 1
    r = max_num*np.arange(a.shape[0])[:,None]
    p = np.searchsorted( (a+r).ravel(), (b+r).ravel() ).reshape(m,-1)
    return p - n*(np.arange(m)[:,None])

def search_indices(a, b):
    sidx = a.argsort(1)
    a_s = np.take_along_axis(a,sidx,axis=1)
    return np.take_along_axis(sidx,searchsorted2d(a_s,b),axis=1)

Sample run -

In [54]: a
Out[54]: 
array([[1, 2, 3, 4],
       [7, 5, 8, 6]])

In [55]: b
Out[55]: 
array([[4, 1],
       [5, 6]])

In [56]: search_indices(a, b)
Out[56]: 
array([[3, 0],
       [1, 3]])

Another vectorized one leveraging broadcasting -

In [65]: (a[:,None,:]==b[:,:,None]).argmax(2)
Out[65]: 
array([[3, 0],
       [1, 3]])

@Navaro Well the usual NumPy way (under the hoods i.e. and the one that is performant) is with implementation in C. So, yes, it would be nice to see these being implemented in native NumPy. For now, I am just building up on existing tools. — Divakar, Dec 28 '19 at 09:40

score 0 · Answer 2 · answered Dec 28 '19 at 09:01

0

If you don't mind using loops, here's a quick solution using np.where:

import numpy as np

a=[[1,2,3,4],
   [7,5,8,6]]
b=[[4,1],
   [5,6]]

a = np.array(a)
b = np.array(b)
c = np.zeros_like(b)

for i in range(c.shape[0]):
    for j in range(c.shape[1]):
        _, pos = np.where(a==b[i,j])
        c[i,j] = pos

print(c.tolist())

answered Dec 28 '19 at 09:01

Mercury

3,417
1
10
35

1

Thanks. But when I process on huge data, there will be much slow. I want it solve in the matrix way, just because of the demand of faster processing speed. – AgaigetS AgaigetS Dec 28 '19 at 09:07

oppressionslayer · Answer 3 · 2019-12-28T10:48:40.570

0

You can do it this way:

np.split(pd.DataFrame(a).where(pd.DataFrame(np.isin(a,b))).T.sort_values(by=[0,1])[::-1].unstack().dropna().reset_index().iloc[:,1].to_numpy(),len(a))                               

# [array([3, 0]), array([1, 3])]

edited Dec 28 '19 at 10:48

answered Dec 28 '19 at 09:09

oppressionslayer

6,942
2
7
24

Search elements of one array in another, row-wise - Python / NumPy

3 Answers3