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.
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]])
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())
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])]
aguaranteed to be unique?