Numpy Lookup (Map, or Point)

Here you go

A = array([[32, 32, 99,  9, 45],  # A   [99, 45,  9, 45, 32],   [45, 45, 99, 99, 32],   [ 9,  9, 32, 45, 99]])B = array([ 99, 32, 45, 9])ii = np.argsort(B)C = np.digitize(A.reshape(-1,),np.sort(B)) - 1

Originally I suggested:

D = np.choose(C,ii).reshape(A.shape)

But I realized that that had limitations when you went to larger arrays. Instead, borrowing from @unutbu's clever reply:

D = np.argsort(B)[C].reshape(A.shape)

Or the one-liner

np.argsort(B)[np.digitize(A.reshape(-1,),np.sort(B)) - 1].reshape(A.shape)

Which I found to be faster or slower than @unutbu's code depending on the size of the arrays under consideration and the number of unique values.

import numpy as npA=np.array([[32, 32, 99,  9, 45],              [99, 45,  9, 45, 32],            [45, 45, 99, 99, 32],            [ 9,  9, 32, 45, 99]])B=np.array([ 99, 32, 45, 9])cutoffs=np.sort(B)print(cutoffs)# [ 9 32 45 99]index=cutoffs.searchsorted(A)print(index)# [[1 1 3 0 2]#  [3 2 0 2 1]#  [2 2 3 3 1]#  [0 0 1 2 3]]    

index holds the indices into the array cutoff associated with each element of A. Note we had to sort B since np.searchsorted expects a sorted array.

index is almost the desired answer, except that we want to map

1-->13-->00-->32-->2

np.argsort provides us with this mapping:

print(np.argsort(B))# [3 1 2 0]print(np.argsort(B)[1])# 1print(np.argsort(B)[3])# 0print(np.argsort(B)[0])# 3print(np.argsort(B)[2])# 2print(np.argsort(B)[index])# [[1 1 0 3 2]#  [0 2 3 2 1]#  [2 2 0 0 1]#  [3 3 1 2 0]]

So, as a one-liner, the answer is:

np.argsort(B)[np.sort(B).searchsorted(A)]

Calling both np.sort(B) and np.argsort(B) is inefficient since both operations amount to sorting B. For any 1D-array B,

np.sort(B) == B[np.argsort(B)]

So we can compute the desired result a bit faster using

key=np.argsort(B)result=key[B[key].searchsorted(A)]