How to replace each element of an array with another array in Python?

Question

MWE: Let us consider the following example.

L0=[[b,0],[b,b]], L1=[[b,b],[b,1]], L2=[[b,b],[2,b]]

S=[[0,1,2],[2,0,1]]

Is there any way the replace each element of S by L0 for 0 and L1 for 1 and L2 for 2 in S to get S1 like in the image? enter image description here

Actually, I want a python program which will check: if the element of S is zero then it will replace 0 by the predefined 2D array and so on.

So, how many such L- arrays do you have in your actual use case? — Divakar
– Divakar, Commented Jan 2, 2018 at 18:03
@Divakar: There are three L-arrays, but size of S is very large. Here I have shown small size of S. — MKS
– MKS, Commented Jan 4, 2018 at 2:06

willeM_ Van Onsem · Accepted Answer · 2018-01-02 18:30:42Z

5

Yes. We can first construct a numpy array that contains L0, L1 and L2:

A = np.array([L0, L1, L2])

Next we construct a numpy array of S:

B = np.array(S)

now we have for C = A[B] (or C = np.take(A,B,axis=0) as suggested by @Divakar):

>>> C = np.take(A,B,axis=0)
>>> C
array([[[[b, 0],
         [b, b]],

        [[b, b],
         [b, 1]],

        [[b, b],
         [2, b]]],


       [[[b, b],
         [2, b]],

        [[b, 0],
         [b, b]],

        [[b, b],
         [b, 1]]]])

This is of course not exactly what we intended: we want to obtain a 2D-array. We can do this by first transposing (or swapaxes, like @PaulPanzer suggests) and then reshaping, we obtain:

>>> C.transpose(0,2,1,3).reshape(4,6)
array([[b, 0, b, b, b, b],
       [b, b, b, 1, 2, b],
       [b, b, b, 0, b, b],
       [2, b, b, b, b, 1]])

Since 4 and 6 of course depend on the size of the dimensions of L0, L1, L2 and S, we can also calculate them based on that size:

A = np.array([L0, L1, L2])
B = np.array(S)
m, n = B.shape
_, u, v = A.shape
np.take(A,B,axis=0).swapaxes(1,2).reshape(u*m, v*n)

Like @DSM says, from Numpy-1.13, there is np.block function for this purpose, and we can write it as:

>>> np.block([[A[i] for i in row] for row in S])
array([[b, 0, b, b, b, b],
       [b, b, b, 1, 2, b],
       [b, b, b, 0, b, b],
       [2, b, b, b, b, 1]])

edited Jan 2, 2018 at 18:30

answered Jan 2, 2018 at 17:44

willeM_ Van Onsem

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8 Comments

MKS Over a year ago

Thank you for your answer. Is it sophisticated process if the dimension of S is large, suppose 400x500? @Willem Van Onsem

AGN Gazer Over a year ago

@MKS Why don't you try and you will see if it is a "sophisticated process"... [whatever you might mean by "sophisticated process"]

MKS Over a year ago

@AGN Gazer: I mean, is there any other way which can do the same thing quickly. May I use bmat for this

Paul Panzer Over a year ago

Instead of transpose(0,2,1,3), swapaxes(1,2) is perhaps slightly more readable. Still, very nice answer.

DSM Over a year ago

While the listcomp isn't nearly as clean as A[B], I think np.block([[A[i] for i in row] for row in S]) (in numpy 1.13) is so much cleaner than the reshape that I'd go that direction.

|

Paul Panzer · Accepted Answer · 2018-01-02 18:15:51Z

1

If the number of distinct building blocks Li is not large we can use the Kronecker product np.kron:

import numpy as np

L0=[[b,0],[b,b]]; L1=[[b,b],[b,1]]; L2=[[b,b],[2,b]]

S=[[0,1,2],[2,0,1]] 

S1 = sum(np.kron(np.equal(i, S), L) for i, L in enumerate((L0, L1, L2)))

Value of S1 assuming b = 3:

[[3 0 3 3 3 3]
 [3 3 3 1 2 3]
 [3 3 3 0 3 3]
 [2 3 3 3 3 1]]

answered Jan 2, 2018 at 18:15

Paul Panzer

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