How to arrange elements in one array to match the position of elements in a second array?

If I have two arrays A and B as displayed below, how can I rearrange B such that a new boolean array Ba which corresponds to the values of array B can be resorted according to the values of array A instead?

For example, if I have:

A =

1 5 3 2 4 4 2 2 3 5

2 1 1 2 0 1 1 1 0 0

30 22 23 25 -1 30 18 25 -1 -1

B =

1 2 2 2 3 3 4 4 5 5

2 1 1 2 0 1 0 1 0 1

30 18 25 25 -1 23 -1 30 -1 22

Ba =

0 0 1 0 1 0 0 0 0 0

Note that array B is simply the sorted version of array A according to the first row in an ascended order. It is resorted twice according to the second row in an ascended order for the values in row 1 that equal one another (e.g., if there are multiples elements equal to 3 in row 1, then the array will sort by the second row instead, so if the second row has values 3 3 3 3 3; 2 6 3 9 1, it will then be sorted to 3 3 3 3 3; 1 2 3 6 9)

I only mentioned this in case reverse sorting might work easier than entirely rearranging the position of the array.

I want to rearrange array B and Ba such that it corresponds to the positioning of array A.

So the new arrays would look like this:

A =

1 5 3 2 4 4 2 2 3 5

2 1 1 2 0 1 1 1 0 0

30 22 23 25 -1 30 18 25 -1 -1

Aa =

0 0 0 0 0 0 0 1 1 0

The new array would essentially match A, and the array Aa is the same array Ba but now rearranged so that it follows the arrangement of A rather than B.

I am not entirely sure how to write the MATLAB code to accomplish this. All help is appreciated!

Answer 1

[Stephen23]

You could find the mapping with ISMEMBER:

https://www.mathworks.com/help/matlab/ref/double.ismember.html

A = [1,5,3,2,4,4,2,2,3,5; 2,1,1,2,0,1,1,1,0,0; 30,22,23,25,-1,30,18,25,-1,-1]

A = 3×10

1 5 3 2 4 4 2 2 3 5 2 1 1 2 0 1 1 1 0 0 30 22 23 25 -1 30 18 25 -1 -1

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B = [1,2,2,2,3,3,4,4,5,5; 2,1,1,2,0,1,0,1,0,1; 30,18,25,25,-1,23,-1,30,-1,22]

B = 3×10

1 2 2 2 3 3 4 4 5 5 2 1 1 2 0 1 0 1 0 1 30 18 25 25 -1 23 -1 30 -1 22

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Ba = [0,0,1,0,1,0,0,0,0,0]

Ba = 1×10

0 0 1 0 1 0 0 0 0 0

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Obtain the 2nd output from ISMEMBER:

[~,idx] = ismember(A.',B.','rows');

Reorder using indexing:

Aa = Ba(idx)

Aa = 1×10

0 0 0 0 0 0 0 1 1 0

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●

Anew = B(:,idx)

Anew = 3×10

1 5 3 2 4 4 2 2 3 5 2 1 1 2 0 1 1 1 0 0 30 22 23 25 -1 30 18 25 -1 -1

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[Tatiana]

Yes, this worked perfectly for me! Thank you for the help!

Answer 2

[Steven Lord]

Here's your sample data.

A = [1     5     3     2     4     4     2     2     3     5
     2     1     1     2     0     1     1     1     0     0
    30    22    23    25    -1    30    18    25    -1    -1];
B = [1     2     2     2     3     3     4     4     5     5
     2     1     1     2     0     1     0     1     0     1
    30    18    25    25    -1    23    -1    30    -1    22];
Ba =[0     0     1     0     1     0     0     0     0     0];
Aa =[0     0     0     0     0     0     0     1     1     0];

Call sortrows on the transpose of A then transpose the result back.

[B1, indices] = sortrows(A.');

Let's check that the output from sortrows matches the hard-coded expected result from the sample data.

check = isequal(B, B1.')

check = logical
   1

You can use the second output to reorder Aa to generate Ba.

isequal(Ba, Aa(indices))

ans = logical
   1