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how to find common values in two matrix for particular column?

Latest activity Edited by Bruno Luong
on 25 Oct 2018
if true
% code
A=[0 1 1 0; 1 0 0 1; 1 1 0 0; 0 0 1 1]
B=[0 1 0 1; 1 0 1 0; 0 0 1 1; 1 1 0 0]
end
for 1st column of A and ALL columns of b
if we check
the expected answer is
1 0 2 1
these are the total number of instances they are matching

  3 Comments

1 0 2 1 explain the logic please
Please put care on
  • code formatting
  • explanation
  • formulation of synthetic question (Why give the whole A and ask just result that depends only on first column)
in the first column of A, there are two ones at 2nd row and 3 rd row
so considering this it will check all columns of B
in the first column of B, there is 1 in 2nd row that match and it did not match with the 3rd row
so for 1st column of B answer is 1
for 2nd column of B and 1st column of A nothing match so the answer is zero
Further for the 3rd column of B and 1st column A the second and 3rd row of both are 1 so the answer is 2
and so on...

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2 Answers

Answer by Bruno Luong
on 24 Oct 2018
 Accepted Answer

>> sum(A(:,1)+B==2)
ans =
1 0 2 1
>>

  9 Comments

The reshape() just moves the 2nd dimension (column) of A to the 3rd dimension
So each origin column A(:,j) now can be addressed as A(:,:j).
The explanation for SUM(... & B) you already know, but now use in the context of auto-expansion. Excepted that the result now is of the size (1 x size(B,2) x size(A,2)): each number of common 1-values of B and A(:,j) is in a slide XX(1,:,j) before SQUEEZE is invoked.
The squeeze command removes the 1st singleton dimension, so XX(:,j) is common 1-values of B and A(:,j).
NOTE: You might transpose the result so each row corresponds to result of a column of A with you prefer.

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Answer by Stephen Cobeldick on 24 Oct 2018

You don't need to use a loop:
>> sum(permute(A,[3,2,1])&permute(B,[2,3,1]),3)
ans =
1 0 1 2
0 1 2 1
2 1 0 1
1 2 1 0

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