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

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

[madhan ravi]

 1 0 2 1 explain the logic please

[Bruno Luong]

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)

[Shubham Mohan Tatpalliwar]

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...

Answer 1

[Bruno Luong]

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

ans =

     1     0     2     1

>>

[Shubham Mohan Tatpalliwar]

Actually i want it sor every column but for simplicity just asked for only one

thanx for the tip and i would keep that into mind while asking next time

[Shubham Mohan Tatpalliwar]

Can u please explain me the code

i cant understand what is this +B==2

[Bruno Luong]

2=1+1 so the expression returns TRUE (=1) before summing if there is a 1 at the same row position in A(:,1) and in a column of B.

[Shubham Mohan Tatpalliwar]

and how can i now repeat in a loop for every column of A?

[Stephen23]

Simpler (fewer operations):

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

[Bruno Luong]

"and how can i now repeat in a loop for every column of A?"

squeeze(sum(reshape(A,size(A,1),1,[])&B))

[Shubham Mohan Tatpalliwar]

thanx a lot for that....

[Shubham Mohan Tatpalliwar]

can u explain a bit about this code?

[Bruno Luong]

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.

Answer 2

[Stephen23]

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

[Shubham Mohan Tatpalliwar]

Really this was very helpful for me