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I have 2 vectors A and B. Both are column vectors and both contain 1000 values.

For example:

A=[1,2,5,1,6,2,8,2,9];

B=[3,4,2,3,7,4,5,4,8];

I wish to find out how many times that (A,B) match so that I can create a contour plot of (A,B) which is colour coded in accordance to the number of matching occurrences:

e.g. (1,3) = 2 occurrences (2,4) = 3 occurrences (5,2) = 1 Occurrence ....etc

I know that I need to create a for loop and I was hoping the unique command would work but as yet I have been unable to generate a working code.

Walter Roberson
on 30 Mar 2011

accumarray([A(:), B(:)])

Provided that the values are positive integers. Otherwise,

[bA, mA, nA] = unique(A);

[bB, mB, nB] = unique(B);

accumarray([nA(:), nB(:)])

Then entry (I,J) counts the match bA(I) to bB(J)

Teja Muppirala
on 31 Mar 2011

What Walter wrote should work. You have to feed in nA and nB into accumarray. Are you sure you wrote it correctly?

[bA, mA, nA] = unique(A);

[bB, mB, nB] = unique(B);

accumarray([nA(:), nB(:)],1)

Todd Flanagan
on 31 Mar 2011

A = [1,2,5,1,6,2,8,2,9];

B = [3,4,2,3,7,4,5,4,8];

c =

0 0 2 0 0 0 0 0

0 0 0 3 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0

0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 1

looking at unique rows

pairs = unique([A(:), B(:)], 'rows')

pairs =

1 3

2 4

5 2

6 7

8 5

9 8

To find the count for row 1

>> c(1,3)

ans =

2

If your rows don't contain positive integers, you can still do the problem by converting it to indeces using unique.

Unique can create a vector of the unique elements and also a vector of indeces to reconstruct the original vector based on these values. For example:

>> [bA, mA, nA] = unique(A)

bA =

1 2 5 6 8 9

mA =

4 8 3 5 7 9

nA =

1 2 3 1 4 2 5 2 6

bA(nA) gives you back the original vector but lets you operate in terms of something (indeces) that ensure you are using positive integers.

>> bA(nA)

ans =

1 2 5 1 6 2 8 2 9

Now to use accumarray:

[bA, mA, nA] = unique(A);

[bB, mB, nB] = unique(B);

accumarray([nA(:), nB(:)],1)

ans =

0 2 0 0 0 0

0 0 3 0 0 0

1 0 0 0 0 0

0 0 0 0 1 0

0 0 0 1 0 0

0 0 0 0 0 1

To find the rows again,

your pairs are now in terms of indeces:

>> pairs = unique([nA(:) nB(:)], 'rows')

pairs =

1 2

2 3

3 1

4 5

5 4

6 6

So, c(1,2) = 2

and the original pair is

>> [bA(1) bB(2)]

ans =

1 3

Hassan
on 15 Mar 2018

Hi all, The suggested method accumarray([nA(:), nB(:)],1) is good to have counts of pairs in 2D representation. What if I interested in a square matrix, I mean full representation of length n by n?

let's say I have two vectors x=1:10, and y=1:8. accumarray([nA(:), nB(:)],1) will result in 10 by 8 matrix. How can I make the result 10 by 10 with two lines of zeros?

Thanks a lot, HM

Hassan
on 15 Mar 2018

I found the answer... just add a dimension argument next to the 1

accumarray([nA(:), nB(:)],1,[10 10])

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