How can I check that a newly generated sequence is already in the existing structure list or not?

Suppose I have a sequence list as follows:

pp(1).x=[1 2 3 4];

pp(2).x=[3 2 1 4];

pp(3).x=[1 2 3 4];

pp(4).x=[2 1 3 4];

How can I check that a newly generated sequence pp(5).x=randperm(4) is already in the existing structure list 'pop' or not?

Answer 1

[Johannes Fischer]

newPerm = [2 1 3 4];

% this line basically runs over the elements of your struct array and checks whether the array in field 'x' is equal to the new permutation
any(cellfun(@(x) isequal(x, newPerm), {pp(:).x}))

If you want to have all possible permutations of an array you could also use

perms(1:4)

[Md. Asadujjaman]

How can I understand that which number of sequence matches with the new sequence.

Here, 4th sequence of the structure 'pp' matches with the new sequence.

How can get the result (4th sequence) that matches with the new sequence.

[Bjorn Gustavsson]

You should use data-structures that makes your programming easy.

In your case I would store the existing permutations in a matrix, ppx, because then something like this would be easier for me to understand:

i_existing = @(newperm) ( ppx(:,1) == newperm(1)) &  ( ppx(:,2) == newperm(2)) &  ( ppx(:,3) == newperm(3))

which will return an array with 1's and 0'.

Or you could solve it with

find(cellfun(@(x) isequal(x, newPerm), {pp(:).x}))

But then you'll have to check for empty arrays for the case where newPerm doesn't exist.

Answer 2

[Bjorn Gustavsson]

For example this can be used:

intersect([1 2 3;1 3 2;2 1 3;2 3 1;3 1 2],[3 2 1],'rows')
%
%ans =
%
%  0x3 empty double matrix
%
intersect([1 2 3;1 3 2;2 1 3;2 3 1;3 1 2],[3 1 2],'rows')
%
%ans =
%
%     3     1     2

Or you can simply check if the minimum Euclidian distance is zero between the new and the existing points:

l_min = min((pop(:,1)-pp.x(1)).^2+(pop(:,2)-pp.x(2)).^2+(pop(:,3)-pp.x(3)).^2+(pop(:,4)-pp.x(4)).^2)

If that is zero the sequence already exists. With the newer matlab-versions you could use the implicit expansion of "+" instead of the

explicit one here. There surely is a large number of ways to do this, others might come up with ways to use strfind (or findstr) or

other variants. You'll have to time these to see which is best for your case.

HTH