Grouping repetead values in cells

Hi, I would like to group repeated values in a new matrix with four cells. As an simple example lets say that I have these two arrays (a,B) plus a new one (NewE) which is the matrix to my outputs:

a=[1;2;1;2]

NewB=cell(1,length(a))

B=[560800 x 20] %large numerical matrix where the first column contain the same values as in vector "a"

so I think to get the NewB in this simple form whitout using a counter:

for i = 1:length(B)

for j= 1: length(a)

if B(i,1) == a(j,1)

NewB{j}(end+1,:)=B(i,:);

end

end

end

It is obvious that the algoritm doesnt work but the idea was that.

Thank you for your help in advance,

/Fernando

Answer 1

[Ameer Hamza]

If you want to are trying to separate the rows of matrix B on the basis of values in its column 1, then just use splitapply(),

NewB = splitapply(@(x) {x}, B, findgroups(B(:,1)))

NewB will have the same number of cells as the number of unique elements in B(:,1).

Edit: As stated by @Fernando, that the matrix must be divided into different portions according to the grouping of 1 and 2 in column 1. The following code will work in that case:

index = [0; find(diff(B(:,1))); 20];
portionSize = diff(fliplr(index));

dividedData = mat2cell(B, portionSize, size(B, 2));

you don't need to specify a separate a, the method split the matrix on the basis of consecutive portions in column 1.

Answer 2

[Jan]

% Some test data:
a    = [1;2;1;2]
B    = rand(560800, 20);
B(:, 1) = randi([1,2], 560800, 1);

% Distribute B over cells of NewB:
NewB = cell(1, length(a))
for k = 1:length(a)
  NewB{k} = B(B(:, 1) == a(k), :);
end

While a has repeated values, the resulting NewB contains redundant vectors.

Answer 3

[Cathal Cunningham]

Is the matrix B in this case a cell array or a matrix? The comment ""%large numerical matrix" would suggest that it's a numeric array but that contradicts "the first column contain the same values as in vector "a" ". For the first column in B to match the vector a it would require that B is a cell array with the first column having a structure such that

B{1,1} = [1;2;1;2]

for the script to consider it a match. The problem statement is a little unclear.

Answer 4

[Fernando]

Hi everybody! ..and thanks for your comments! The answer of Jan comes closer but I forgot to comment that for the matrix B the first column of B(:,1) contains, for this exemple, only two values; 1 and 2 BUT in a sequence (group) of data, i.e:

B(:,1)= [1;1;1;1;1;2;2;2;2;2;1;1;1;1;;2;2;2;2;2;2]

I mean that the third and fourth values/rows in the vector "a" are the second repetition in a measurements series for the first and the second values. In this case I would like to have four cells in the matrix NewB like this:

NewB=[{first_1,:},{first_2,:},{second_1,:},{second_2,:}]

I appreciate very much all your comments Kind regards Nando

[Ameer Hamza]

Please view the updated answer .

[Jan]

Does this mean, that a is an alternating sequence of 1 and 2 in every case?

The more precise the question is at the beginning, the less time is wasted with given not matching answers.

[Fernando]

You are right @Jan but I´m working with a large database containing specifically matrices with these two types of "problems" in my code. Your answer help me to solve one of the problems when I dont need to check for a "par" of alternated values like in a. The answer of @Ammer is the complement that I was looking for in order to run my code correctly....and It works now!

Thank you very much for your help!!

regards/Saludos

Fernando