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Grouping repetead values in cells

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Fernando
Fernando 2018-5-10
评论： Fernando 2018-5-10
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

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采纳的回答

Ameer Hamza
Ameer Hamza 2018-5-10
编辑：Ameer Hamza 2018-5-10
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.

更多回答（3 个）

Jan
Jan 2018-5-10
% 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.
Cathal Cunningham
Cathal Cunningham 2018-5-10
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.
Fernando
Fernando 2018-5-10
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
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Jan
Jan 2018-5-10
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
Fernando 2018-5-10
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

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