MATLAB: Remove rows based on length of column value

colummatrixmatrix arraymatrix manipulation

I have a matrix , 1531*3 double.

At a first column, numbers represent same particles from 1 to 14. I have 14 particles and each of them has different time points(second column) with intensity value(third column).

I'd like to take only particles which the number of time points is over 2, and make a new matrix. In this example, I need to remove rows of 14th particle and make the new matrix which doesn't have 14th particle, because it has only 2 time points. Should I use 'loop'?

for j = 1:14
    if length(find(A(:,1)==j)) > 2
      
    
    end
end

Best Answer

Instead of implicit expansion (which can take a lot of memory for bigger arrays), you can use histogram counts:

S=load('practice.mat');
A=S.A;
c=histcounts(A(:,1),numel(unique(A(:,1))));
remove_ID=find(c<=2);
L=ismember(A(:,1),remove_ID);
B=A(~L,:);%keep only IDs with >2 time points

Related Solutions

MATLAB: Align different starting point to zero

try with find and index

for i=1:tn
    ind=find(B(:,1)==number(i))
    y=min(B(ind,2));
    B(ind,2)=b(ind,2)-y;
end

MATLAB: How to Sum Certain Values in Matrix A Based on Values of Matrix B in a Loop

Im not a 100% sure if this is what you asked for but ill have a go at it =)

A=[0:23;...  %hours 0 to 23
   rand(15,24)]; %15x24 submatrix with random values.
B=randi([0,23],15,2)]; % 15x2 matrix with numbers between 0 and 23
B=sort(B,2);           % now the first value in each row is the lowest.

Im not sure what "files" and length(files) is, but ill make an example based on your first A and B.

J=zeros(size(B,1)); %pre allocate J to avoid having to reallocate during loop.
for i=1:size(B,1)
   J(:,i)=sum( A(2:end,(B(i,1):B(i,2))+1),2);  % sum each row in a from B(i,1) 
                                               % to B(i,2).
                                               % add one to the index since 
                                               % "hour 0" in A
                                               % is located in column 0.
end

Is this what you wanted? If e.g B was equal to

B=[0,6;
   ..
   ..
   ..

J(1,1) will contain sum( A (2, 1:7) ) . While J(2,1) contains sum( A (3,1:7) ) and so forth.

To realete to your first example:

J=[];
A = [0 1 2 3 4 5; 
     5 6 7 8 9 2];
B = [0 3];
J(:,1)=sum( A(2:end , (B(1,1):B(1,2))+1 ,2)
J =
      26

Note that the sum is 26, because i assume that when B is [0,3] you want to use colums of A that contains 0 , 1 ,2 AND 3. If you only want 0 1 2 remove

   (B(i,1):B(i,2))+1

and replace it with

    B(i,1)+1:B(i,2)

In your specific example:

sum( A(2:end , B(1,1)+1:B(1,2) ) ,2)
ans =
      18

Best Answer

Related Solutions

MATLAB: Align different starting point to zero

MATLAB: How to Sum Certain Values in Matrix A Based on Values of Matrix B in a Loop

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