How to spread elements in a cell based on vector?

Dear all, I have cell b12 with number of elements. R is the first element after the value in vector D1. X is all values after the first element in R, however, X value does not give me the right elements. it only shows the first value after the first value in R? I appreciate if someone help me with this issue.

b12={[3,282,103,271,123,1],[3,282,103,280,260,9],[3,282,55,52,90,12],[3,282,103,280,76,13],[3,282,195,255,23]};%,[3,282,249,143,34],[3,282,103,52,90,41],[3,282,103,280,45],[3,282,103,52],[3,282,195,255,58],[3,282,103,52,63],[3,282,195,73],[3,282,103,280,76],[3,282,249,117,82],[3,282,249,84],[3,282,103,52,90],[3,282,249,143,93],[3,282,108,100],[3,282,103],[3,282,108],[3,282,103,280,110],[3,282,249,117],[3,282,103,271,123],[3,282,103,280,76,125],[3,282,249,143,126],[3,282,108,127],[3,282,128],[3,282,103,52,90,41,134],[3,282,249,143],[3,282,103,280,260,184],[3,282,195],[3,282,103,280,260,197],[3,282,216],[3,282,103,52,217],[3,282,103,280,222],[3,282,103,52,224],[3,282,249,239],[3,282,103,52,90,41,245],[3,282,249],[3,282,103,280,110,254],[3,282,195,255],[3,282,103,280,260],[3,282,103,280,222,262],[3,282,103,271],[3,282,103,280],[3,282],[3,282,103,280,260,296]}

%D1 = many numbers but on here I put only 2 elements.

D1=[282,55];

      N = numel(b12);
      R = cell(1,N);
      X = cell(1,N);
      for k = 1:N
          idx = ismember(b12{k},D1);
          idx = [false,idx(1:end-1)];
          if any(idx)

              R{k} = b12{k}(find(idx,1,'first'));
          end
          idx = [false,idx(1:end-1)];
          if any(idx)

              X{k} = b12{k}(find(idx,1,'first'));
          end
      end

      %Expected results:
      R={103,103,55,103,195}
      X={271,123,1,280,260,9,52,90,12,280,76,13,195,255,23}

Answer 1

[the cyclist]

For X, I think you want something like

X{k} = b12{k}(find(idx,1,'first')+2:end);

The following code vectorizes the whole thing using the cellfun function.

b12={[3,282,103,271,123,1],[3,282,103,280,260,9],[3,282,55,52,90,12],[3,282,103,280,76,13],[3,282,195,255,23]};

D1 = [282,55];

ND1 = numel(D1);

R = cell(1,ND1);
X = cell(1,ND1);

for nd = 1:ND1
    R{nd} = cellfun(@(x)x(find(x==D1(nd),1)+1),b12,'Uniform',false);
    X{nd} = cell2mat(cellfun(@(x)x(find(x==D1(nd),1)+2:end),b12,'Uniform',false));
end

[laith Farhan]

Dear Cyclist, Thanks for your reply but the code is not work.As I said I have many elements in vector D1. Is there anyway to modify my previous code?

[the cyclist]

Sorry, I misunderstood how you defined your index. My suggestion should be

X{k} = b12{k}(find(idx,1,'first'):end)

Your code is just grabbing one value. Mine is getting all the values until the end.

[the cyclist]

Also, it was not clear to me whether you wanted each value of R and X in its own element of the cell array, or all combined into one long vector. Your "expected result" shows them as one long vector, but your code has separate cells.

[laith Farhan]

Dear Cyclist,

This code "X{k} = b12{k}(find(idx,1,'first'):end)", totally works. I really appreciate it .

Many thanks