Loop in vector of cells

Hello,

I have a Vector A, containing 1X2 cell. in each cell there are 1X252 data.

I want to creat a Vector B, containing 1X2 cells, which in each cell I'll take the value of the "n+1" from Vector A and have it in cell "n" vector B for each cell. How can I do that?

[Adam Danz]

Could you provide a short example? What would be the outcome of this?

A{1,1} = [1,2,3,4,5]; 
A{1,2} = [11,22,33,44,55]; 

[Yossi]

B{1,1}=[2,3,4,5];
B{1,2}=[22,33,44,55];

[Adam Danz]

So you just want to remove the first digit of each vector? What does this mean below?

I'll take the value of the "n+1" from Vector A and have it in cell "n" vector B for each cell.

[Yossi]

I want to take the second digit of one vector in a loop and place it as a first digit in other Vector (like in the example) using loop.

[Adam Danz]

I don't see that pattern in your example. In your example, the first number of each vector is removed.

If you're taking the 2nd number of one vector and putting it in position 1 of the next vector, wouldn't that look like this:

A{1,1} = [1,2,3,4,5]; 
A{1,2} = [2,22,33,44,55]; 

[Yossi]

I want the code to start and take each argument in each cell and place it in the second matrix, so I can extend it and have another algorithm of NN inside.

[Adam Danz]

I'd like to help but I can't picture what you're describing.

What does it mean to "place [each argument] in the second matrix"? What does it mean to "extend" a matrix? What's "NN inside"?

Could you give a few examples? In the example you already gave, the only pattern I see is that you removed the first element of each vector.

[Yossi]

I thought of something like that (but it didn't work):

A={[1,2,3,4,5],[11,12,13,14,15]};
for iA=1:numel(A)        
    a=A{iA};
    for iB=2:numel(a)
        B{iA,iB-1}=A{iA,[1,iB]};
    end
end

[Adam Danz]

Given that input, what do you expect B to look like? This is what I need to see. I can only help if you provide some examples of inputs / outputs that show a clear pattern.

[Yossi]

Input:

A={[1,2,3,4,5],[11,12,13,14,15]};

Output:

B={[2,3,4,5],[12,13,14,15]};

[Adam Danz]

Again, you're just removing the first element of each vector and that's a lot simpler than what you're describing.

Are all of these correct (below)?

% #1
A={[9 8 7 6 5 4 3],[4 5 6 7 8 9 10]};
B={[  8 7 6 5 4 3],[  5 6 7 8 9 10]};

% #2
A={[22 44 77 92],[999 777 555 222]};
B={[   44 77 92],[    777 555 222]};

% #3
A={[0 0],[8 8]};
B={[  0],[  8]};

[Yossi]

yes, but the question is how to do that in loop

[Adam Danz]

Why do you need to use a loop? The answers below demonstrate how to produce "B" without a loop and these are more efficitent, cleaner, and easier to read than using a loop.

[Jan]

@Yossi: I have posted a loop method several hours ago already. Simply use it: It removes the first element from each vector, exactly as in all of your examples.

[madhan ravi]

@Yossi if you are too stubborn to not respond to any comments, next time don't expect help in this forum.

[Yossi]

@madhan The reason for not responding is a different time zone.

[Jan]

@Yossi: This is your question and nobody but you suffers, if it is not solved. Take your time.

The argument of different time zones is not needed, because you do not have to justify yourself. But if you mention it, remember that you have even posted some comments after I've posted a solution and comments - without taking notice of the given hints and solutions.

The description of what you want to do still confuses me, but the given examples look like all you want to do is to remove the first element.

[Yossi]

Assume that I want to take each cell and at matrix A and multiply it by number and add another scalar, how can I do that?

for example: assume that I want to take each cell and multiply by 2 and add 3 for a given Matrix and to delete the first column:

A = {[1,2,3,4,5], [11,22,33,44,55]};

The answer should be:

B={[7,9,11,13],[47,69,91,113]}

I thought doing that by loop. is there any other option?

[Jan]

This is another question. You can implement this as:

A = {[1,2,3,4,5], [11,22,33,44,55]};
B = cell(size(A));
for iA = 1:numel(A)
    a     = A{iA};
    B{iA} = a(2:end) * 2 + 3;
end

Or if you want to do this with a slower loop for any good reason (I do not know any):

B = cell(size(A));
for iA = 1:numel(A)
    a = A{iA};
    b = zeros(size(a));  % Important
    for ib = 1:numel(a) - 1
        b(ib) = a(ib + 1) * 2 + 3;
    end
    B{iA} = b;
end

Matlab is designed to operate on vectors and matrices efficiently. Such loops are rarely needed and the "vectorized" approach is nicer and faster. Matlab is not C, and it is worth to use its power.

Or with a cellfun call without a loop:

B = cellfun(@(x) x(2:end)*2+3, A, 'UniformOutput', false)

cellfun looks nice and compact, but usually the first method with the loop over the cell elements is faster.

These 3 methods are very similar to the already posted solutions of the original question. All I've added is "*2 + 3" . I assume you are able to implement this transfere by your own.

[Yossi]

Thank you very much, that works!

Can I use this function to multiple it instead of "2" by array C={[2,3}] .

so that the first row will multiply by "2" andf the second row will multiply by "3"?

[Adam Danz]

Just FYI, I've moved on to other questions in the forum and consider this one answered.

[Jan]

@Yossi: The thread is easier to read without trailing blank lines under a message. I'm removing them in all of your messages.

If the provided code works, please choose the answer, which helped, as "accepted".

{[2,3}] is no valid Matlab syntax, therefore I cannot guess, what it means. Use either curly braces {} or square brackets [].

The input data in your example are A = {[1,2,3,4,5], [11,22,33,44,55]} . So I cannot guess what "the first" and "the second" row means. Of course you can muliply each vector with another number:

A = {[1,2,3,4,5], [11,22,33,44,55]};
B = cell(size(A));
mult = [2, 3];
for iA = 1:numel(A)
    a     = A{iA};
    B{iA} = a(2:end) * mult(iA) + 3;  % Or whatever
end

If you do not have any experiences with programming, the forum is not the right place to learn the fundamental basics. Please read the "Getting started" chapters and Matlab's onramp (ask Google for these terms).

You are a member of this forum for over 2 years now. Actually I assume, that you know the basics already.

[Yossi]

I meant a cell 1X2 i.e C={2,3} .

I want to multiply 2 in each cells at the first row: [1,2,3,4,5] and mutiply 3 with the second row [11,22,33,44,55].

[Jan]

@Yossi: I have fixed your message by removing the trailing blank lines to increase the readability. I've explained this to you and it seems like you do not want to care about this advice.

If you want mult to be a cell, simply define it as cell and use the curly braces:

A = {[1,2,3,4,5], [11,22,33,44,55]};
B = cell(size(A));
mult = {2, 3};
%      ^    ^
for iA = 1:numel(A)
    B{iA} = A{iA}(2:end) * mult{iA};
    %                      ^  ^
end

Again, a cellfun appraoch is still possible:

A = {[1,2,3,4,5], [11,22,33,44,55]};
mult = {2, 3};
B = cellfun(@(x,y) x(2:end) * y, A, mult, 'Uniformoutput', false);

I consider this question as solved, although you do not want to mark an accepted answer.

Answer 1

[Jan]

Maybe:

A = {[1,2,3,4,5], [11,22,33,44,55]};
B = cellfun(@(x) x(2:end), A, 'UniformOutput', false)

Or with a loop:

B = cell(size(A));
for iA = 1:numel(A)
    a     = A{iA};
    B{iB} = a(2:end); 
end

[Yossi]

I want the code to start and take each argument in each cell and place it in the second matrix, so I can extend it and have another algorithm of NN inside. Thus, i need another loop rather than:

B{iB} = a(2:end); 

[madhan ravi]

B = cellfun(@(x) x(2:end), A,'un',0)

[Jan]

@Yossi: I cannot follow you. Maybe you mean:

B = cell(size(A));
for iA = 1:numel(A)
    a = A{iA};
    b = NaN(1, numel(a) - 1);  % Pre-allocate
    for ib = 1:numel(a) - 1
        b(ib) = a(ib + 1);
    end
    B{iB} = b; 
end

But this is very inefficient and unmatlabish.

[Yossi]

Thank you Jan,

I know its ineffective, but as I told, I need it for cell manipulation in order to use it for NN.

[Jan]

@Yossi: So did this loop at least work? Then please mention this and mark a helpful answer as "accepted", such that the readers know that the problem is solved.

I cannot know what "use it for NN" means, but I really assume, that what ever it is, it will work with efficient methods also.

Answer 2

[Adam Danz]

Based on several examples provided in comments under the question, it appears that the goal is to remove the first element of each vector stored in a cell array. If that is the case, loops are not necessary.

A={[1,2,3,4,5],[11,12,13,14,15]};   %desired outout: B={[2,3,4,5],[12,13,14,15]};
B = cellfun(@(x)x(2:end), A, 'UniformOutput', false); 
celldisp(B)
B{1} =
     2     3     4     5
B{2} =
    12    13    14    15

You mentioned you'd like to "take each argument in each cell and place it in the second matrix". I interpret that as putting the elements of the cell array into a matrix.

% This puts the elements of B into a single row vector. 
Bv = [B{:}];
%Bv =
%     2     3     4     5    12    13    14    15

% This puts the elements of B into a matrix but will only work if
% each element of B is the same size. 
Bm = cell2mat(B');
%Bm =
%     2     3     4     5
%    12    13    14    15

Answer 3

[Yossi]

I was managed to solve it:

A={[1,2,3,4,5],[11,12,13,14,15]};
for iA=1:numel(A)        
    a=A{iA};
    for iB=2:numel(a)      
        C(1,iB-1)=a(iB);
    end
    B(1,iA)={C};
end

[Adam Danz]

That produces the same results as the solutions proposed here a while ago. If you don't need to use the for-loop, you should just do this in 1 line of code:

B = cellfun(@(x)x(2:end), A, 'UniformOutput', false); 

[Jan]

@Yossi: This will fail, if a vector is shorter than another one before, because then the former elements of C and not overwritten. As said already, your code does exactly the same as e.g. my answer, which has been given 7 hours ago:

    a=A{iA};
    for iB=2:numel(a)      
        C(1,iB-1)=a(iB);
    end
    % is the same as:
    a = A{iA}
    C = a(2:end);
    % except that your loop does not crop a formerly existing C