Add a zero element to the beginning of each existing cell array

I have C={[-8;-5],[1;-4;-5;-5;-5],[-3;-5]}

I want to get this result CC={[0;-8;-5],[0;1;-4;-5;-5;-5],[0;-3;-5]}

I use this code, but do not works c = cellfun(@(x)(x(0)==0), a, 'UniformOutput', false);

Answer 1

[Stephen23]

cellfun(@(v)[0;v],C,'uni',0)

And checking:

>> C = {[-8;-5],[1;-4;-5;-5;-5],[-3;-5]};
>> CC = cellfun(@(v)[0;v],C,'uni',0);
>> CC{:}
ans =
   0
  -8
  -5
ans =
   0
   1
  -4
  -5
  -5
  -5
ans =
   0
  -3
  -5

[NA]

I have index ={[2,3,4],[4,5,6,9,10,11],[4,7,9],[7,8]} and value as CC

CC={[0;-8;-5],[0;1;-4;-5;-5;-5],[0;-3;-5],[0;0.1]};

I also have this array

B=[0, -5.1, -12.2, -10.1, -9.1, -14.1, -13.1, -13.1, -15.1,-15.1,-15.1]

index_B=[1,2,3,4,5,6,7,8,9,10,11]

I want to write a code that compare CC to B. for this reason need to change scale of CC according to B

CC{1}(index2)=0 B(2)=-5.1 0-5.1=-5.1

CC{1}(index3)=-8 B(3)=-12.2 -8-5.1=-13.1

CC{1}(index4)=-5 B(4)=-10 -10-5.1=-15.1

5.1 changes for CC{2}

so my result should be

New_B=[0, -5.1, -13.1,-10.1,-9.1,-14.1,-13.1,0,-15.1,-15.1,-15.1]

[Stephen23]

@Naime Ahmadi: please explain these four lines, they are not clear to me:

CC{1}(index2)=0      B(2)=-5.1         0-5.1=-5.1
CC{1}(index3)=-8     B(3)=-12.2      -8-5.1=-13.1
CC{1}(index4)=-5     B(4)=-10         -10-5.1=-15.1
5.1 changes for CC{2}

You write that for the first line:

CC{1}(index2)=0      B(2)=-5.1    from 0 it changes to -5.1      0-5.1=-5.1

Then why is the second line not?:

CC{1}(index3)=-8     B(3)=-12.2   from -8 it changes to -20.2    -8-12.2=-20.2

What are index2, index3, and index4? You do not define them anywhere, nor do they seem to be required to get those values from CC.

EDIT: Your previous three comments seem to give different algorithms. It is even less clear which, if any, of these comments correctly clarifies what you are trying to do. Here you wrote

CC{2}(index11)=-5     B(11)=-15.1          -5-10.1=-15.1
%              -5           -15.1   <->    -5      -15.1

whereas here you wrote:

CC{1}(index4)=-5     B(4)=-10         -10-5.1=-15.1
%             -5          -10   <?>   ???     ?????
%                                      -5     -10   following above example.

[NA]

CC={[0;-8;-5],[0;1;-4;-5;-5;-5],[0;-3;-5],[0;0.1]};

when I put zero in the beginning of each array in CC, means that I put that measurement as base.

so I choose 2 to be my base (as [0;-8;-5] corespond to [2,3,4])

for [0;1;-4;-5;-5;-5] I choose 4 to be my base (as [0;1;-4;-5;-5;-5] corespond to [4,5,6,9,10,11]).

So I have to scale up all measuremnet in each array according the base of each array.

[Stephen23]

 index2 means first element of [2,3,4] means 2
 index3 means second element of [2,3,4] means 3
 index4 means third element of [2,3,4] means 4 

This was confusing because you wrote the same index in two different ways. It helps to be consistent, ratther than using different names for the same thing:

 CC{2}(index5)=1       B(5)=- -9.1                1-10.1=-9.1
 %     ^^^^^^     <->    ^

In any case, here is one way to get those values that you wanted:

>> tmp = cellfun(@(c,i)c+B(i(1)),CC,index,'uni',0);
>> idx = [index{:}];
>> newB = zeros(1,max(idx));
>> newB(idx) = vertcat(tmp{:})
newB =
    0.00000   -5.10000  -13.10000  -10.10000   -9.10000  -14.10000  -13.10000  -13.00000  -15.10000  -15.10000  -15.10000

[Stephen23]

Lets split this line:

tmp = cellfun(@(c,i)c+B(i(1)),CC,index,'uni',0);

into two lines:

fun = @(c,i)c+B(i(1));
tmp = cellfun(fun,CC,index,'uni',0);

where the first line defines an anonymous function:

https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html

For each of the corresponding cell contents in CC and index, that anonymous function will calculate

c+B(i(1))

where c and i are the function inputs, as provided by cellfun, i.e. are the cell contents of CC and index. So this is equivalent to:

CC{1}+B(index{1}(1))
CC{2}+B(index{2}(1))
CC{3}+B(index{3}(1))
... etc.

[NA]

Thank you so much.

Answer 2

[Image Analyst]

Granted, cellfun() is a bit cryptic, so if you want a simple, easy to understand intuitive method, just use for loops:

C = {[-8;-5],[1;-4;-5;-5;-5],[-3;-5]};
for col = 1 : size(C, 2)
	for row = 1 : size(C, 1)
		existingContents = C{row, col}; % Get existing vector
		C{row, col} = [0; existingContents]; % Prepend 0
	end
end
celldisp(C)

Sure, it's not as compact, and slightly slower (a millsecond or so for that array) but it's more intuitive.