Modify elements in arrays stored in cells at specified rows/columns

Hi, I have a 2d cell array, where each cell contains a 1x2 double,

E.g. X = { [1,0], [1,0], [1,0], [1,0]; [1,0], [1,0], [1,0], [1,0] };

I want to assign the value 1 to the second element in each cell at row r=2, columns c=1:2. Because my actual data is large I want to do this without a for loop. In concept something like:

cellfun( @(c) (X{r,c}(2) = 1), 1:2 );

Thanks in advance

Answer 1

[Michael Van de Graaff]

I modified X to make it a bit clearer where each element is going, but I believe this does what you want:

X = { [1,0.1], [2,0.2], [3,0.3], [4,0.4]; [11,0.11], [12,0.12], [13,0.13], [14,0.14] };

szeX = size(X);
y = cell2mat(X);
y =reshape(y',[],2);
y(2,:) = 1;
y = reshape(y,[],2)';
szy = size(y);

cellsize = size(X{1});
d1 = cellsize(1)*ones(szeX(1),1);
d2 = cellsize(2)*ones(1,szeX(2));
newX = mat2cell(y,d1,d2)
cell2mat(newX)
cell2mat(X)

The last two lines just for comparison

[Michael Van de Graaff]

But I share your intuition that there must be a more elegant way of doing this

[Elissa]

Thank you, this did the job!

Answer 2

[Image Analyst]

Your cells do not contain arrays with two rows. Look

X = { [1,0], [1,0], [1,0], [1,0]; [1,0], [1,0], [1,0], [1,0] }

X = 2×4 cell array
    {[1 0]}    {[1 0]}    {[1 0]}    {[1 0]}
    {[1 0]}    {[1 0]}    {[1 0]}    {[1 0]}

Your cells contain a 1-row row vector with 2 columns.

But let's say you did have two ross in the arrays in the cells. I think the best way is to use a for loop:

X = { randi(9, 4, 3), randi(9, 4, 3), randi(9, 4, 3), randi(9, 4, 3);
    randi(9, 4, 3), randi(9, 4, 3), randi(9, 4, 3), randi(9, 4, 3)}

X = 2×4 cell array
    {4×3 double}    {4×3 double}    {4×3 double}    {4×3 double}
    {4×3 double}    {4×3 double}    {4×3 double}    {4×3 double}

% Show the starting values:
celldisp(X)

 
X{1,1} =
 
     7     2     6
     2     9     1
     5     5     3
     4     3     3

 
 
X{2,1} =
 
     6     4     1
     9     2     6
     9     1     5
     7     6     8

 
 
X{1,2} =
 
     6     6     9
     6     5     5
     2     8     1
     6     8     4

 
 
X{2,2} =
 
     7     9     5
     2     5     1
     1     4     1
     6     5     1

 
 
X{1,3} =
 
     5     9     9
     5     9     8
     1     9     5
     5     4     3

 
 
X{2,3} =
 
     3     6     7
     2     6     6
     9     9     2
     9     1     7

 
 
X{1,4} =
 
     7     1     6
     8     4     2
     8     3     8
     5     9     6

 
 
X{2,4} =
 
     7     9     7
     7     3     3
     7     6     8
     4     4     5

 

for k = 1 : numel(X)
    % Get cell contents.
    thisCellContents = X{k};
    % Change it.
    thisCellContents(2, 1:2) = [1, 2];
    % Put it back into the cell.
    X{k} = thisCellContents;
end
% See the result:
celldisp(X)

 
X{1,1} =
 
     7     2     6
     1     2     1
     5     5     3
     4     3     3

 
 
X{2,1} =
 
     6     4     1
     1     2     6
     9     1     5
     7     6     8

 
 
X{1,2} =
 
     6     6     9
     1     2     5
     2     8     1
     6     8     4

 
 
X{2,2} =
 
     7     9     5
     1     2     1
     1     4     1
     6     5     1

 
 
X{1,3} =
 
     5     9     9
     1     2     8
     1     9     5
     5     4     3

 
 
X{2,3} =
 
     3     6     7
     1     2     6
     9     9     2
     9     1     7

 
 
X{1,4} =
 
     7     1     6
     1     2     2
     8     3     8
     5     9     6

 
 
X{2,4} =
 
     7     9     7
     1     2     3
     7     6     8
     4     4     5

 

See? the first and second columns of the second row of each array get changed to 1 and 2 respectively.

Now I know you didn't want a for loop, probably for speed and efficiency, but let's face it - you threw away all possibility of getting speed and efficiency when you decided to use a cell array in the first place.

And it's a myth that cell arrays are horribly slow. Many years ago they've been speeded up a great deal. How many elements are in your cell array anyway? Hundred of millions or billions? For a few million I doubt you'd notice any difference. Any slowness would be caused not by the for loop but by your decision to use a cell array. Look, here is 10 million for loop iterations where it basically does nothing but iterate the for loop:

tic

for k = 1 : 1e7

;

end

toc

Elapsed time is 0.010463 seconds.

How many iterations would you do? See? The for loop alone is not the problem. Takes only a hundredth of a second. It's the data structure and what operations you do that is what eats up the time, not the for loop.