How do I edit certain cells of a cell-array with a logical index (without using for-loops)

Hi everyone,

lets say I have a cell array and a numeric array like:

Old_Array = {20 , '2' ; 21 , '2' ; 22 , '2,3' ; 25 , '3' ; 28 , '2'};

and

new_Numbers = [18 ; 21 ; 22 ; 32];

What I want, is to some kind of mix those two. If a number of new_Numbers matches with the first column of old_Array, i want to add '4' to the cell in the 2nd column on just that row. If a number of new_Numbers is not matched, i want to add it, with the 2nd column containing the'4' String. The result should look like

New_Array = {18 , '4' ; 20 , '2' ; 21 , '2,4' ; 22 , '2,3,4' ; 25 , '3' ; 28 , '2' ; 32 , '4'};

I wonder if there is a solution not using for loops but [i,j]=ismember(new_Numbers,cell2mat(Old_Array(:,1))) and the logical indexes.

Thanks in advance. Any Help is appreciated! :)

[Mr Anderson]

the for loop is indeed not that big, but perhaps anyone has a "logical-only" solution.

in the meantime if anyone else comes across this, this for-loop does the trick for me:

rs='4'
for n=1:length(new_Numbers)
    [i,j] = ismember(new_Numbers(n),cell2mat(Old_Array(:,1)));
    if i
        Old_Array{j,2} = [Old_Array{j,2} ',' num2str(rs)];
    else
        pos_new_entry = length(Old_Array)+1;
        Old_Array{pos_new_entry,1} = new_Numbers(n);
        Old_Array{pos_new_entry,2} = num2str(rs);
    end;
end;

Answer 1

[the cyclist]

This is fairly inelegant, but it is a vectorized solution:

Old_Array = {20 , '2' ;
             21 , '2' ;
             22 , '2,3' ;
             25 , '3' ;
             28 , '2'};

new_Numbers = [18 ; 21 ; 22 ; 32];

% Find which numbers are really new
[tr,loc] = ismember(new_Numbers,cell2mat(Old_Array(:,1)));

% Initialize new array as a copy
New_Array = Old_Array;

% Append ',4' to rows that match
New_Array(loc(tr),2) = cellfun(@(x)[x,',4'],New_Array(loc(tr),2),'UniformOutput',false)

% Add new numbers that don't appear in old
Number_New_Numbers = sum(not(tr));
New_Block = [num2cell(new_Numbers(not(tr))),repmat({'4'},Number_New_Numbers,1)]
New_Array = [New_Array; New_Block]

% Sort the new array
[~,sortingIndex] = sort([New_Array{:,1}]')
New_Array = New_Array(sortingIndex,:)

[Mr Anderson]

that works perfectly! thank you so much! :)

I was suspecting something like

New_Array(loc(tr),2) = cellfun(@(x)[x,',4'],New_Array(loc(tr),2),'UniformOutput',false)

But i couldnt figure it out myself, because i cant handle cellfun pretty good, even worse with "user-defined" functions :(

just for me to learn what your code is doing:

@(x) is reading the value of New_Array(loc(tr),2) , [x,',4'] is editing it and like usual New_Array(loc(tr),2) assigns it back to its position. is my explanation right, or did i get that code wrong? :S Thanks again for your help and your time!

[the cyclist]

I think you have it basically correct, but I would explain it like this.

@(x)[x,',4'] should be considered as one functional unit. It is the definition of an anonymous function.

I could have defined it separately, like this:

append_4_func = @(x)[x,',4'];
New_Array(loc(tr),2) = cellfun(append_4_func,New_Array(loc(tr),2),'UniformOutput',false)

What cellfun does, in general, is apply a function to each element of the input cell array, and output the result. In this particular case, that function is "append ',4' to a string".

Then, as you say, then output of cellfun is assigned back into New_Array again.

Finally, I would just mention that I have some doubts that this will be faster than a for loop. You might want to do some testing. The vectorized code is pretty obscure, and future-you might find it confusing.

The best form of "thanks" is upvoting and/or accepting a solution that you found useful. That rewards the contributor, and may point future users to helpful solutions.

[Mr Anderson]

Those anonymous functions have been a mistery to me since beginning with matlab and cell-arrays. Now thanks to your help I kinda found out how they work, and even though I might use the for loop on this task, I already can use the new knowledge for other parts of my work.

Again, thank you so much and have great christmas holidays! :)