why dosen't this work ? i need a (:,4) cell. but 90 % of my result is empty spaces.

PLS = [PIDSorted; NameSorted; ExamSorted; DateSorted];

NewPL = cell(size(PLS,1),4); %Patients sorted 

for n=1:size(PLS,1);

      if size(PLS{n,:},1)==1 ;
          NewPL{1,n} = PIDSorted(n,:);
          NewPL{2,n}= NameSorted{n};
          NewPL{3,n}=ExamSorted{n};
          NewPL{4,n}=DateSorted{n};
       end           
  end

I tried to use NewPL = NewPL(~cellfun('isempty',NewPL)) but then I have my result in only one column

Answer 1

[Guillaume]

There's not much in your code that makes sense.

First issue:

if size(PLS{n, :}, 1) == 1

PLS{n, :} implies that PLS has more than one column (since the column says "return all columns"), yet if the PLS cell array has more than one column, size will error (invalid number of arguments). So, we're assuming that PLS is just one column, in which case

if size(PLS{n}, 1) == 1

would be a lot better.

Second issue:

NewPL = cell(size(PLS,1),4)
for n=1:size(PLS,1)
    if  ...
       NewPL{1,n} = ...
       NewPL{2,n} = ...
       NewPL{3,n} = ...
       NewPL{4,n} = ...

You create a h x 4 cell array, yet your code in the if block fills a 4 x h cell array (with h being size(PLS, 1)). This would cause an 'index exceeds matrix dimension' error as soon as n is greater than 4. If you don't get an error that means the if block is never entered, which must mean that the matrices in PLS have more than one row.

Third issue:

PLS = [PIDSorted; NameSorted; ExamSorted; DateSorted];
for n=1:size(PLS,1)
    if  ...
       ... = PIDSorted(n,:);
       ... = NameSorted{n};
       ... = ExamSorted{n};
       ... = DateSorted{n};

n goes from 1 to the number of rows in PLS. The number of rows in PLS is the sum of the number of rows in your 4 variables. So, unless the variables are empty, n is guaranteed at some point to be more than the number of rows in any of these variable. So clearly, if the if block gets entered, var{n} is going to error with 'index exceeds matrix dimension' again.

I'm guessing that you're trying to concatenate a PIDSorted matrix with n rows with cell arrays with also n rows. In that case, all that is needed to create your cell array is:

NewPL = [num2cell(PIDSorted, 2), NameSorted, ExamSorted, DateSorted]

[Guillaume]

Anne's comment copied here from Answer:

Thanks. I kind of need

PID1 PID2 PID3 PID4

Name1 Name2 Name3 Name4

Exam1 Exam2 Exam3 Exam4

Date1 Date2 Date3 Date4

PID5 PID6 PID7 PID8

Name5 Name6 Name7 Name8

Exam5 Exam6 Exam37 Exam8

Date5 Date6 Date7 Date8

Is that possible?

[Guillaume]

It probably is possible, but to answer that accurately, I need to know the output of:

size(PIDSorted)
size(NameSorted)
size(ExamSorted)
size(DateSorted)

[Guillaume]

Anne's comment copied here from Answer:

size(PIDSorted) = 147 10

size(NameSorted) = 147 1

size(ExamSorted) = 147 1

size(DateSorted) = 147 1

but 147 is the number of patients which can vary day to day

[Guillaume]

Pleas use comment on this Answer to reply.

Isn't

NewPL = [num2cell(PIDSorted, 2), NameSorted, ExamSorted, DateSorted]'

what you're after?

I'm confused by what you're asking. If you have 147 elements that you reshape into 4 columns:

PID1    PID2    PID3    PID4
Name1   Name2   Name3   Name4
...
PID145  PID146  PID147  ???
Name145 Name146 Name147 ???

What should go in ???, empty cells?

[Anne]

I Wish to get this

[Anne]

it all come from a patientlist (one column) that was sorted by name but had to be sorted by exam instead. Now it have to fit an A4 sheet with 4 labels in a row. One patient pr lable; including information PID,name, exam and date. maybe i make more sense now ?

[Guillaume]

Ok. But you haven't said what should go in the last row when the number of labels is not a multiple of 4. I assume just blanks:

NewPL = reshape([num2cell(PIDSorted, 2), NameSorted, ExamSorted, DateSorted; cell(mod(-numel(NameSorted), 4), 4)]', [], 4);

I'm not sure why PIDSorted has 10 columns, so not entirely sure that num2cell is what is required.

[Anne]

thanks!!

but it makes an Error: Unbalanced or unexpected parenthesis or bracket.

then i tried to fix it but then:

Dimensions of matrices being concatenated are not consistent.

[Guillaume]

For some reason, it was missing the mod call. Now fixed:

NewPL = reshape([num2cell(PIDSorted, 2), NameSorted, ExamSorted, DateSorted; cell(mod(-numel(NameSorted), 4), 4)]', [], 4);

[Anne]

AMAZING!!!!!!