Convert cell-array to array-of-cells, for array-of-struct creation?

Hi, how can I convert a cell-array to an array-of-cells?

I have (following a file textscan) a cell array of format:

    Fr = [{uint8([1;2;3])}, {uint32([4;5;6])}, {[7;8;9]}]

And need to get to this array of cells:

    To = [{Fr{1,1}(1), Fr{1,2}(1), Fr{1,3}(1)} ;...
          {Fr{1,1}(2), Fr{1,2}(2), Fr{1,3}(2)} ;...
          {Fr{1,1}(3), Fr{1,2}(3), Fr{1,3}(3)}]

..without having to loop (Fr can be large) through Fr.

The reason is that I want an array-of-struct, rather than struct-of-array. Reference:

Incorrect  = cell2struct(Fr,{'f1','f2','f3'},2)
Correct    = cell2struct(To,{'f1','f2','f3'},2)

Answer 1

[Andrei Bobrov]

Fr = [{uint8([1;2;3])}, {uint32([4;5;6])},...
     {[7;8;9]}, {uint8([11,12;13,14;15,16])}];

To = cellfun(@(x)num2cell(x,2),Fr,'un',0);
out = [To{:}];

[Azzi Abdelmalek]

You mean

   To = cellfun(@(x)num2cell(x),Fr,'un',0);

instead of num2cell(x,2)

[Cedric]

No actually, NUM2CELL can take a second "dim" argument. It's neat, I didn't realize!

[Azzi Abdelmalek]

For this case num2cell(x,2) doesn't give the expected result

[Andrei Bobrov]

Hi Azzi! Please see solution on my laptop (R2014a).

[Azzi Abdelmalek]

Sorry Andrei, I didn't read completly the explanation given by Bjoern. There is no doubt about your result, I misunderstood the new question.

[Bjoern]

Excellent, thanks a lot! Just FYI for other users:

This proposal is about 60% quicker than Cedric's proposal. Both produced the same result :)

Answer 2

[Cedric]

 To = cellfun( @(cc)num2cell(cc), Fr, 'UniformOutput', false ) ;
 To = [To{:}] ;

[Bjoern]

Wow, very nice and really fast!

I realize missed one important factor in my original question, maybe you you would know how to handle that as well?

There's actually also an array within Fr:

 Fr = [{uint8([1;2;3])}, {uint32([4;5;6])},...
       {[7;8;9]}, {uint8([11,12;13,14;15,16])}];

The problem is that the array becomes "multiple columns".

[Cedric]

Is it the final structure (2 columns in the last array) or do you have more arrays and columns? Note that my solution treats the last array properly according to your definition of To in the statement (with no internal structure). So what output do you expect/need with this extra array?

 >> To
 To = 
    [1]    [4]    [7]    [11]    [12]
    [2]    [5]    [8]    [13]    [14]
    [3]    [6]    [9]    [15]    [16]

 >> for cId = 1 : size( To, 2), disp( class( To{1,cId} )) ;  end
 uint8
 uint32
 double
 uint8
 uint8

[Bjoern]

You're right.

And I should have supplied a clearer response :) Here:

Fr = [{uint8([1;2;3])}, {uint32([4;5;6])},...
     {[7;8;9]}, {uint8([11,12;13,14;15,16])}]
Fr = 

    [3x1 uint8]    [3x1 uint32]    [3x1 double]    [3x2 uint8]

To = [{Fr{1,1}(1), Fr{1,2}(1), Fr{1,3}(1), Fr{1,4}(1,:) } ;...
      {Fr{1,1}(2), Fr{1,2}(2), Fr{1,3}(2), Fr{1,4}(2,:) } ;...
      {Fr{1,1}(3), Fr{1,2}(3), Fr{1,3}(3), Fr{1,4}(3,:) }]

To = 

      [1]    [4]    [7]    [1x2 uint8]
      [2]    [5]    [8]    [1x2 uint8]
      [3]    [6]    [9]    [1x2 uint8]

[Cedric]

Then here is a solution

 To = cellfun( @(cc) mat2cell( cc, ones( size(cc, 1), 1 )), Fr, ...
               'UniformOutput', false ) ;
 To = [To{:}] ;

[Cedric]

See Andrei's answer, NUM2CELL can take a second "dim" argument and I didn't know that!

[Bjoern]

Alright, thanks!!

Answer 3

[Azzi Abdelmalek]

To=num2cell([Fr{:}])

[Bjoern]

That almost works but it casts all variables to the same type.. In this case they're all uint8 following that line..

[Azzi Abdelmalek]

Fr = [{uint8([1;2;3])}, {uint32([4;5;6])}, {[7;8;9]}]
n=numel(Fr{1})
m=numel(Fr)
[jj,ii]=meshgrid(1:n,1:m)
out=arrayfun(@(x,y) Fr{x}(y),ii,jj,'un',0)'

[Bjoern]

Thanks once again! Please see the comment (for Cedric) on me missing to mention an important fact. It looks like your proposal removes the last column.

[Azzi Abdelmalek]

Fr = [{uint8([1;2;3])}, {uint32([4;5;6])},...
     {[7;8;9]}, {uint8([11,12;13,14;15,16])}]
mm=cellfun(@(x) size(x,2),Fr) 
nn=1:numel(mm)
r=size(Fr{1},1)
dd=cell2mat(arrayfun(@(x,y) [1:x;y*ones(1,x)],mm,nn,'un',0))
[i1,j1]=meshgrid(dd(1,:),1:r)
[i2,j2]=meshgrid(dd(2,:),1:r)
out=arrayfun(@(x,y,z) Fr{x}(y,z),i2,j1,i1,'un',0)