Reshaping cell array having element with different sizes

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I have a variable named A which is a 5000×4 cell array containing M×2 matrices. M varies for each element. 5000 is the number of different measurement for 4 profiles. The following shows only two out of 5000 measurements. First row shows measurement#1 and the second row is measurement#2:
[3413x2 double] [3082x2 double] [3186x2 double] [3143x2 double]
[3344x2 double] [3044x2 double] [3200x2 double] [3143x2 double]
I am trying to have all the profile for each measurement in one matrix of size N×2, named B. for example, for the first measurement N=3413+3082+318+3143=12824. So, B is vertcat of the first row of A, it means B is constructed by putting each profile under the previous profile. finally B{i} will be all the profiles for ith measurement.
Could you help me with that?

Risposta accettata

Andrei Bobrov
Andrei Bobrov il 15 Giu 2011
A1 = A';
B = arrayfun(@(x)cat(1,A1{:,x}),1:size(A1,2),'un',0)
  3 Commenti
AP
AP il 15 Giu 2011
The result is the following:
A1 =
[3413x2 double] [3344x2 double]
[3082x2 double] [3044x2 double]
[3186x2 double] [3200x2 double]
[3143x2 double] [3143x2 double]
B =
[12824x2 double] [12731x2 double]
Andrei Bobrov
Andrei Bobrov il 16 Giu 2011
Hi! Having seen Your solution to <http://www.mathworks.com/matlabcentral/answers/9600-subtracting-a-constant-from-all-the-array-elements-in-a-structure>, think that the explanation is not needed?

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Più risposte (1)

Walter Roberson
Walter Roberson il 15 Giu 2011
I think this might work:
B = arrayfun(@(idx) [A{idx,:}], 1:size(A,1), 'un', 0);
  3 Commenti
AP
AP il 15 Giu 2011
Another question that I have is that why did you use arrayfun while A is a cell array? Can't cellfun do the same?
Walter Roberson
Walter Roberson il 15 Giu 2011
Ah, try
B = arrayfun(@(idx) vertcat(A{idx,:}), 1:size(A,1), 'un', 0);
You use arrayfun instead of cellfun because cellfun processes each cell individually whereas you want to process groups of cells.

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