Using accumarry and changing size of output

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susman
susman el 21 de Feb. de 2021
Comentada: susman el 22 de Feb. de 2021
I am using the accumarray by using the following command on the first and second column of struct array. The first part of the code works good.
but the last two commands "went_from_to_countAge" and "went_from_to_probAge", does not work. any idea?
sequence = [1 1 1 1 2 2 2 2 2 3;...
1 1 1 1 1 1 1 1 1 1;...
1 1 1 1 1 1 1 1 1 1;...
1 1 1 1 1 1 1 2 2 2;...
1 1 1 1 1 1 1 1 1 1;...
2 2 2 2 2 2 2 2 2 2; ...
2 2 2 2 2 2 2 3 3 3; ...
2 3 3 3 3 3 2 2 2 2; ...
2 2 2 2 2 2 4 4 4 4]
% first part of the code
k = 1:size(sequence)
Age (k) = 25 + k
States = unique([sequence(:,k); sequence(:,k+1)]); % Find the unique rows based on the data in the first two columns.
[TF, fromstateAge] = ismember(sequence(:,k), States); % logical arrays
[TF, tostateAge] = ismember(sequence(:,k+1), States);
% second part of the code
went_from_to_countAge(:,k) = accumarray( [fromstateAge(:), tostateAge(:)], 1, []) % frequency of each unique number
went_from_to_probAge(k)= went_from_to_countAge ./ sum(went_from_to_countAge); % pribability of each unique number
  1 comentario
dpb
dpb el 21 de Feb. de 2021
Attach a .mat file with a sample of the struct and an example of what the desired output should be.
Almost impossible to visualize such complexity without having an example at hand.

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Walter Roberson
Walter Roberson el 21 de Feb. de 2021
I am not clear if you are saying that you want different sizes to be returned? If so then you either need to use a cell array or else you have to pack the results into a larger array with padding.
If instead the point is that the size returned is sort of by chance but you want a fixed size, then change the place you used [] to give the required output size, such as
A = accumarray( [data(1).fromstate(:,2), data(1).tostate(:,3)], 1, [5 5]);
  7 comentarios
susman
susman el 22 de Feb. de 2021
and this remains as it is?
went_from_to_countAge(:,k) = accumarray( [fromstateAge(:), tostateAge(:)], 1, []) % frequency of each unique number
went_from_to_probAge(k)= went_from_to_countAge ./ sum(went_from_to_countAge); % pribability of each unique number
Sorry for asking these stupid questions, as I already spend enough time to figure it out.
susman
susman el 22 de Feb. de 2021
Thanks alot, This worked for me
A{i} = accumarray( [data(1).fromstate(:,i), data(1).tostate(:,i)], 1, [5 5]);

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