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Convert 2-D array to multidimensional array

3 vues (au cours des 30 derniers jours)
Jim
Jim le 6 Avr 2016
Commenté : Guillaume le 6 Avr 2016
I have a 2-D array with 7 columns that I want to convert to a 6-D array of dimensions 16*20*22*6*3*2. The values in the first 5 columns of the 2-D array are the first 5 indices of the 6-D array and the values in the 6th and 7th columns of the 2-D array shall be contained in the 6th dimension of the 6-D array.
I tried the following
misc(rslts1(:,1:5),1:2) = rslts1(:,6:7)
where misc is the 6-D array and rslts1 is the 2-D array but got the error "Subscripted assignment dimension mismatch"

Réponse acceptée

Guillaume
Guillaume le 6 Avr 2016
Modifié(e) : Guillaume le 6 Avr 2016
Using accumarray, you would have to duplicate your columns 1 to 5, once with 1 appended, then with 2 appended, to create the destination subscripts, and reshape the last two columns into one column:
misc = accumarray([rslts1(:, 1:5), 1; rslts1(:, 1:5), 2], reshape(rslts1(:, 6:7), [], 1));
Alternatively, you can use your syntax twice, and concatenate along the 6th dimension:
misc = cat(6, accumarray(rslts1(:, 1:5), rslts1(:, 6)), accumarray(rslts1(:, 1:5), rslts1(:, 7)));
Using sub2ind, while technically more correct, is more complicated:
misc = zeros(16, 20, 22, 6, 3, 2);
idxs = [repmat(rslts1(:, 1:5), 2, 1), repelem([1; 2], size(rslts1, 1))]; %another way to create the same matrix used in accumarray
misc(sub2ind(size(misc), idxs(:, 1), idxs(:, 2), idxs(:, 3), idxs(:, 4), idxs(:, 5), idxs(:, 6)) = rslts1(:, 6:7)
or to avoid all of these arguments in the sub2ind call (but the conversion to cell array will make the execution slower):
misc = zeros(16, 20, 22, 6, 3, 2);
idxs = num2cell([repmat(rslts1(:, 1:5), 2, 1), repelem([1; 2], size(rslts1, 1))], 1);
misc(sub2ind(size(misc), idxs{:})) = rslts1(:, 6:7)
edit: I also suggested using sparse, but that only works for 2D matrices.
  2 commentaires
Jim
Jim le 6 Avr 2016
I get an error with the first example |??? Error using ==> horzcat CAT arguments dimensions are not consistent.| and I'm trying to understand the source of it. Seems that there should be another grouping of terms in the generation of the destination subscripts.
Guillaume
Guillaume le 6 Avr 2016
Do'h! Of course, trying to horzcat a column vector with a matrix is not going to work.
You can either use the construct I used in the sub2ind example:
[repmat(rslts1(:, 1:5), 2, 1); repelem([1; 2], size(rslts1, 1))];
or
[rslts1(:, 1:5), repmat(1, size(rslts1, 1), 1); rslts1(:, 1:5), repmat(2, size(rslts1, 1), 1)]
or
[[rslts1(:, 1:5); rslts1(:, 1:5)], repelem([1; 2], size(rslts1, 1))]

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Plus de réponses (2)

Steven Lord
Steven Lord le 6 Avr 2016
Either use accumarray or sub2ind depending whether there are repeated values for the same coordinate that you want to accumulate or not.
  2 commentaires
Jim
Jim le 6 Avr 2016
I don't understand how to use sub2ind for this. I don't see how to make use of the linear index that it would provide.
Jim
Jim le 6 Avr 2016
Modifié(e) : Jim le 6 Avr 2016
I understand how I can use accumarray a little better. I'm thinking of using it something like
misc = accumarray(rslts1(:,1:5), rslts1(:,6));
but I think that's not quite right. This will be assigning the values in | rslts1(:,6)| to the 5th dimension of misc instead of the 1st column of the 6th dimension.

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Kuifeng
Kuifeng le 6 Avr 2016
How about the function reshape? Take one column for example, Assume there is 126720 data. Make some changes for more columns.
a = rand([126720 1]);
a_5D = reshape(a,[16 20 22 6 3])
  1 commentaire
Guillaume
Guillaume le 6 Avr 2016
Modifié(e) : Guillaume le 6 Avr 2016
The only way that reshape would work is in the unlikely case that the indices in the first five columns happened to be ordered linearly with no gaps and no repetition. In that case, there actually would be no need for the 5 index columns.

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