How do I reshape a large matrix?
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I am trying to reshape a matrix so that it keeps a consistent amount of columns (8 columns) and the number of rows is ever changing. These rows change because of modifications to number of iterations of a for loop. I know that the correct syntax to use is the reshape function but is it possible to not explicitly state the number of rows and/or columns so that the matrix will automatically reshape to 9 columns and "X" rows? Thanks!
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James Clinton
2018-7-12
I believe you can still use the reshape function for what you are describing. The documentation shows you can input an empty matrix ("[]") to the function so that it will automatically reshape the matrix given a value for the other dimensions of the matrix. Reshaping the 2D matrix "A" into 9 columns would be accomplished by the following:
B = reshape(A,[],9);
nrows = size(B,1);
Where nrows will give you the number of rows the matrix was reshaped into.
Of course this only works when the number of elements is divisible by 9, however. If you'd like to reshape when the number of elements is not divisible by 9 you'll need to pad the matrix with extra numbers. I would do this in the following way (using trailing zeros to pad the matrix):
B = reshape(A,1,[]); % convert A to vector stored in column major format (transpose for row major)
nzeros = rem(size(B,1),ncols); % find the number of zeros needed to pad on the end
C = [B zeros(1,nzeros)]; % add the padding zeros
D = reshape(C,[],ncols); % reshape to have ncols columns
2 个评论
Jewel
2018-7-12
Walter Roberson
2018-7-12
In another recent question I recently showed this for someone who needed a multiple of 256 rows:
pad_needed = 256 - (mod(size(points,1) - 1, 256) + 1);
if pad_needed > 0
points(end+pad_needed,:) = 0;
end
What I recommend, though, is that if you have the Communications System Toolbox, that you use buffer()
更多回答（1 个）
B=reshape(A,[],9);
The kicker is that mod(numel(A),9)==0 or will fail or will have to pad/truncate A.
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