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Find the exact multiple of a number by zero padding values.

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CalebJones
CalebJones 2020 年 2 月 5 日
コメント済み: CalebJones 2020 年 2 月 5 日
Say
A = 13848 x 1
and I want to reshape into a 173(matters a lot) by 80 or 81(this doesn't matter) but that isn't possible as 13848 isn't divible by 173.
So I want a way where it reads A and calculates the next perfect value which is 14013 by zero padding A.
I could just remove 8 from A but I don't loose data but I don't mind zero padding to A towards the end.
  1 件のコメント
CalebJones
CalebJones 2020 年 2 月 5 日
編集済み: CalebJones 2020 年 2 月 5 日
I could just manually zero pad A, but I want it to be as automated as possible.

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Stephen23
Stephen23 2020 年 2 月 5 日
編集済み: Stephen23 2020 年 2 月 5 日
Method One: create the output matrix first, then use indexing to add the values:
R = 173;
B = zeros(R,ceil(numel(A)/R));
B(1:numel(A)) = A(:);
Method Two: pad some extra zeros and then remove the superfluous elements:
R = 173;
X = 1:R*fix((1+numel(A))/R)
B = [A;zeros(R,1)];
B = reshape(B(X),R,[]);
  4 件のコメント
Stephen23
Stephen23 2020 年 2 月 5 日
編集済み: Stephen23 2020 年 2 月 5 日
Why did you convert this index into a table?:
X = array2table(1:R*fix((1+A)/R))
X is supposed to be a numeric index vector (just like in my answer). Converting it into a table is unlikely to work, and just makes the code pointlessly complex.
Apparently your data are in a table, something that you omitted to mention in your question. I doubt that you have really considered what it means to reshape a table: a table is a container class which contains multiple different variables, it is not a simple array class that can be reshaped, as the variables are not one contiguous array stored in memory. In any case, to reshape you should probably get the appropriate array out of the table first:
CalebJones
CalebJones 2020 年 2 月 5 日
A = size(right_pc,1);
R = 173;
X = 1:R*fix(1+A/R);
B = [right_pc.HbO ; zeros(R,1)];
B = reshape(B(X),R,[]);
Ah yes, SWEEET!
Fixed !
Thanks bud.

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その他の回答 (1 件)

KSSV
KSSV 2020 年 2 月 5 日
編集済み: KSSV 2020 年 2 月 5 日
A = rand(13848,1) ;
n = length(A) ; % length of A
r = 80 ; % to reshape by this dimension
N = mod(-mod(n,r),r); % get the number of zeros to be appended
A = [A ; zeros(N,1)] ; % append zeros
iwant = reshape(A,r,[]) ; % reshape
  2 件のコメント
Stephen23
Stephen23 2020 年 2 月 5 日
CalebJones's "Answer" moved here:
getting an error.
n = size(right_pc,1) ; % length of A
r = 80; % to reshape by this dimension
N = mod(-mod(n,r),r); % get the number of zeros to be appended
A = [right_pc(:,1) ; zeros(N,1)] ; % append zeros
iwant = reshape(A,r,[]) ; % reshape
when i tried to run...this
N = mod(-mod(n,r),r);
Index in position 1 exceeds array bounds (must not exceed 173).
I have atteched a mat file as well.
Stephen23
Stephen23 2020 年 2 月 5 日
You have created a variable named mod. Clear that variable from the workspace.

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