How can I subtract columns for each row by using a for loop

조회 수: 7 (최근 30일)
Jimmy
Jimmy 2014년 9월 29일
편집: Andrei Bobrov 2014년 9월 30일
Hi,
I have a matrix like this:
[1.011 1.004 1.054
1.008 0.998 1.042
0.984 0.988 1.024
1.026 1.006 1.016
1.000 0.996 0.977]
I would like to subtract each column for each row and store these results in a new matrix. How can I do this?
Thanks!
  댓글 수: 2
Joseph Cheng
Joseph Cheng 2014년 9월 29일
can you expand on what you mean by subtract each column for each row? I do not understand what you're subtracting with.
Jimmy
Jimmy 2014년 9월 29일
Ok.
In this example I use 3 columns. I want to subtract the column for each possible combinations for each row. So lets say for row 1, I want subtract col 1 - 2, col 1 - 3, col 2 - 3 and so on. I want for row 2 the same subtractions.

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채택된 답변

dpb
dpb 2014년 9월 29일
편집: dpb 2014년 9월 29일
nk=nchoosek(1:size(x,2),2);
dx=zeros(size(x,1),size(nk,1));
for i=1:size(nk,1)
dx=x(:,nk(i,2))-x(:,nk(i,1));
end

추가 답변 (3개)

Joseph Cheng
Joseph Cheng 2014년 9월 29일
편집: Joseph Cheng 2014년 9월 29일
you can use combnk() or nchoosek to determine the combination of column subtraction and perform a for loop for each combination.
X = randi(10,4,3);
combin = combnk(1:size(X,2),2);
for ind = 1:size(X,2)
newX(:,ind) = X(:,combin(ind,1))-X(:,combin(ind,2));
end

Guillaume
Guillaume 2014년 9월 29일
Use nchoosek to get all possible combinations of columns, and use that to calculate your differences:
m = [1.011 1.004 1.054
1.008 0.998 1.042
0.984 0.988 1.024
1.026 1.006 1.016
1.000 0.996 0.977];
colcomb = nchoosek(1:size(m, 2), 2);
coldiff = zeros(size(m, 1), size(colcomb, 1));
for comb = 1:size(colcomb, 1)
coldiff(:, comb) = diff(m(:, colcomb(comb, :)), 1, 2);
end
  댓글 수: 7
dpb
dpb 2014년 9월 29일
For what working definition of small? But, basic idea is one of two choices...
a) go ahead and generate all pairs and then compute the comparison statistic and choose the N smallest of those, or,
b) compute each pair and the statistic at same time; after N replace the largest of those kept with the last if new is less; update the maxValue comparison value.
Jimmy
Jimmy 2014년 9월 30일
I choose for option a. Since the 10 smallest pairs are different for each row (the rows are basically the time line (t).) In the end I want to trade each pair for 125 days on a rolling window. For that reason I want to know which pairs I am trading.

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Andrei Bobrov
Andrei Bobrov 2014년 9월 30일
편집: Andrei Bobrov 2014년 9월 30일
X = [1.011 1.004 1.054
1.008 0.998 1.042
0.984 0.988 1.024
1.026 1.006 1.016
1.000 0.996 0.977];
n = nchoosek(1:size(X,2),2);
out = squeeze(diff(reshape(X(:,n'),[],2,3),1,2));

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