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How to divide large data in small intervals?

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Anderson
Anderson 2016-4-22
评论： Guillaume 2016-4-22
How to divide a large matrix into small intervals?
For example, take the matrix [1 1; 1 2; 2 3; ...;5 100; 6 100; ...; 1 1.0e4; ...; 5 1.0e9] with many entries.
How to efficiently divide the second column into intervals of length 1, e.g. [0,1], [1,2]...[1000,1001] and so on such that, for each interval, the elements of the first column sum to 1.
Small example: M = [1 1; 2 1; 1 10; 3 10];
The output should be:
M_out = [0.333 1; 0.6667 1; 0.25 10; 0.75 10]
Take M=[1 1; 2 1; 1 1.5; 1 10; 1 10.6; 3 10];
M_out = [0.25 1; 0.5 1; 0.25 1.5; 0.20 10; 0.20 10.6; 0.6 10]
The solution could be
for i=1:1.0e9; j=find (i <= M(:,2) & M(:,2) < i+1); M(j,1) = M(j,1)./sum(M(j,1)); end
However, is it an efficiently way to do it?
  2 个评论
Anderson
Anderson 2016-4-22
编辑：Jan 2016-4-22
No.
The solution could be
for i=1:1.0e9;
j=find (i <= M(:,2) & M(:,2) < i+1);
M(j,1) = M(j,1)./sum(M(j,1));
end
However, is it an efficiently way to do it?

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采纳的回答

Guillaume
Guillaume 2016-4-22
编辑：Guillaume 2016-4-22
It's not very clear from your question that you want to bin the second column in bins of width 1, and you haven't given a criteria for the bin edges. Should the edges always be integer?
Anyway, find out which bin your second column falls into with discretize and use these bins as input to accumarray as per Star's or the cyclist's answer:
M = [1 1; 2 1; 1 1.5; 1 10; 1 10.6; 3 10];
%compute bins. Assume integer edges
binlow = floor(min(M(:, 2)));
binhigh = ceil(max(M(:, 2)));
if binhigh == max(M(:, 2)), binhigh = binhigh + 1; end %otherwise if max is integer it'll be included in the previous bin
binidx = discretize(M(:, 2), binlow : binhigh);
%apply to accumarray
binsum = accumarray(binidx, M(:, 1));
%normalise
M_out = [M(:, 1) ./ binsum(binidx), M(:, 2)]
  2 个评论
Anderson
Anderson 2016-4-22
编辑：Anderson 2016-4-22
Thank you for your reply.
Yes, the edges are integer and the range is large: from 0 to 1.0e9.
Is there a way to not use discretize function? This function is not available for previous MATLAB versions.
Guillaume
Guillaume 2016-4-22
If you're not using up to date matlab, please mention it in your question.
Any histogram function will do, the second return value of histc will work. Since histc behaves differently for the last edge, the code becomes:
binlow = floor(min(M(:, 2)));
binhigh = ceil(max(M(:, 2))) + 1; %always add an extra bin
[~, binidx] = histc(M(:, 2), binlow:binhigh);
%accumarray code as before

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更多回答（1 个）

the cyclist
the cyclist 2016-4-22
编辑：the cyclist 2016-4-22
M = [1 1; 2 1; 1 10; 3 10];
[~,~,idx] = unique(M(:,2));
S = accumarray(idx,M(:,1),[]);
M_out = [M(:,1)./S(idx),M(:,2)]
  1 个评论
Anderson
Anderson 2016-4-22
This solution does not divide the second column into intervals of length 1.
Take M=[1 1; 2 1; 1 1.5; 1 10; 1 10.6; 3 10];
M_out = [0.25 1; 0.5 1; 0.25 1.5; 0.20 10; 0.20 10.6; 0.6 10]

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