How to divide large data in small intervals?

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?

[Image Analyst]

Is this your homework?

[Anderson]

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?

Answer 1

[Guillaume]

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)]

[Anderson]

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]

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

Answer 2

[the cyclist]

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)]

[Anderson]

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]