How can I find matching rows from three out of four colomns

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Joanie
Joanie 2014-5-15
评论： Henric Rydén 2014-5-19
I have a matrix with 4 columns, x y z and D, and a lot of rows. I want the program to find the rows where x, y and z are the same (so the first three colomns match) and add the corresponding D value of these matching rows. When the matching rows are found only one stays in the matrix with the new D value. To make it a little more complicated I like to attach a tolerance for what the matching values can diverge.
For example:
x y z D
1 2 3 4
2 4 6 8
1 2 3 5
New one:
x y z D
1 2 3 9
2 4 6 8

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采纳的回答

Henric Rydén
Henric Rydén 2014-5-15
编辑：Henric Rydén 2014-5-15
This solution might not be as simple and elegant as the one Andrei Bobrov provided, but it allows for a tolerance level.
% Example data
A=[ 1 2 3 ; ...
2 4 6 ; ...
1.05 2 3 ; ...
1 52 3 ; ...
2 4 6 ; ...
1.05 2 3 ; ...
5 2 3 ; ...
1 52 3 ; ...
5 2 3 ; ...
];
D = [4; 8; 5; 1; 2; 3; 5; 7; 8];
% Vector to track which have been processed and how they are grouped
processed = zeros(size(A,1),1);
% Tolerance level
tol = .1;
row = 1;
while ~all(processed)
% Only check rows that havent been matched yet
if ~processed(row)
% Take out the row you want to compare to the others
curRow = A(row,:);
% Subtract that row from all rows in A
subtractedA = A - repmat(curRow,size(A-1,1), 1);
% Check to see if they are within the tolerance level
matchingRows = all((subtractedA <= tol & subtractedA >= -tol),2);
% Mark the rows as processed
processed(matchingRows) = row;
% Update A to the mean of all matches
A(matchingRows,:) = repmat(mean(A(matchingRows,:),1),sum(matchingRows),1);
% Sum the D values that matches
D(matchingRows) = sum(D(matchingRows));
end
% Check the next row
row = row + 1;
end
% Results
[A(unique(processed),:) D(unique(processed))]
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更多回答（1 个）

Andrei Bobrov
Andrei Bobrov 2014-5-15
编辑：Andrei Bobrov 2014-5-17
d = [1 2 3 4
2 4 6 8
1 2 3 5];
[a,b,c] = unique(d(:,1:end-1),'rows');
out = [a, accumarray(c,d(:,end))];
add
d0 =[[ 1.1 2 3.2
2 4 6
1.2 2 3.3
2.09 3.96 6.05
1.05 1.99 3.25],randi(25,5,1)];
eps1 = .1;
d = d0(:,1:end-1);
l = bsxfun(@(x,y)abs(x-y)<=eps1,permute(d,[1 3 2]),permute(d,[3 1 2]));
[ii,jj] = find(triu(all(l,3),1));
a = num2cell(unique([[ii;ii],[ii;jj]],'rows'),1);
out = [cell2mat(accumarray(a{:},[],@(x){mean(d0(x,1:end-1))})),...
accumarray(a{1},d0(a{2},end))];
  2 个评论
Joanie
Joanie 2014-5-15
This works.
But what if the values diverge a little so I can add a tolerance and take the average of the row:
d = [1.1 2 3.2 4; 2 4 6 8; 1.2 2 3.3 5]
and the tolerance is 0.1 so the outcome will be:
dout = [1.15 2 3.25 9; 2 4 6 8]

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