How to calculate the median of a column depending on the value of another column?

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INTRODUCTION: I have two columns of values. The values of the first column are partially constant and the values of the second column are arbitrary ones.
GOAL: I want to build a third column with values of median for each group of constant value of the first column.
EXAMPLE:
A=[1 3;
1 2;
1 3;
2 4;
2 4;
2 3;
2 4;
3 5;
3 1;
3 1;
3 1;
3 2;
4 3;
4 2];
B1=median(A(1:3,2));
B2=median(A(4:7, 2));
B3=median(A(8:12, 2));
B4=median(A(13:14, 2));
B=[B1 B2 B3 B4]';
PROBLEM: The number m of rows are typically much larger than only 14 and makes impossible to write the commands B1 until BN per hand.
I wonder if someone could tell me how to write some command lines that makes this automatically.
Thank you in advance for your help
Emerson

Accepted Answer

Andrei Bobrov
Andrei Bobrov on 16 Sep 2012
B = accumarray(A(:,1),A(:,2),[],@median);
out = [A, B(A(:,1))];
  2 Comments
Andrei Bobrov
Andrei Bobrov on 17 Sep 2012
yes,
[c,c,c] = unique(A(:,1));
B = accumarray(c,A(:,2),[],@median);
out = [A, B(A(:,1))];

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More Answers (1)

Azzi Abdelmalek
Azzi Abdelmalek on 16 Sep 2012
Edited: Azzi Abdelmalek on 16 Sep 2012
B=squeeze(median(reshape(A(:,2),3,1,size(A,1)/3)))
%A must contains a multiple of 3 rows, if not, we have to complete with nan or zero values, write at the begening this code
nc=mod(size(A,1),3);
if nc>0;
A=[A;nan(3-nc,2)]
end
  5 Comments
Azzi Abdelmalek
Azzi Abdelmalek on 17 Sep 2012
for our example
idx1 =
1 3 1 is repeating from index 1 to 3
4 7 2 is repeating from index 4 to 7
8 12 3 is repeating from index 8 to 12
13 14 4 is repeating from index 13 to 14
B is changing a size because it's in the loop, B(1), B(2),... then B(k)
to make preallocation,( in case we work with big array)
B=zeros(1,numel(idx))

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