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

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Emerson De Souza on 16 Sep 2012
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.
Emerson

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

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
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))