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

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Emerson De Souza
Emerson De Souza 2012-9-16
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

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采纳的回答

Andrei Bobrov
Andrei Bobrov 2012-9-16
B = accumarray(A(:,1),A(:,2),[],@median);
out = [A, B(A(:,1))];
  2 个评论
Emerson De Souza
Emerson De Souza 2012-9-16
Hi Andrei, I have a problem:
The values of the first column may be 1.5 instead of 1 and so on for the others rows. I this case I obtain the following error:
Error using accumarray First input SUBS must contain positive integer subscripts.
Do you know what to change in the command to make it more general?
Thank you
Emerson

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更多回答（1 个）

Azzi Abdelmalek
Azzi Abdelmalek 2012-9-16
编辑：Azzi Abdelmalek 2012-9-16
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 个评论
显示 3更早的评论
Emerson De Souza
Emerson De Souza 2012-9-16
Thank you Azzi,
your suggestion works now. I only don't understand what idx1 is doing.
I also get a red line for B(k)=..... with the comment:
The variable B appears to change size on every iteration (within a script). Consider preallocating for speed. I don't understand what that means.
Thank you again for your help
Emerson
Azzi Abdelmalek
Azzi Abdelmalek 2012-9-17
编辑：Azzi Abdelmalek 2012-9-17
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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