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

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

Answer 1

[Andrei Bobrov]

B = accumarray(A(:,1),A(:,2),[],@median);
out = [A, B(A(:,1))];

[Emerson De Souza]

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

[Andrei Bobrov]

yes,

[c,c,c] = unique(A(:,1));
B = accumarray(c,A(:,2),[],@median);
out = [A, B(A(:,1))];

Answer 2

[Azzi Abdelmalek]

 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]

if you want to complete with zeros use

zeros(3-nc,2)

instead of

nan(3-nc,2)

[Emerson De Souza]

Thank you Azzi Abdelmalek,

unfortunately your suggestion is not doing what I want because the number of rows with constant values ARE NOT MULTIPLE of three.

The example that I gave displays the rows of the first column with:

1 three times, 2 four times, 3 five times and 4 two times. The number of rows with constant values are not multiple of any particular number.

I hope there is a way to modify your suggestion to do what I need.

Thank you again for your attention

Emerson

[Azzi Abdelmalek]

ok try this

[~,idx]=unique(A(:,1),'stable');
idx1=[idx [diff(idx)-1+idx(1:end-1); size(A,1)]];
for k=1:size(idx1,1);
   B(k)=median(A(idx1(k,1):idx1(k,2),2));
end
B

[Emerson De Souza]

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]

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))