How to find the median of one part of a column being in a while loop?

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Hello, I have the following problem: I have one matrix in which I want the output to be presented and which looks like this:
A1=[1950 0;1951 0;1952 0;1953 0;1954 0;1955 0];
There is also this second matrix which contains my input:
A2=[1960 1;1961 2;1962 3;1963 4;1950 5;1951 6;1952 7;1950 8;1950 9;1952 10;1951 11;1952 12];
In the second column of A1 I want to print the median of the second column of A2 for each year. That is I want to print in A1(1,2) the median of A2(5,2),A2(8,2) and A2(9,2).
This is the code I have written and I know it is wrong I just can't figure the correct approach out.
A1=[1950 0;1951 0;1952 0;1953 0;1954 0;1955 0];
A2=[1960 1;1961 2;1962 3;1963 4;1950 5;1951 6;1952 7;1950 8;1950 9;1952 10;1951 11;1952 12];
Year=1950;
for k=1:1:12;
for l=1:1:6;
while A2(k,1)==Year;
A1(l,2)=median(A2(k,2));
Year=Year+1;
end
end
end
Thanks!

Answers (2)

Nobel Mondal
Nobel Mondal on 16 Sep 2015
Is this what you're looking for?
A1=[1950 0;1951 0;1952 0;1953 0;1954 0;1955 0];
A2=[1960 1;1961 2;1962 3;1963 4;1950 5;1951 6;1952 7;1950 8;1950 9;1952 10;1951 11;1952 12];
thisYear = 1950;
thisRow = 1;
while (thisYear < 1956)
temp = find(A2(:,1) == thisYear);
if ~isempty(temp)
A1(thisRow,2) = median(A2(temp,2));
end
thisYear = thisYear + 1;
thisRow = thisRow + 1;
end
  1 Comment
saravan1992
saravan1992 on 16 Sep 2015
It works but unfortunately I am not even aware of what the temp structure is about. So I will first search and study a bit about it before using the code you proposed.
Thank you very much though!

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Thorsten
Thorsten on 16 Sep 2015
Edited: Thorsten on 16 Sep 2015
A solution without a while loop; if now values are given for the year, NaN is assigned:
for i = 1:size(A1,1)
A1(i,2) = median(A2(A2(:,1) == A1(i,1),2));
end
The expression
A2(:,1) == A1(i,1)
are the logical indices of those rows in A2 that have the matching year (given by A1(i,1)) in the first column.

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