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

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!

Answer 1

[Nobel Mondal]

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

[saravan1992]

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!

Answer 2

[Thorsten]

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.