Find Mean from Duplicate Entries

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Chris
Chris 2012-4-24
评论： John Yang 2013-11-17
Hello,
I have a matrix with 2 columns (X,Y) and a large number (hundreds or thousands) of rows. It's a time series, so the first column has year values and the second column, Y, corresponds to an observation at that time. However, some of the dates repeat with different values in Y (due to the fact that I did a "floor" function on the dates, some of which originally had monthly resolution).
What I want to do is locate all dates that are found more than once and take the mean of the corresponding Y values. For example, if I have
1900 8
1901 6
1902 4
1902 3
1903 5
.
.
.
I want the output to be,
1900 8
1901 6
1902 3.5
1903 5
.
.
.
Any help would be appreciated! Thanks.

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回答（2 个）

Walter Roberson
Walter Roberson 2012-4-24
[uyear, yearjunk, yearidx] = unique( Matrix(:,1) );
OutMatrix = [uyear(:), accumarray( yearidx, Matrix(:,2) ) ];
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Walter Roberson
Walter Roberson 2012-4-25
OutMatrix = [uyear(:), accumarray( yearidx, Matrix(:,2), [], @mean ) ];

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T. Guillod
T. Guillod 2013-5-13
Thanks for this solution. It helped me a lot.
The solution by Walter is OK but very slow (call to mean()). A better one:
Instead: accumarray( yearidx, Matrix(:,2), [], @mean )
Use: accumarray( yearidx, Matrix(:,2))./accumarray( yearidx, ones(size(Matrix,1),1))./
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Sean de Wolski
Sean de Wolski 2013-5-13
You could skip the accumarray call in the denominator and instead use histc() on the yearidx with edges being uyear. histc() should be faster than accumarray()
John Yang
John Yang 2013-11-17
Find this old thread. Your solution is really fast, 10 times faster of using histc than calling @mean.
Just curious, why using histc so fast? In my eye, calling a basic function such as mean should use the same order of time as histc.

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