Find Mean from Duplicate Entries
4 ビュー (過去 30 日間)
古いコメントを表示
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.
0 件のコメント
回答 (2 件)
Walter Roberson
2012 年 4 月 24 日
[uyear, yearjunk, yearidx] = unique( Matrix(:,1) );
OutMatrix = [uyear(:), accumarray( yearidx, Matrix(:,2) ) ];
3 件のコメント
Walter Roberson
2012 年 4 月 25 日
OutMatrix = [uyear(:), accumarray( yearidx, Matrix(:,2), [], @mean ) ];
T. Guillod
2013 年 5 月 13 日
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))./
2 件のコメント
Sean de Wolski
2013 年 5 月 13 日
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
2013 年 11 月 17 日
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.
参考
カテゴリ
Help Center および File Exchange で Logical についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!