Can I get rid of this for loop?
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Ryan
am 18 Mär. 2016
Bearbeitet: Stephen23
am 18 Mär. 2016
I have a matrix that has 3 columns that consist of -1, 0, and 1 (let's call them the "level columns"), and a fourth column that contains real numbers (call it the "data column"). For the level columns, I want to find the average of the "-1", "0", and "1" terms in the data column. So for the following matrix,
-1 -1 0 837
-1 0 -1 880
-1 0 1 879
-1 1 0 864
0 -1 -1 834
0 -1 1 833
0 0 0 874
0 0 0 875
0 0 0 876
0 1 -1 860
0 1 1 859
1 -1 0 829
1 0 -1 872
1 0 1 874
1 1 0 856
the -1 average for column 1 would be 865. Is there a way that I can do that all at once, for each level column, and each level column value? Right now, I loop over the level columns and do this:
[mean(testMat(testMat(:,col) == -1,4)) mean(testMat(testMat(:,col) == 0,4)) mean(testMat(testMat(:,col) == 1,4))]
Is there a better way?
Thanks!
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Akzeptierte Antwort
Stephen23
am 18 Mär. 2016
Bearbeitet: Stephen23
am 18 Mär. 2016
You cannot get rid of the for loop, but you can hide it inside a cellfun call. Note that my answer works for any number of columns, unlike the other solutions.
M = [...
-1 -1 0 837
-1 0 -1 880
-1 0 1 879
-1 1 0 864
0 -1 -1 834
0 -1 1 833
0 0 0 874
0 0 0 875
0 0 0 876
0 1 -1 860
0 1 1 859
1 -1 0 829
1 0 -1 872
1 0 1 874
1 1 0 856 ];
%
[~,~,X] = cellfun(@unique,num2cell(M(:,1:end-1),1),'Uni',0);
N = cellfun(@(s)accumarray(s,M(:,end),[],@mean),X, 'Uni',0);
out = [N{:}]
creates this output matrix, where the rows correspond to [-1;0;1]:
>> out
out =
865 833.25 861.5
858.71 875.71 858.71
857.75 859.75 861.25
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Weitere Antworten (2)
Star Strider
am 18 Mär. 2016
I would use three concatenated accumarray calls:
M = [-1 -1 0 837
-1 0 -1 880
-1 0 1 879
-1 1 0 864
0 -1 -1 834
0 -1 1 833
0 0 0 874
0 0 0 875
0 0 0 876
0 1 -1 860
0 1 1 859
1 -1 0 829
1 0 -1 872
1 0 1 874
1 1 0 856];
Result = [accumarray(M(:,1)+2, M(:,4), [], @mean), accumarray(M(:,2)+2, M(:,4), [], @mean), accumarray(M(:,3)+2, M(:,4), [], @mean)];
Result =
865 833.25 861.5
858.71 875.71 858.71
857.75 859.75 861.25
Here, the columns correspond to those in ‘M(:,1:3)’ and the rows correspond to ‘[-1; 0; 1]’ respectively.
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Image Analyst
am 18 Mär. 2016
If you have the Statistics and Machine Learning Toolbox, you can use grpstats(), but I don't know if it would be any more compact that what you're already doing.
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