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Splitapply command and merge results

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Ivan Mich
Ivan Mich on 26 Jul 2021
Commented: Rik on 27 Jul 2021
In general I am using this command "splitapply" in order to find mean (average) of a group of data.
edges=1:0.5:10
[N, edges, bin] = histcounts(B, edges);
mean_B=splitapply(@mean, B, bin) %mean
%B is 100x1 double
But I would like to make a code that will :
1) Group my data into bins from edges : 1:0.5:10 (min=1, max=10 with step equal to 0.5)
2) Compute the means of the values in each bin (lets call it set1).
3) Group my data into bins from edges : 2:1:10 (min=2, max=10 with step equal to 1)
4) Compute the means of the values in each bin (set2).
5) Merge two sets of data
Could you please help me in order to make it?
Thanking you in advance

Answers (1)

Rik
Rik on 26 Jul 2021
The question for you is what you mean by merging, but steps 1 to 4 are below.
%generate example data
B=10*rand(1,100);
edges1=[1 0.5 10];
edges2=[2 1 10];
mean_B_set1=mean_group_data(B,edges1)
mean_B_set1 = 1×18
1.1522 1.6931 2.2359 2.7160 3.2395 3.8087 4.3454 4.7591 5.2552 5.7937 6.2050 6.7348 7.2927 7.6282 8.2723 8.7246 9.2756 9.7841
mean_B_set2=mean_group_data(B,edges2)
mean_B_set2 = 1×8
2.5560 3.3270 4.5522 5.6142 6.4405 7.5062 8.4984 9.5299
function mean_B=mean_group_data(B,msm)
B(B<msm(1) | B>msm(3))=[];%remove data outside of bounds
edges=msm(1):msm(2):msm(3);
[~, ~, bin] = histcounts(B, edges);
mean_B=splitapply(@mean, B, bin);
if numel(mean_B)<(numel(edges)-1)
mean_B((end+1):(numel(edges)-1))=NaN;%extend to fill all bins
end
end
  4 Comments
Ivan Mich
Ivan Mich on 27 Jul 2021
Edited: Ivan Mich on 27 Jul 2021
Excuse me my final output is :
0.5
0.75
1
1.25
1.6
1.7
1.9
2.5
3.2
that corresponds to bins
[2-3]
[2.5-3.5]
[3-4]
[3.5-4.5]
[4-5]
[4.5-5.5]
[5-6]
[5.5-6.5]
[6-7]
Rik
Rik on 27 Jul 2021
Your desired output will be the third column of the array below.
edges1=[2 1 7 ];
edges2=[2.5 1 6.5];
mean_B_set1=[0.5;1.25;1.6;1.9;3.2];
mean_B_set2=[0.75;1;1.7;2.5];
e1=edges1(1):edges1(2):edges1(3);
e2=edges2(1):edges2(2):edges2(3);
%col 1 contains bin starts, col 2 contains bin ends
%col 3 contains the sets
tmp=[e1(1:(end-1)).' e1(2:end).' mean_B_set1;...
e2(1:(end-1)).' e2(2:end).' mean_B_set2];
output=sortrows(tmp)
output = 9×3
2.0000 3.0000 0.5000 2.5000 3.5000 0.7500 3.0000 4.0000 1.2500 3.5000 4.5000 1.0000 4.0000 5.0000 1.6000 4.5000 5.5000 1.7000 5.0000 6.0000 1.9000 5.5000 6.5000 2.5000 6.0000 7.0000 3.2000

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