Splitapply command and merge results

1 回表示 (過去 30 日間)
Ivan Mich
Ivan Mich 2021 年 7 月 26 日
コメント済み: Rik 2021 年 7 月 27 日
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

回答 (1 件)

Rik
Rik 2021 年 7 月 26 日
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 件のコメント
Ivan Mich
Ivan Mich 2021 年 7 月 27 日
編集済み: Ivan Mich 2021 年 7 月 27 日
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 2021 年 7 月 27 日
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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