Splitapply: Returning multiple output values
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Maximilian Hauschel
el 24 de Jun. de 2021
Comentada: dpb
el 25 de Jun. de 2021
Hello everyone. I am currently stuck at a problem which refers to the application of the splitapply function for a function with multiple output values. More specifically, i am trying to get the correlation coefficient r, as well as the corresponding p-value for the subsets of multiple columns of a table, specified by group numbers. Here is an example:
A = array2table(rand(100,1));
A.Properties.VariableNames{1} = 'Row1';
A.Row2 = rand(100,1);
B = [1:1:10]';
A.Groups = repmat(B,10,1);
Given is a table A with 2 columns, containing 10 observations in chronological order for 10 different groups. The following formula returns the r coefficients i am looking for.
R = splitapply(@(a,b) {corrcoef(a,b)},A.Row1,A.Row2,A.Groups);
But the syntax for multiple outputvalues, [R,P] = corrcoef (a,b), does not work if i simply put it in front of the splitapply funcion as follows:
[R,P] = splitapply(@(a,b) {corrcoef(a,b)},A.Row1,A.Row2,A.Groups);
I guess i have to use a local function to recieve both of the values. Unfortunately i am not really experienced in using them, and after doing a little research, i have still absoluty no idea what to type. It would be a blessing if somebody could help me out.
Thank you very much for reading
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Respuesta aceptada
dpb
el 24 de Jun. de 2021
Editada: dpb
el 24 de Jun. de 2021
This is harder because corrcoef returns a square maxtrix, not just the 2-sample coefficient, p-values.
Easiest is to write a helper function to do that part --
function [r,p]=mycorr(x,y)
% two-sample correlation, p values for x,y
[r,p]=corrcoef(x,y);
r=r(2,1);
p=p(2,1);
end
tA=table(rand(100,1),rand(100,1),'VariableNames',{'A','B'}); % sample data table
tA.G=fix([0:height(tA)-1]/10).'; % create grouping variable
% the engine
>> rowfun(@(a,b) (mycorr(a,b)),tA,"InputVariables",{'A','B'},'SeparateInputs',1,'NumOutputs',2,'GroupingVariables','G')
ans =
10×4 table
G GroupCount Var3 Var4
_ __________ _________ _______
0 10 0.19267 0.59383
1 10 0.39138 0.26338
2 10 0.13497 0.71007
3 10 -0.11946 0.74237
4 10 0.11982 0.74163
5 10 0.27951 0.43414
6 10 0.01039 0.97728
7 10 -0.087027 0.81106
8 10 -0.039839 0.91299
9 10 -0.078133 0.83012
>>
>> % Illustrate get same result for second group of 10
>> [r,p]=corrcoef(tA.A([1:10]+10),tA.B(10+[1:10]))
r =
1.0000 0.3914
0.3914 1.0000
p =
1.0000 0.2634
0.2634 1.0000
>>
2 comentarios
dpb
el 25 de Jun. de 2021
Guess all depends on what want for the output...the cell of redundant data or just the actual values without redundancy...
Más respuestas (1)
Adam Danz
el 24 de Jun. de 2021
Use arrayfun instead.
A = array2table(rand(100,1));
A.Properties.VariableNames{1} = 'Row1';
A.Row2 = rand(100,1);
B = [1:1:10]';
A.Groups = repmat(B,10,1);
[R,P] = arrayfun(@(i) corrcoef(A.Row1(A.Groups==i), A.Row2(A.Groups==i)), ...
unique(A.Groups), 'UniformOutput',false)
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