# Splitapply: Returning multiple output values

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Maximilian Hauschel on 24 Jun 2021
Commented: dpb on 25 Jun 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

dpb on 24 Jun 2021
Edited: dpb on 24 Jun 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 CommentsShowHide 1 older comment
dpb on 25 Jun 2021
Guess all depends on what want for the output...the cell of redundant data or just the actual values without redundancy...

### More Answers (1)

Adam Danz on 24 Jun 2021
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)
R = 10×1 cell array
{2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double}
P = 10×1 cell array
{2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double} {2×2 double}
##### 1 CommentShowHide None
Maximilian Hauschel on 25 Jun 2021
That's an even more elegant solution. Never thought of applying arrayfun, because i've never had to use it anytime before. Looks like i should really get used to it. Also thank you for your fast and helpfull reply.

R2021a

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