Splitapply: Returning multiple output values

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

Answer 1

[dpb]

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
>> 

[Maximilian Hauschel]

Didn't expect those fast replys. Seems like splitapply isn't the most suitable function to handle this case. But your answer really helps me out. Many thanks, dpb.

[dpb]

Guess all depends on what want for the output...the cell of redundant data or just the actual values without redundancy...

Answer 2

[Adam Danz]

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)

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}

[Maximilian Hauschel]

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.