How to calculate mean of standard deviation (mean deviation) in a table

With the table below how do I find the players mean_score and standard deviation.

what I tried. now quite sure how to handle the mean deviations.

Average standard deviation = √ (s12 + s22 + … + sk2) / k. ref here

summary = groupsummary(table,{'player'},{'mean'},{'mean_score','mean_rate'})

DATA

week = [1 2 1 2 1 2]';
  player = [1 1 2 2 3 3]';
  mean_score = [3.4 5.2 4.5 5.5 2.8 3.8]';
  std_score = [2.1 2.4 2.8 2.3 2.2 2.3]'; %standard deviation 
  mean_rate = [5.4 8.2 4.5 15.5 22.8 12.8]';
  std_rate = [0.1 0.16 0.14 0.12 0.3 0.16]'; % standard deviation 
  T = table(week,player,mean_score,std_score,std_rate);

Answer 1

[Image Analyst]

You forgot to attach your data, so we can't try anything with your actual data.

DON'T use "table" as the name of your table because table is a built in function!

Did you try

meanStd = mean(summary.std_score)

If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:

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[Frederick Awuah-Gyasi]

I want it grouped by player that's why I used the groupsummary function.

[Image Analyst]

My answer remains the same.

[Frederick Awuah-Gyasi]

The code I added now should help in creating table T. In the end it should have 3 rows one for each player. Thank you.

[Image Analyst]

Is there something wrong with my suggested solution?

week = [1 2 1 2 1 2]';
player = [1 1 2 2 3 3]';
mean_score = [3.4 5.2 4.5 5.5 2.8 3.8]';
std_score = [2.1 2.4 2.8 2.3 2.2 2.3]';
std_rate = [0.1 0.16 0.14 0.12 0.3 0.16]';
T = table(week,player,mean_score,std_score,std_rate)

T = 6×5 table
    week    player    mean_score    std_score    std_rate
    ____    ______    __________    _________    ________

     1        1          3.4           2.1          0.1  
     2        1          5.2           2.4         0.16  
     1        2          4.5           2.8         0.14  
     2        2          5.5           2.3         0.12  
     1        3          2.8           2.2          0.3  
     2        3          3.8           2.3         0.16  

meanStd = mean(T.std_score)

meanStd = 2.3500

Answer 2

[Star Strider]

The 'mean_rate' variable does not exist in ‘T’.

Giving groupsummary the correct variable names (or at least variable names that exist) works —

week = [1 2 1 2 1 2]';
  player = [1 1 2 2 3 3]';
  mean_score = [3.4 5.2 4.5 5.5 2.8 3.8]';
  std_score = [2.1 2.4 2.8 2.3 2.2 2.3]';
  std_rate = [0.1 0.16 0.14 0.12 0.3 0.16]';
  T = table(week,player,mean_score,std_score,std_rate)

T = 6×5 table
    week    player    mean_score    std_score    std_rate
    ____    ______    __________    _________    ________

     1        1          3.4           2.1          0.1  
     2        1          5.2           2.4         0.16  
     1        2          4.5           2.8         0.14  
     2        2          5.5           2.3         0.12  
     1        3          2.8           2.2          0.3  
     2        3          3.8           2.3         0.16  

  
  summary = groupsummary(T,{'player'},{'mean'},{'mean_score','std_score','std_rate'})

summary = 3×5 table
    player    GroupCount    mean_mean_score    mean_std_score    mean_std_rate
    ______    __________    _______________    ______________    _____________

      1           2               4.3               2.25             0.13     
      2           2                 5               2.55             0.13     
      3           2               3.3               2.25             0.23     

.

[Frederick Awuah-Gyasi]

this is what I tried but turns out the average of standard devitation has this formular

Average standard deviation = √ (s12 + s22 + … + sk2) / k.

in my table std_ are the standard deviation values.

[Image Analyst]

Do you really mean, or want:

rmsStdDev = sqrt(s1^2 + s2^2 + ....sk^2) / k

or just use rms.

rmsStdDev = rms([s1, s2, ....,sk])

I don't know that I've heard of taking the RMS of the Std Deviation. It almost seems redundant. Are you 100% sure you don't just want the average of the standard deviations of all the players (like I showed in my answer)?

[Star Strider]

That doesn’t lend itself to groupsummary, and the means of the standard deviations is not the same as rms, so a different approach is necessary.

Try this —

week = [1 2 1 2 1 2]';
  player = [1 1 2 2 3 3]';
  mean_score = [3.4 5.2 4.5 5.5 2.8 3.8]';
  std_score = [2.1 2.4 2.8 2.3 2.2 2.3]';
  std_rate = [0.1 0.16 0.14 0.12 0.3 0.16]';
  T = table(week,player,mean_score,std_score,std_rate)

T = 6×5 table
    week    player    mean_score    std_score    std_rate
    ____    ______    __________    _________    ________

     1        1          3.4           2.1          0.1  
     2        1          5.2           2.4         0.16  
     1        2          4.5           2.8         0.14  
     2        2          5.5           2.3         0.12  
     1        3          2.8           2.2          0.3  
     2        3          3.8           2.3         0.16  

mean3 = accumarray(T{:,2}, (1:size(T,1)).', [], @(x)mean(T{x,3}));                              % Mean Of Mean Values
mean45c = accumarray(T{:,2}, (1:size(T,1)).', [], @(x){sqrt(sum(T{x,[4 5]}.^2))/numel(x)});     % MEan Of Standard  Deviations
mean45 = cell2mat(mean45c);
VN = T.Properties.VariableNames;

Player_Means_Table = table(unique(T{:,2},'stable'),mean3,mean45(:,1),mean45(:,2), 'VariableNames',VN(2:end))

Player_Means_Table = 3×4 table
    player    mean_score    std_score    std_rate
    ______    __________    _________    ________

      1          4.3         1.5945       0.09434
      2            5         1.8118      0.092195
      3          3.3         1.5914          0.17

Those results appear to be correct (I checked some manually), if their calculations (as I understand them) are correct.

.

[Star Strider]

@Image Analyst — I don’t believe this is the same as rms.

As I understand it (although the expression posted is a bit ambiguous, and would benefit from appropriately-placed parenthesees) —

square root of the mean of the squares,

while

mean of the square root of the summed variaces

Anyway, that’s how I coded it.