Why R-squares are different between "fitglm" and "fitglme"? or how do "fitglme" and "fitglm" calculate R-squared?

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Heeseung Lee 2020-10-22

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评论： Heeseung Lee 2020-11-17

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Hi,

I am using "fitglme" for fitting a mixed-effect logistic regression model.

I could have R-squared from the fitted model.

glme.Rsqaured.Adjusted

Then, I tried to have individual-subject R-squared by using "fitglm" for each subject.

But, the subjectwise-averaged R-squared from "fitglm" was so different from the R-squared of "fitglme".

Why are they so different?

I found that the "fitglm"s R-squared can be derived by the definition of R-squred:

y = glm.Variable.y;
yhat = glm.predict;
ybar = mean(y);
SST = sum((y-ybar).^2);
SSR = sum((yhat-ybar).^2);
R_square = SSR/SST;
R_square_glm = glm.Rsquared.Ordinary;
R_square = R_square_glm;

However, "fitglme"s R-squred is different from the derived R-squared;

y = glme.Variable.y;
yhat = glme.predict;
ybar = mean(y);
SST = sum((y-ybar).^2);
SSR = sum((yhat-ybar).^2);
R_square = SSR/SST;
R_square_glm = glme.Rsquared.Ordinary;
R_square ~= R_square_glm;

How shoud I understand this inconsistency?

I will highly appreciate for you help!

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Aditya Patil 2020-11-17

How different is it? Some small differences might be present due to floating point accuracy. Can you provide complete code to reproduce the issue?

Heeseung Lee 2020-11-17

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Hi Aditya!

Here is the code

x = rand(100,1);
y = (x*2 + normrnd(0,1,100,1))>1;
 
Reg = cell2table(num2cell([x y]),'variablenames',{'x','y'});
 
glm = fitglm(Reg,'y~1+x','distribution','binomial','link','logit');
glme = fitglme(Reg,'y~1+x','Distribution','Binomial','Link','Logit');
 
glm.Rsquared
glme.Rsquared

and the output printed in the command window

>> glm.Rsquared
ans = 
  struct with fields:
          Ordinary: 0.1766
          Adjusted: 0.1682
               LLR: 0.1309
          Deviance: 0.1309
    AdjGeneralized: 0.2199
>> glme.Rsquared
ans = 
  struct with fields:
    Ordinary: 0.1968
    Adjusted: 0.1886