Trouble with a correlation analysis code

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Alexandra Brian
Alexandra Brian el 28 de En. de 2017
Editada: John Chilleri el 1 de Feb. de 2017
I'm writing a code to run a correlation analysis between two variables under certain conditions. I want to see if there is a correlation between the second column and the first if the second's value is greater than 0. I wrote the following code, but MATLAB continues to print the error, "Index exceeds matrix dimensions".
clear all
indicators = xlsread('largedata1.xlsx');
X = xlsread('largedata1.xlsx', 1, 'A:A');
rows = indicators(:,2)>0; %Logical vector with the rows which satisfy all conditions.
if any(rows) % True if there is at least 1 row which meats the condition.
md1 = fitlm(indicators(rows, [2]), X(rows)); % Fit with the rows which satisfy condition.
end
Can you please help?
Best, A
  2 comentarios
Image Analyst
Image Analyst el 28 de En. de 2017
Can you attach 'largedata.xlsx' so people can try your code?
Alexandra Brian
Alexandra Brian el 29 de En. de 2017
I added it! Thanks for the suggestion.

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Respuesta aceptada

John Chilleri
John Chilleri el 28 de En. de 2017
Editada: John Chilleri el 1 de Feb. de 2017
Hello,
I downloaded your data and ran your code and it didn't encounter any errors for me (rows was a logical vector so rows==1 actually isn't needed like I suggested - I was testing with a double vector).
clear all
indicators = xlsread('largedata1.xlsx');
X = xlsread('largedata1.xlsx', 1, 'A:A');
rows = indicators(:,2)>0; %Logical vector with the rows which satisfy all conditions.
if any(rows) % True if there is at least 1 row which meats the condition.
md1 = fitlm(indicators(rows, [2]), X(rows)); % Fit with the rows which satisfy condition.
end
>> md1
md1 =
Linear regression model:
y ~ 1 + x1
Estimated Coefficients:
Estimate SE tStat pValue
________ _________ ______ __________
(Intercept) 0.056209 0.0049663 11.318 4.6459e-28
x1 0.18966 0.10075 1.8825 0.060056
Number of observations: 1022, Error degrees of freedom: 1020
Root Mean Squared Error: 0.12
R-squared: 0.00346, Adjusted R-Squared 0.00249
F-statistic vs. constant model: 3.54, p-value = 0.0601
  3 comentarios
Alexandra Brian
Alexandra Brian el 31 de En. de 2017
Thanks John!
John Chilleri
John Chilleri el 1 de Feb. de 2017
Changed solution, but it doesn't help you it would seem.

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