Trouble with a correlation analysis code

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

[Image Analyst]

Can you attach 'largedata.xlsx' so people can try your code?

[Alexandra Brian]

I added it! Thanks for the suggestion.

Answer 1

[John Chilleri]

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

[Alexandra Brian]

No luck, but thanks for the suggestion!

[Alexandra Brian]

Thanks John!

[John Chilleri]

Changed solution, but it doesn't help you it would seem.