Bug using LinearModel.fit in a loop?

2 次查看(过去 30 天)
Everytime i try using my datas and LinearModel.fit it works perfectly, but when i try to use it in a loop i always end up having a matrix (of the correct size) of empty cells... Somebody who has matlab R2013a on a MAC said my said was running perfectly but it doesn't seem to run on a PC with R2012a. Thanks in advance!
V = nchoosek(6,4)
for i = 1:15
A{i}¨= V(i,:)
end
for i = 1:15
C{i} = horzcat(A{1,i}{1,1},A{1,i}{1,2},A{1,i}{1,3},A{1,i}{1,4})
end
for i = 1:15
DATA{i} = [C{1,i} Y] %Y is my independent variable
end
for i = 1:15
REG{i} = LinearModel.fit(DATA{1,i})

采纳的回答

Sean de Wolski
Sean de Wolski 2013-5-29
Well this part doesn't work:
V = nchoosek(6,4)
for i = 1:15
A{i}= V(i,:)
end
because V is a scalar. Start using the debugger, run the following:
dbstop if error
Then run your script. It will stop and you'll be able to inspect what's causing the error.
  1 个评论
Simon
Simon 2013-5-30
Good thinking. I made it work because i realised that LinearModel.fit is an object so i can't make many object in a matrix... but using only Varname.Rsquared.Adjusted can give me all the Rsquared im looking for in a matrix. The only problem is that it doesn't give me the var name anymore, only the answer :S. Is there a way to generate the Rsquared AND having the coefficient names for each answer at the same time? That would be great. I know the ''.Coefficients'' gives the coeff names but i don't know how to put it together WITH Rsquared :/
Thank you!

请先登录,再进行评论。

更多回答(1 个)

Tom Lane
Tom Lane 2013-5-30
I don't really understand the question. But with the hint from Sean, I will show you the following code that will collect R-squared values from all of the fits into a single matrix.
V = nchoosek(1:6,4)
Y = [1.1;2.5;3.2;1.5];
for i = 1:15
REG{i} = LinearModel.fit(V(i,:)',Y)
end
rsq = zeros(15,1);
for i=1:15
rsq(i,1) = REG{i}.Rsquared.Ordinary;
end

类别

Help CenterFile Exchange 中查找有关 Data Distribution Plots 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by