I have to conduct anova1 on log2ratios (microarray) that I am binning in to various segments based on biological selection. I have managed to set this up though it was tedious to do. I basically have 89 segments some with only 30 elements and some with upwards of 5000. The value of each element is totally independent and usually different from all others.
The problem is constructing a 'group array' in order to do anova1 on my segments.
result=anova1(dataVector,groupVector)
Currently I am building a for loop, ie. specifying elements between which row numbers of the whole log2ratio file are to be categorized as group 1, group2 and so on.
But this means I have to manually type in 89 for-end loops.
Does anyone know an easier way to do this?
Any help would be much appreciated. Thanks!
Subject: setting up groups for anova1 -large for loop issue
"Farah " <fr.rasheed@gmail.com> wrote in message
news:gvvuda$437$1@fred.mathworks.com...
>I have to conduct anova1 on log2ratios (microarray) that I am binning in to
>various segments based on biological selection. I have managed to set this
>up though it was tedious to do. I basically have 89 segments some with only
>30 elements and some with upwards of 5000. The value of each element is
>totally independent and usually different from all others.
> The problem is constructing a 'group array' in order to do anova1 on my
> segments.
> result=anova1(dataVector,groupVector)
>
> Currently I am building a for loop, ie. specifying elements between which
> row numbers of the whole log2ratio file are to be categorized as group 1,
> group2 and so on.
> But this means I have to manually type in 89 for-end loops.
> Does anyone know an easier way to do this?
Post a small sample of your code that shows how you generate your
groupVector. I'm willing to bet there's an easier way to do what you want
than 89 FOR-loops. [In particular, if you're generating 89 variables group1
through group89, there's DEFINITELY a better way -- see Q4.6 in the
newsgroup FAQ (http://matlabwiki.mathworks.com/MATLAB_FAQ) if that's what
you're doing.]
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