[BioC] No $df.residual
Johnson, Franklin Theodore
franklin.johnson at email.wsu.edu
Wed May 8 02:16:45 CEST 2013
Dear Sean Davis,
Thanks for the input.
I took the average of the RMA normalized values for each genotype. Then, log2 transformed as below:
The raw data is from NimbleGen array custom-designed for apple genome.
It was normalized using RMA method. This was downloaded from NCBI/GEO,
then I took log2(exprs(eset), after reading the papers and making phenoData.txt...Or, I did "log2(matrix(assayData))", then made an eset with this input file.
Anyhow, I will go back and start from the RMA normalized values and replicates.
However, for 3 reps each for 3 genotypes:
After making an eset,
This contrast matrix:
contrastMatrix=c(0,0,1, 0,1,0, 1,0,0, 0,0,2, 0,2,0, 2,0,0, 0,0,3, 0,3,0, 3,0,0)
should work, right?
Or, is another format, i.e. -1,.. ; {1,1,1}, a better option for comparision?
If its 1,1,1, for one genotype, contrast with the other 2 genotypes<- still cannot get $df.residuals correct?
Ambrosia Gala Melrose
> 1 1 0 0
> 2 1 -1 0
> 3 1 0 -1
---didn't work either---
Is there a way to coerce limma to accept a variable with only 2 levels?
I can write an ANOVA that accepts a factor with just 1 level...it's kinda of odd I can't do this with limma.
For example, sex is M an F...only 2 levels. If I'm not mistaken, limma require 'more than 2 levels', right?
But, limma is very useful, and efficient.
Hope this helps...I'm here all night!
Regards,
Franklin
________________________________________
From: seandavi at gmail.com [seandavi at gmail.com] on behalf of Sean Davis [sdavis2 at mail.nih.gov]
Sent: Tuesday, May 07, 2013 3:12 PM
To: FRANKLIN JOHNSON [guest]
Cc: bioconductor at r-project.org; franklin.johnson at wsu.edu
Subject: Re: [BioC] No $df.residual
Hi, Franklin. See below.
On Tue, May 7, 2013 at 6:04 PM, FRANKLIN JOHNSON [guest]
<guest at bioconductor.org> wrote:
>
> Hello,
>
> I want to determine differences between three genotypes.
> I'm using an exprs(eset).
> I made:
> contrastMatrix
> Ambrosia Gala Melrose
> 1 1 0 0
> 2 0 1 0
> 3 0 0 1
> However, in fit2()
> $df.residual
> [1] 0 0 0 0 0
> 68660 more elements ...
>
> $sigma
> [1] NA NA NA NA NA
> 68660 more elements ...
> I looked back to fit() and had the same output.
> However, my design matrix should allow me to compute differences. Does it say somewhere reps are necessary, other times seems to suggest need based on hypothesis.
>
Yes, you need replicates to have residual degrees of freedom and to
perform a statistical test.
> I already computed the average.
Might be obvious, but the average of what?
> I can also compute the stats for this as well, if needed.
> Can I create an object of this to use with limma?
Again, can your clarify what data you mean to use with limma?
Sean
> On the other hand,
> $qr
> Ambrosia Gala Melrose
> 1 -1 0 0
> 2 0 -1 0
> 3 0 0 1
>
> Also, this contrastMatrix doesn't fit.
>
> Regards,
> Franklin
>
>
>
>
>
>
> -- output of sessionInfo():
>
> R 2.15.1
>> objects()
> [1] "constrast.matrix" "contrastNames" "contrastsMatrix"
> [4] "design" "eFBestN" "eFBestNlog2t"
> [7] "eSetlog2_eFBestN" "exprs" "fit"
> [10] "fit2" "geneID" "HistogramPlot"
> [13] "limmaGUIenvironment" "log2_eFBestN" "mydesign"
> [16] "myfit" "NEOffsetDefault" "numParameters"
> [19] "parameterNames" "pDataN" "pDatarootstocKlog2"
> [22] "pDatarootstocNlog2" "pDatascioNlog2" "phenoDatN"
> [25] "SampleNames" "scion.phenoData" "ss.rootstock.eFBestNlog2t"
> [28] "ss.rootstock.log2" "ss.rootstock.log2t" "ss.scion.eFBestNlog2t"
> [31] "ss.scion.log2" "ss.scion.log2t" "targets"
> #####################
> function (package = NULL)
> {
> z <- list()
> z$R.version <- R.Version()
> z$platform <- z$R.version$platform
> if (nzchar(.Platform$r_arch))
> z$platform <- paste(z$platform, .Platform$r_arch, sep = "/")
> z$platform <- paste(z$platform, " (", 8
>
>
> --
> Sent via the guest posting facility at bioconductor.org.
>
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