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Last seen 10.3 years ago
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.
I already computed the average.
I can also compute the stats for this as well, if needed.
Can I create an object of this to use with limma?
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
--
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