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@jordi-altirriba-gutierrez-682

Last seen 3.5 years ago

Thank you very much for your prompt answer Denise!
Yes, you are right, the command FUN=mainES is wrong, I did this
mistake when
I copied the commands to the mail (sorry). Therefore, the other
commands are
correct.
But there is something that I don't understand (sorry if it is a very
basic
question), because when I want to get the genes that are
differentially
expressed due to diabetes I "fit" my data to the functions: lm(y ~
DIABETES
+ TREATMENT + DIABETES * TREATMENT) and lm(y ~ TREATMENT). Therefore
the
genes that "fit" better to the first function are differentially
expressed
due to diabetes, but why don't I fit my data to the functions: lm(y ~
DIABETES + TREATMENT + DIABETES * TREATMENT) and lm(y ~ TREATMENT +
DIABETES
* TREATMENT)? I know that the parameter DIABETES * TREATMENT is the
intersection of the other two parameters, but it should be independent
of
these parameters.
Thank you very much for your comments and help! (and your patience)
Jordi Altirriba
IDIBAPS - Hospital Clinic (Barcelona, Spain)
>
>Hello Jordi,
>
>My comments to your questions are below. I hope this helps. -Denise
>
>_____________________________________________________________________
_____
>Denise Scholtens
>Department of Biostatistics
>Harvard School of Public Health
>dscholte@hsph.harvard.edu
>
>Hi all!
>I?ve been using RMA and LIMMA to analyse my data and I am currently
trying
>to analyse it with the package factDesign. My design is a 2x2
factorial
>design with 4 groups: diabetic treated, diabetic untreated, health
treated
>and health untreated with 3 biological replicates in each group. I
want to
>know what genes are differentially expressed due to diabetes, to the
>treatment and to the combination of both (diabetes + treatment).
>My phenoData is:
> >pData(eset)
> DIABETES TREATMENT
>DNT1 TRUE FALSE
>DNT2 TRUE FALSE
>DNT3 TRUE FALSE
>DT1 TRUE TRUE
>DT2 TRUE TRUE
>DT3 TRUE TRUE
>SNT1 FALSE FALSE
>SNT2 FALSE FALSE
>SNT3 FALSE FALSE
>ST1 FALSE TRUE
>ST2 FALSE TRUE
>ST3 FALSE TRUE
>
>Are these commands correct to get the results listed below? Where are
the
>errors?
> >lm.full<-function(y) lm(y ~ DIABETES + TREATMENT + DIABETES *
TREATMENT)
> >lm.diabetes<-function(y) lm(y ~ DIABETES)
> >lm.treatment<-function(y) lm(y ~ TREATMENT)
> >lm.diabetestreatment<-function(y) lm(y ~ DIABETES + TREATMENT)
> >lm.f<-esApply(eset, 1, lm.full)
> >lm.d<-esApply(eset, 1, lm.diabetes)
> >lm.t<-esApply(eset, 1, lm.treatment)
> >lm.dt<-esApply(eset, 1, lm.diabetestreatment)
>
>#####
># Yes, these commands look correct for making the linear models and
># running them for the exprSet called eset.
>######
>
>## To get the genes characteristics of the treatment:
> >Fpvals<-rep(0, length(lm.f))
> >for (i in 1:length(lm.f)) {Fpvals[i]<-anova(lm.d[[i]],
lm.f[[i]])$P[2]}
> >Fsub<-which(Fpvals<0.01)
> >eset.Fsub<-eset[Fsub]
> >lm.f.Fsub<-lm.f[Fsub]
> >betaNames<-names(lm.f[[1]] [["coefficients"]])
> >lambda<-par2lambda(betaNames, c("TREATMENTTRUE"), c(1)) ## I get
the same
> >genes if I write : > lambda2 <- par2lambda (betaNames,
> >list(c("TREATMENTTRUE" , "DIABETESTRUE:TREATMENTTRUE")),list(
c(1,1)))
> >mainTR<-function(x) contrastTest(x,lambda,p=0.1) [[1]]
> >mainTRgenes<-sapply(lm.f.Fsub, FUN=mainES)
>
>#####
># I think the problem is the use of mainES rather than mainTR in the
last
># sapply. mainES is a function that is defined in the factDesign
vignette
># - your own function should be used here instead. If you define the
># function differently for different contrasts, my guess is you will
see
># different gene lists for the lambda and lambda2 defined above.
>#####
>
>
>## To get the genes characteristics of the diabetes:
> >for (i in 1:length(lm.f)) {Fpvals[i]<-anova(lm.t[[i]],
lm.f[[i]])$P[2]}
> >Fsub<-which(Fpvals<0.01)
> >eset.Fsub<-eset[Fsub]
> >lm.f.Fsub<-lm.f[Fsub]
> >betaNames<-names(lm.f[[1]] [["coefficients"]])
> >lambda<-par2lambda(betaNames, c("DIABETESTRUE"), c(1)) ## I get
also the
> >same genes if I consider the intersection
DIABETESTRUE:TREATMENTTRUE.
> >mainDI<-function(x) contrastTest(x,lambda,p=0.1) [[1]]
> >mainDIgenes<-sapply(lm.f.Fsub, FUN=mainES)
>
>#####
># See above comments.
>#####
>
>## To get the genes characteristics of the diabetes+treatment:
> >for (i in 1:length(lm.f)) {Fpvals[i]<-anova(lm.dt[[i]],
lm.f[[i]])$P[2]}
> >Fsub<-which(Fpvals<0.01)
> >eset.Fsub<-eset[Fsub]
> >lm.f.Fsub<-lm.f[Fsub]
> > betaNames<-names(lm.f[[1]] [["coefficients"]])
> >lambda<-par2lambda(betaNames, c("DIABETESTRUE:TREATMENTTRUE"),
c(1))
> >mainDT<-function(x) contrastTest(x,lambda,p=0.1) [[1]]
> >mainDTgenes<-sapply(lm.f.Fsub, FUN=mainES) ## I don?t get any ?fail
to
> >reject? gene.
>
>#####
># Again, I think changing mainES to mainDT will do the trick.
>#####
>
>
>When I get the ?rejected? and the ?failed to reject? genes, can I
classify
>them by their Fvalues? How?
>
>#####
># Currently, the contrastTest function only returns the contrast
estimate
># (cEst), the pvalue from the F-test (pvalue), and a statement of
either
># "REJECT" or "FAIL TO REJECT" based on the p-value cutoff you
specify.
># This can be changed to return the F-value as well, and I'm happy to
put
># this change into the package. Then you can use the Fvalues for
whatever
># you would like.
>#
># One thing to consider if you are going to use p-values from the F
tests
># to select genes - you will want to corrent for multiple testing.
The
># multtest package is very useful for this.
>######
>
>
>Thank you very much for your comments and help.
>Yours sincerely,
>
>Jordi Altirriba
>IDIBAPS-Hospital Clinic (Barcelona, Spain)
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