[R] doing 1000 permutations and doing test statistics distribution

[Bert Gunter]

I am not going to do your programming for you. If the following doesn't
suffice, maybe someone else will provide you something that will.

m = your matrix

code = your code that uses m

your list of results <- lapply(seq_len(1000), FUN = function(m){
  m <- m[, sample.int(132)]
 code
} )

or use an explicit for() loop to populate a list or vector with your
results.

Again, if I have misunderstood what you want to do, then clarify, and
perhaps someone else will help.

-- Bert



Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Tue, Feb 4, 2020 at 12:41 PM Ana Marija <sokovic.anamarija using gmail.com>
wrote:

> Hi Bert,
>
> thanks for getting back to me. I have to permute those 132 columns
> 1000 times and perform the code given in the previous email.
>
> Can you please show me how you would do that in the loop? This is also
> a huge data set ...
>
> Thanks
> Ana
>
> On Tue, Feb 4, 2020 at 2:34 PM Bert Gunter <bgunter.4567 using gmail.com> wrote:
> >
> > If you just want to permute columns of a matrix,
> >
> > ?sample
> > > sample.int(10)
> >  [1]  9  2 10  8  4  6  3  1  5  7
> >
> > and you can just use this as an index into the columns of your matrix,
> presumably within a loop of some sort.
> >
> > If I have misunderstood, just ignore.
> >
> > Cheers,
> > Bert
> >
> >
> >
> >
> > On Tue, Feb 4, 2020 at 12:23 PM Ana Marija <sokovic.anamarija using gmail.com>
> wrote:
> >>
> >> Hello,
> >>
> >> I have a matrix
> >> > dim(dat)
> >> [1] 15568   132
> >>
> >> It looks like this:
> >>
> >>                    NoD_14381_norm.1 NoD_14381_norm.2 NoD_14381_norm.3
> >> NoD_14520_30mM.1 NoD_14520_30mM.2 NoD_14520_30mM.3
> >> Ku8QhfS0n_hIOABXuE             4.75             4.25             4.79
> >>            4.33             4.63             3.85
> >> Bx496XsFXiAlj.Eaeo             6.15             6.23             6.55
> >>            6.26             6.24             5.99
> >> W38p0ogk.wIBVRXllY             7.13             7.35             7.55
> >>            7.37             7.36             7.55
> >> QIBkqIS9LR5DfTlTS8             6.27             6.73             6.45
> >>            5.39             4.75             4.96
> >> BZKiEvS0eQ305U0v34             6.35             7.02             6.76
> >>            5.45             5.25             5.02
> >> 6TheVd.HiE1UF3lX6g             5.53             5.02             5.36
> >>            5.61             5.66             5.37
> >>
> >> So it is a matrix with gene names ex. Ku8QhfS0n_hIOABXuE, and subjects
> >> named ex. NoD_14381_norm.1
> >>
> >>
> >> How to do 1000 permutations of these 132 columns and on each created
> >> new permuted matrix perform this code:
> >>
> >> subject="all_replicate"
> >> targets<-readTargets(paste(PhenotypeDir,"hg_sg_",subject,"_target.txt",
> sep=''))
> >> Treat <- factor(targets$Treatment,levels=c("C","T"))
> >> Replicates <- factor(targets$rep)
> >> design <- model.matrix(~Replicates+Treat)
> >> corfit <- duplicateCorrelation(dat, block = targets$Subject)
> >> corfit$consensus.correlation
> >> fit
> <-lmFit(dat,design,block=targets$Subject,correlation=corfit$consensus.correlation)
> >> fit<-eBayes(fit)
> >> qval.cutoff=0.1; FC.cutoff=0.17
> >> y1=topTable(fit, coef="TreatT",
> n=nrow(genes),adjust.method="BH",genelist=genes)
> >>
> >> y1 for each iteration of permutation would  have P.Value column and
> >> these I would have plotted on the end to find the distribution of all
> >> p values generated in those 1000 permutations.
> >>
> >> Please advise,
> >> Ana
> >>
> >> ______________________________________________
> >> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
>

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