Yes, a clear thinko... Thanks for the correction.
-- Bert
On Tue, Feb 4, 2020 at 1:41 PM Duncan Murdoch <murdoch.dun...@gmail.com>
wrote:
> On 04/02/2020 4:28 p.m., Ana Marija wrote:
> > I tired your code on this simplified data just for say 10 permutations:
> >
> > dat <- read.table(text = " code.1 code.2 code.3 code.4
> > 1 82 93 NA NA
> > 2 15 85 93 NA
> > 3 93 89 NA NA
> > 4 81 NA NA NA",
> > header = TRUE, stringsAsFactors = FALSE)
> >
> > dat2=data.matrix(dat)
> >
> >> result<- lapply(seq_len(10), FUN = function(dat2){
> > + dat2 <- dat2[, sample.int(4)]
> > + print(colnames(dat2))
> > + } )
> > Error in dat2[, sample.int(4)] : incorrect number of dimensions
>
> Yes, Bert did suggest that, but it's obviously wrong. The argument to
> FUN is an element of seq_len(10), it's not the full dataset. Try
>
> result<- lapply(seq_len(10), FUN = function(i){
> dat <- dat2[, sample.int(4)]
> print(colnames(dat))
> } )
>
> Duncan Murdoch
>
> >
> >
> > On Tue, Feb 4, 2020 at 3:10 PM Bert Gunter <bgunter.4...@gmail.com>
> wrote:
> >>
> >> 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.anamar...@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.4...@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.anamar...@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@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.
> >
> > ______________________________________________
> > R-help@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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