I tried to solve the task via following code:
manyorders <- replicate(100, sample(colnames(dat)), simplify=FALSE)
all_results <- lapply(manyorders, function(ord) {
tmpdat <- `colnames<-`(dat, ord) # copies and renames in one line
corfit <- duplicateCorrelation(dat, block = targets$Subject)
corfit$consensus.correlation
fit
<-lmFit(dat,design,block=targets$Subject,correlation=corfit$consensus.correlation)
fit <- eBayes(fit)
y1 <- topTable(fit, coef="TreatT", n=nrow(genes),
adjust.method="BH", genelist=genes)
list(fit = fit, y1 = y1)
})
and I wrote all_results in a file
write.table(all_results, file="all_res", sep = " ", row.names = FALSE,
col.names = TRUE,quote=FALSE)
when I tried to open the file:
> a=read.table("all_res", header=T)
Error in read.table("all_res", header = T) :
more columns than column names
also with fread:
> a=fread("all_res")
Warning messages:
1: In fread("all_res") :
Detected 35300 column names but the data has 35700 columns (i.e.
invalid file). Added 400 extra default column names at the end.
2: In fread("all_res") :
Stopped early on line 5. Expected 35700 fields but found 35400.
Consider fill=TRUE and comment.char=. First discarded non-empty line:
<<5.73583235032546 0.182418204566858 0.178323702841331
-1.69234503648485 -1.83571739423166 -1.72136694103431 1.35210840970636
1.37698365727967 1.65643614366521 1.33750809366081 1.22116614774455
1.07000318265432 -1.13789084968265 -1.0757049956716 -1.24824627469937
-0.208441151406013 -0.246971068064524 -0.272463550264579
-0.561691522816148 -0.407713100376217 -0.538071283637418
-0.979274581542649 -0.909855772342568 -0.974827213844384
0.21700425934175 0.134586251989901 0.0818947419096577 -0.6788584605>>
my original dat matrix has 132 columns and around 15000 rows
Can you please advise on this?
Thanks
Ana
On Tue, Feb 4, 2020 at 4:45 PM Bert Gunter <bgunter.4...@gmail.com> wrote:
>
> 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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