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
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______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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