[R] doing 1000 permutations and doing test statistics distribution
Duncan Murdoch
murdoch@dunc@n @end|ng |rom gm@||@com
Tue Feb 4 22:41:53 CET 2020
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.4567 using 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.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.
>
> ______________________________________________
> 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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