See ?lm.fit.
I must be missing something, because:
results <- sapply(74:35164, \(i) residuals(lm.fit(purity2,
gem751be.rpkm[, i] )))
would give you a 751 x 35091 matrix of the residuals from each of the
regressions.
I assume it will be considerably faster than all the overhead you are
carrying in your current code, but of course you'll have to try it and
see. ... Assuming that I have interpreted your request correctly.
Ignore if not.
Cheers,
Bert
On Fri, Aug 9, 2024 at 4:50 PM Yuan Chun Ding via R-help
<r-help@r-project.org> wrote:
>
> Dear R users,
>
> I am running the following code below, the gem751be.rpkm is a dataframe with
> dim of 751 samples by 35164 variables, 73 phenotypic variables in the furst
> to 73rd column and 35091 genomic variables or genes in the 74th to 35164th
> columns. What I need to do is to calculate the residuals for each gene using
> the simple linear regression model of genelist[i] ~ purity2;
>
> The following code is running, it takes long time, but I have an expensive
> ThinkStation window computer.
> Can you provide a fast way to do it?
>
> Thank you,
>
> Ding
>
> ---------------------------------------------------------------------------------
>
>
> gem751be.rpkm <-merge(gem751be10, as.data.frame(t(rna849.fpkm2)),
> + by.x="id2",by.y=0)
> > row.names(gem751be.rpkm)<-gem751be.rpkm$id3
> >
> > colnames(gem751be.rpkm)<-gsub(colnames(gem751be.rpkm),pattern="-",replacement="_")
> > genelist <- gem751be.rpkm %>% dplyr::select(74:35164)
> > residuals <- NULL
> > for (i in 1:length(genelist)) {
> + #i=1
> + formula <- reformulate("purity2", response=names(genelist)[i])
> + model <- lm(formula, data = gem751be.rpkm)
> + resi <- as.data.frame(residuals(model))
> + colnames(resi)[1]<-names(genelist)[i]
> + resi <-as.data.frame(t(resi))
> + residuals <- rbind(residuals, resi)
> + }
>
>
>
> ----------------------------------------------------------------------
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