[R] a fast way to do my job

[Bert Gunter]

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 using 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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