Need Help Understanding A Programming Statment Written In R

Hi,

I have a data frame (selData) on which I'm using a for loop to collect the first 100 rownames and data into a samller data frame (exprs100.row) as follows:

data frame eaxmple:

            hr0       hr6      hr24      hr48      hr72      hr80
 1   0 0.9879942 0.9833003 0.9737801 0.9737969 0.9587732
 2   0 0.9928526 0.9860887 0.9812401 0.9840965 0.9737232
 3   0 0.9956888 1.0116334 1.0155620 1.0247329 1.0252338

for loop:

for(i in 0:99){
indx = rank(modF) == nrow(selData) - i
exprs.row = selDataMatrix_ave.log.ratio[indx,]
 exprs100.row[[i]] <- exprs.row[i]
 }

The statement that I have problem with is :

exprs100.row[[i]] <- exprs.row[i]

Could someone please help with the correct way of writing it. Thanks

r bioconductor microarray

• 3,320 views • link

updated 12.8 years ago by Istvan Albert

• written 12.8 years ago by robjohn7000

A brief introduction to “apply” in R

• 199 views • link

written 12.8 years ago by zx8754

You can try these also: 
1) head(selData,100) 
2) selData[1:100,]

• 100 views • link

written 12.8 years ago by always_learning

please read ?'[' and read a basic introduction to R first. R is different from many other programming languages. As a rule of thumb in R you don't need to use for loops almost anywhere, but instead vectorized functions, sub setting, or apply. Using loops is clumsy and extremely slow compared to other options.

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written 12.8 years ago by Michael Dondrup

In addition to all other comments: R is 1 indexed

• 149 views • link

written 12.8 years ago by Ido Tamir

1 answer

This is actually just an R usage question, it has nothing to do with bioinformatics.
What exactly isn't working (i.e., what is the current output and what is the expected output)?
If you just want to subset selDataMatrix_ave.log.ratio, wouldn't it be easier to just exprs100.row <- selDataMatrix_ave.log.ratio[c(1:100),] without a for loop (which is slow in R)?

• 125 views • link

written 12.8 years ago by Devon Ryan

Thanks. The trouble is that the output has to be based on index "indx = rank(modF)" .

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written 12.8 years ago by robjohn7000

Then something like

indx <- which(rank(modF) %in% c(nrow(selData):(nrow(selData)-99)))
exprs100.row <- selDataMatrix_ave.log.ratio[indx,]

without any loops or such is a better way of going. BTW, in the future, please say exactly what you're trying to do, that way we can give more appropriate feedback.

• 199 views • link

written 12.8 years ago by Devon Ryan