For each row in an R dataframe

Question

You can use the by() function:

by(dataFrame, seq_len(nrow(dataFrame)), function(row) dostuff)

But iterating over the rows directly like this is rarely what you want to; you should try to vectorize instead. Can I ask what the actual work in the loop is doing?

You can try this, using apply() function

> d
  name plate value1 value2
1    A    P1      1    100
2    B    P2      2    200
3    C    P3      3    300

> f <- function(x, output) {
 wellName <- x[1]
 plateName <- x[2]
 wellID <- 1
 print(paste(wellID, x[3], x[4], sep=","))
 cat(paste(wellID, x[3], x[4], sep=","), file= output, append = T, fill = T)
}

> apply(d, 1, f, output = 'outputfile')

First, Jonathan's point about vectorizing is correct. If your getWellID() function is vectorized, then you can skip the loop and just use cat or write.csv:

write.csv(data.frame(wellid=getWellID(well$name, well$plate), 
         value1=well$value1, value2=well$value2), file=outputFile)

If getWellID() isn't vectorized, then Jonathan's recommendation of using by or knguyen's suggestion of apply should work.

Otherwise, if you really want to use for, you can do something like this:

for(i in 1:nrow(dataFrame)) {
    row <- dataFrame[i,]
    # do stuff with row
}

You can also try to use the foreach package, although it requires you to become familiar with that syntax. Here's a simple example:

library(foreach)
d <- data.frame(x=1:10, y=rnorm(10))
s <- foreach(d=iter(d, by='row'), .combine=rbind) %dopar% d

A final option is to use a function out of the plyr package, in which case the convention will be very similar to the apply function.

library(plyr)
ddply(dataFrame, .(x), function(x) { # do stuff })

I think the best way to do this with basic R is:

for( i in rownames(df) )
   print(df[i, "column1"])

The advantage over the for( i in 1:nrow(df))-approach is that you do not get into trouble if df is empty and nrow(df)=0.