replacing ugly for loops

sq.n <- seq_len(nrow(data.df))
tapply(sq.n,data.df$seq,function(x)with(data.df[x,],

I am not sure you have expressed what you wanjt to do correctly. See inline:

On Wed, Oct 10, 2012 at 9:10 PM, andrewH <ahoerner at rprogress.org> wrote:

I have a couple of hundred American Community Survey Summary Files files
containing rectangular arrays of data, mainly though not exclusively
numeric.  Each file is referred to as a sequence (henceforth "seq").

-- so 1 "seq" (terrible identifier -- see below for why) = 1 file

 From

these files I am trying to extract particular subsets (tables) consisting of
a sets of columns.  These tables are defined by three numbers (now in
columns in a data frame):
1.      a file identifier (seq)
2.      first column position numbers (startNo)
3.      length of table (len)

So your data frame, call it yourframe, has columns named:

seq      startNo       len

so the columns to select for one triple would consist of
startNo:(startNo+length-1).   I am trying to create for each sequence a
vector of all the column numbers for tables in that sequence.

So for each seq id you want to find all the column numbers, right?

sq.n <- seq_len(nrow(yourframe)) ## Just to make it easier to read
colms <-  tapply(sq.n, yourframe$seq,function(x) with(yourframe[x,],
   sort(unique(do.call(c, mapply(seq, from=startNo,
length=len,SIMPLIFY = FALSE)))))

## Comments
In the mapply call, seq is the R function, ?seq.  That's why using it
as a name for a file id is terrible -- it causes confusion.

In the absence of data, this is untested -- and probably not quite
right. But it should be close, I hope. The key idea is the use of
mapply to get the sequence of columns for each row in all the rows for
each seq id. The SIMPLIFY = FALSE guarantees that this yields a list
of vectors of column indices, which are then glopped together and
cleaned up by the sort(unique(do.call(  ...  stuff.

colms should then be a list giving the sorted column numbers to choose
for each "seq" id.

I do not know whether (once cleaned up,) this is either more elegant
or more efficient than what you proposed. And I wouldn't be surprised
if someone like Bill Dunlap comes up with a lot better way, either.
But it is different -- and perhaps amusing.

... If I have properly understood what you wanted. If not, ignore all.

Cheers,
Bert

Obviously I could do this with nested for loops,e.g..

seq <- c(1,1,2,2)
startNo  <- c(3, 10, 3, 15)
len <- c(4, 2, 5, 3)
data.df <- data.frame(seq, startNo, len)

seq.f <- factor(data.df$seq)
data.l <- split(data.df, seq.f)
selectColsList<- vector("list", length(levels(seq.f)))
for (i in seq_along(levels(seq.f))){

   selectCols <- numeric()
       for (j in seq_along(data.l[[i]]$startNo)){
           selectCols <- c(selectCols,
data.l[[i]]$startNo[j]:(data.l[[i]]$startNo[j]
           data.l[[i]]$len[j]-1))
        }
    selectColsList[[i]] <- selectCols
}

selectColsList

[[1]]
[1]  3  4  5  6 10 11
[[2]]
[1]  3  4  5  6  7 15 16 17

But this code strikes me as inelegant and verbose. It seems to me that there
ought to be a way to make the outer loop, (indexed with i) into a tapply
function (which is why I started with a split()), and the inner loop
(indexed with j) into some cute recursive function, but I was not able to do
so. If anyone could suggest some nicer (e.g. shorter, or faster, or just
more sophisticated) way to do this instead, I would be most grateful.

Sincerely, andrewH




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Bert Gunter
Genentech Nonclinical Biostatistics

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