splitting up ranges - r

Question

Say I have some ranges represented by start coordinates start<-c(1,2,3) and end coordiantes end<-c(4,5,4) ;ranges<-data.frame(start,end) How can I split this up into one length intervals?
i.e. I want
this
starts ends
1 1 4
2 2 5
3 3 4
to be transformed into this:
starts ends
1 1 2 |
2 3 4 <-end of original first interval
3 2 3 |
4 4 5 <-end of original second interval
5 3 4 <-end of original third interval
right now I have a for loop iterating through the list and creating a sequence sequence that goes from start to end but this loop takes a very long time to execute for long lists of ranges.

Answer 1

Here's one way. It's a "glorified for-loop" in the disguise of lapply on a sequence.
# Your sample data
ranges<-data.frame(start=c(1,2,3),end=c(4,5,4))
# Extract the start/end columns
start <- ranges$start
end <- ranges$end
# Calculate result data
res <- lapply(seq_along(start), function(i) start[i]+seq(0, end[i]-start[i]))
# Make it into a data.frame by way of a matrix (which has a byrow argument)
newRanges <- as.data.frame( matrix(unlist(res), ncol=2, byrow=TRUE, dimnames=list(NULL, names(ranges))) )
Which gives the correct result:
> newRanges
start end
1 1 2
2 3 4
3 2 3
4 4 5
5 3 4
And then time it on a bigger problem:
n <- 1e5
start <- sample(10, n, replace=TRUE)
end <- start + sample( 3, n, replace=TRUE)*2-1
system.time( newRanges <- as.data.frame( matrix(unlist(lapply(seq_along(start), function(i) start[i]+seq(0, end[i]-start[i]))), ncol=2, byrow=TRUE) ) )
This takes about 1.6 seconds on my machine. Good enough?
...The trick is to work on the vectors directly instead of on the data.frame. And then build the data.frame at the end.
Update #Ellipsis... commented that lapply is no better than a for-loop. Let's see:
system.time( a <- unlist(lapply(seq_along(start), function(i) start[i]+seq(0, end[i]-start[i]))) ) # 1.6 secs
system.time( b <- {
res <- vector('list', length(start))
for (i in seq_along(start)) {
res[[i]] <- start[i]+seq(0, end[i]-start[i])
}
unlist(res)
}) # 1.8 secs
So, not only is the for-loop about 12% slower in this case, it is also much more verbose...
UPDATE AGAIN!
#Martin Morgan suggested using Map, and it is indeed the fastest solution yet - faster than do.call in my other answer. Also, by using seq.int my first solution is also much faster:
# do.call solution: 0.46 secs
system.time( matrix(do.call('c', lapply(seq_along(start), function(i) call(':', start[i], end[i]))), ncol=2, byrow=TRUE) )
# lapply solution: 0.42 secs
system.time( matrix(unlist(lapply(seq_along(start), function(i) start[[i]]+seq.int(0L, end[[i]]-start[[i]]))), ncol=2, byrow=TRUE) )
# Map solution: 0.26 secs
system.time( matrix(unlist(Map(seq.int, start, end)), ncol=2, byrow=TRUE) )

Answer 2

You could try creating text for the vectors, parse-ing and eval-uating and then using a matrix to create the data.frame:
txt <- paste("c(",paste(ranges$start,ranges$end,sep=":",collapse=","),")",sep="")
> txt
[1] "c(1:4,2:5,3:4)"
vec <- eval(parse(text=txt))
> vec
[1] 1 2 3 4 2 3 4 5 3 4
mat <- matrix(vec,ncol=2,byrow=T)
> data.frame(mat)
X1 X2
1 1 2
2 3 4
3 2 3
4 4 5
5 3 4

Answer 3

Here's another answer based on #James great solution. It avoids paste and parse and is a little bit faster:
vec <- do.call('c', lapply(seq_along(start), function(i) call(':', start[i], end[i])))
mat <- matrix(vec,ncol=2,byrow=T)
Timing it:
set.seed(42)
n <- 1e5
start <- sample(10, n, replace=TRUE)
end <- start + sample( 3, n, replace=TRUE)*2-1
# #James code: 6,64 secs
system.time({
for(i in 1:10) {
txt <- paste("c(",paste(start,end,sep=":",collapse=","),")",sep="")
vec <- eval(parse(text=txt))
mat <- matrix(vec,ncol=2,byrow=T)
}
})
# My variant: 5.17 secs
system.time({
for(i in 1:10) {
vec <- do.call('c', lapply(seq_along(start), function(i) call(':', start[i], end[i])))
mat <- matrix(vec,ncol=2,byrow=T)
}
})