Sequence expansion question

Question

I have a sequence of 'endpoints', e.g.:

c(7,10,5,11,15)     

that I want to expand to a sequence of 'elapsed time' between the endpoints, e.g.

c(7,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,1,2,3,4,5,6,7,8,9,10,11,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)

Whats the most efficient way to do this in R? I'm imagining some creative use of the embed function, but I can't quite get there without using a ugly for loop.

Here's the naive way to do this:

expandSequence <- function(x) {
    out <- x[1]
    for (y in (x[-1])) {
        out <- c(out,seq(1,y))
    }
    return(out)
}

expandSequence(c(7,10,5,11,15))

Corpse

Thanks for a nicely worded and reproducible question. – Steam

Answer 1

There is a base function to do this, called, wait for it, sequence:

sequence(c(7,10,5,11,15))

 [1]  1  2  3  4  5  6  7  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  1  2  3
[26]  4  5  6  7  8  9 10 11  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15

In your case it seems your first endpoint is in fact not part of the sequence, so it becomes:

c(7, sequence(c(10,5,11,15)))
 [1]  7  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  1  2  3  4  5  6  7  8  9
[26] 10 11  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15

Leatrice

+1 Nice catch. It is implemented effectively as @gsk and I reinvented. I'd come across sequence before but presumed it was an alias for seq. Great to find out about these less well-used functions! – April

@Andrie: Thanks! For my next trick, I will reverse engineer the embed function. – Corpse

Answer 2

How about this:

> unlist(sapply(x,seq))
 [1]  1  2  3  4  5  6  7  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  1  2
[25]  3  4  5  6  7  8  9 10 11  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15

With the first element added on at the end:

c( x[1], unlist( sapply( x[seq(2,length(x))], seq ) ) )

And a slightly more readable version:

library(taRifx)
c( x[1], unlist( sapply( shift(x,wrap=FALSE), seq ) ) )

Steam

Sadly my solution doesn't involve bolded numbers. ;-) – Steam

Perfect! I just got to do.call(c,mapply(seq,1,x)) but your solution is better. – Corpse

Except that isn't the output shown in the Q. The first element should be 7 according to the Q. You have 1:7. – April

So to be complete about this you'd need to omit the first element of x from the sapply call and then add it back on to the front afterwards, to get the OP's desired output, right? – Devoir

If that is what the OP wants. He has commented here that the output is "perfect", so perhaps the Q is not exactly what was wanted? – April

@Gavin Simpson I added the "omit first element of x and add it back at the end" myself – Corpse

Answer 3

A combination of lapply() and seq_len() is useful here:

expandSequence <- function(x) {
    out <- lapply(x[-1], seq_len)
    do.call(c, c(x[1], out))
}

Which gives for

pts <- c(7,10,5,11,15)

> expandSequence(pts)
 [1]  7  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  1  2  3  4
[21]  5  6  7  8  9 10 11  1  2  3  4  5  6  7  8  9 10 11 12 13
[41] 14 15

(An alternative is:

expandSequence <- function(x) {
    out <- lapply(x[-1], seq_len)
    unlist(c(x[1], out), use.names = FALSE)
}

)

April

I like reinventing the wheel! – April