Producing numeric sequences in R using standard patterns

Question

I am working on a project where I need to enter a number of "T score" tables into R. These are tables used to convert raw test scores into standardized values. They generally follow a specific pattern, but not one that is simple. For instance, one pattern is:

34,36,39,42,44,47,50,52,55,58,60,63,66,68,
71,74,76,79,82,84,87,90,92,95,98,100,103,106

I'd prefer to use a simple function to fill these in, rather than typing them by hand. I know that the seq() function can create a simple seqeuence, like:

R> seq(1,10,2)
[1] 1 3 5 7 9

Is there any way to create more complex sequences based on specific patterns? For instance, the above data could be done as:

c(34,seq(36:106,c(3,3,2)) # The pattern goes 36,39,42,44,47,50,52 (+3,+3,+2)

...however, this results in an error. I thought there would be a function that should do this, but all my Google-fu has just brought me back to the original seq().

Steal

Thanks for all the great answers. cumsum() looks like it will be the best choice! – Steal

Answer 1

This could be done using the cumsum (cumulative sum) function and rep:

> 31 + cumsum(rep(c(3, 2, 3), 9))
 [1]  34  36  39  42  44  47  50  52  55  58  60  63  66  68  71  74  76  79  82
[20]  84  87  90  92  95  98 100 103

To make sure sure the sequence stops at the right place:

> (31 + cumsum(rep(c(3, 2, 3), 10)))[1:28]
 [1]  34  36  39  42  44  47  50  52  55  58  60  63  66  68  71  74  76  79  82
[20]  84  87  90  92  95  98 100 103 106

Eastnortheast

Answer 2

Here is a custom function that should work in most cases. It uses the cumulative sum (cumsum()) of a sequence, and integer division to calculate the length of the desired sequence.

cseq <- function(from, to, by){
  times <- (to-from) %/% sum(by)
  x <- cumsum(c(from, rep(by, times+1)))
  x[x<=to]
}

Try it:

> cseq(36, 106, c(3,3,2))
 [1]  36  39  42  44  47  50  52  55  58  60  63  66  68  71  74  76  79  82  84  87  90  92  95  98
[25] 100 103 106

> cseq(36, 109, c(3,3,2))
 [1]  36  39  42  44  47  50  52  55  58  60  63  66  68  71  74  76  79  82  84  87  90  92  95  98
[25] 100 103 106 108

Acquisitive

Answer 3

Here is a non-iterative solution, in case you need a specific element of the sequence

f <- function(x){
 d <- (x) %/% 3
 r <- x %% 3
 31 + d*8 + c(0,3,5)[r+1]
}

> f(1:10)
 [1] 34 36 39 42 44 47 50 52 55 58

Hannigan