cumsum(rep(c(2,1), 51/3))
probably inefficient though.
Pregunta
I Have a sequence below that has a pattern that does not change. I can create a vector to represent the missing variables to this pattern(below). But I can't seem to figure out a way to print this specific sequence below as a vector. How would one create a vector that shows the sequence below but stops at a specific row(the last row in the pattern), instead of 51? Thanks
bad <- seq(1,51,by=3)
2,3,5,6,8,9,11,12,14,15,17,18,20,21,23,24,26,27,29,30,32,33,35,36,38,39,41,42,44,45,47,48,50,51
Solución 3
cumsum(rep(c(2,1), 51/3))
probably inefficient though.
Otros consejos
The most straightforward way I can think of is to use "recycling" of a logical vector:
(1:51)[c(FALSE, TRUE, TRUE)]
# [1] 2 3 5 6 8 9 11 12 14 15 17 18 20 21 23 24 26 27 29 30 32 33 35 36 38 39
# [27] 41 42 44 45 47 48 50 51
> bad <- 2:51
> bad[!bad %% 3 == 1]
[1] 2 3 5 6 8 9 11 12 14 15 17 18 20 21 23 24 26 27 29 30 32 33 35 36 38 39 41 42 44 45 47 48 50 51
a = 2:51
b = seq(1, 51, by=3)
setdiff(a,b)