Generate an Incrementally Increasing Sequence Like 112123123412345

Generate an incrementally increasing sequence like 112123123412345

Use sequence:

> sequence(1:5)
 [1] 1 1 2 1 2 3 1 2 3 4 1 2 3 4 5

Fastest way to make expanding sequence

You can do

sequence(1:n)

It is short and beats everything else... (so far). I also find

sequence(sequence(n))

somewhat satisfying. It is the same speedwise.

How to create an vector with incremental values based on values of another vector?

You can use sequence to generate the numbers.

sequence(v1)
# [1] 1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 5

And with a loop using lapply:

unlist(lapply(v1, seq))

How can I remove one number each time while replicating a number of sequence in r?

I am sure there is a fancier way, but the following code achieves the goal in base R.

out = integer()
a = 2:6
while( length(a) > 1 ) {
  a = a[-length(a)]
  out = c(out, a)
}
out
#>  [1] 2 3 4 5 2 3 4 2 3 2

^{Created on 2021-03-18 by the reprex package (v1.0.0)}

Generate sequence (1,301,2, 302, 3, 303, 4,304 ...) in R

Use rbind to alternate between the values

as.vector(rbind(1:300, 301:600))

Generate series 1, 2,1, 3,2,1, 4,3,2,1, 5,4,3,2,1

With sequence:

rev(sequence(5:1))
# [1] 1 2 1 3 2 1 4 3 2 1 5 4 3 2 1

From R 4.0.0 sequence takes arguments from and by:

sequence(1:5, from = 1:5, by = -1)
# [1] 1 2 1 3 2 1 4 3 2 1 5 4 3 2 1

Far from the golf minimalism of rev... However, if you wake up one morning and want to create such a sequence with n = 1000 (like in the answer below), the latter is in fact faster (but I can hear Brian Ripley in fortunes::fortune(98))

n = 1000

microbenchmark(
  f_rev = rev(sequence(n:1)),
  f_seq4.0.0 = sequence(1:n, from = 1:n, by = -1))
# Unit: microseconds
#        expr   min     lq     mean  median     uq    max neval
#       f_rev 993.7 1040.3 1128.391 1076.95 1133.3 1904.7   100
#  f_seq4.0.0 136.4  141.5  153.778  148.25  150.1  304.7   100

generate increasing sequence in R

Try:

rep(1:62,each=96)

length(rep(1:62,each=96))
[1] 5952

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