I am trying to generate a vector containing decreasing sequences of increasing length, such as 1, 2,1, 3,2,1, 4,3,2,1, 5,4,3,2,1, i.e.
c(1, 2:1, 3:1, 4:1, 5:1)
I tried to use a loop for this, but I don't know how to stack or concatenate the results.
for (i in 1:11)
{
x = rev(seq(i:1))
print(x)
}
[1] 1
[1] 2 1
[1] 3 2 1
[1] 4 3 2 1
[1] 5 4 3 2 1
[1] 6 5 4 3 2 1
[1] 7 6 5 4 3 2 1
[1] 8 7 6 5 4 3 2 1
[1] 9 8 7 6 5 4 3 2 1
[1] 10 9 8 7 6 5 4 3 2 1
[1] 11 10 9 8 7 6 5 4 3 2 1
I have also been experimenting with the rep, rev and seq, which are my favourite option but did not get far.
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