implement frev() - base::rev() that allocates less #5885
Description
Since we know now how costly negation is, what about reverting in-place when possible? Reverting with copy costs 1/2 of `rev`. Reverting in-place costs 1/10 of `rev` (no parallelization).
x = rnorm(1e8)
system.time(t1 <- rev(x))
# user system elapsed
# 0.481 0.212 0.693
system.time(t2 <- reverse(x, copy=TRUE))
# user system elapsed
# 0.117 0.216 0.334
system.time(t3 <- reverse(x, copy=FALSE))
# user system elapsed
# 0.060 0.000 0.061
identical(t1, t2)
identical(t1, t3)
x = rnorm(1e8)
n = rep(1e2, 1e8)
system.time(a1 <- frollmax(x, n, align="left", adaptive=TRUE))
# user system elapsed
# 14.330 3.147 7.669
system.time(a2 <- reverse(frollmax(reverse(x, copy=TRUE), reverse(n, copy=TRUE), align="right", adaptive=TRUE), copy=FALSE))
# user system elapsed
# 9.092 0.807 2.687
identical(a1, a2)Originally posted by @ben-schwen in #5441 (comment)