How to copy a value in a vector to next position(s

问题:

I have a vector that looks something like this:

c(0.5,0,0,0,0,0.7,0,0,0,0,0.4,0,0,0,0)

Suppose I want to copy the values on positions 1, 6 and 11 (the ones that are not 0) to the four positions following that specific value, to make the vector look like this:

c(0.5,0.5,0.5,0.5,0.5,0.7,0.7,0.7,0.7,0.7,0.4,0.4,0.4,0.4,0.4)

How could I best do that in R?

Many thanks!

回答1:

I'd probably loop through every single element greater 0 using lapply, then apply rep function to repeat each of these values 5 times and merge the resulting list entries via do.call("c", ...).

do.call("c", lapply(which(tmp > 0), function(i) rep(tmp[i], 5)))
[1] 0.5 0.5 0.5 0.5 0.5 0.7 0.7 0.7 0.7 0.7 0.4 0.4 0.4 0.4 0.4

回答2:

Another possibility:

vec <- c(0.5,0,0,0,0,0.7,0,0,0,0,0.4,0,0,0,0)

library(zoo)
vec[vec==0] <- NA
na.locf(vec)
#[1] 0.5 0.5 0.5 0.5 0.5 0.7 0.7 0.7 0.7 0.7 0.4 0.4 0.4 0.4 0.4

回答3:

Here's one way:

zero.locf <- function(x) {
    if (x[1] == 0) stop('x[1] should not be 0')
    with(rle(x), {
        no.0 <- replace(values, values == 0, values[(values == 0) - 1])
        rep(no.0, lengths)
    })
}
x <- c(0.5,0,0,0,0,0.7,0,0,0,0,0.4,0,0,0,0)
zero.locf(x)
#  [1] 0.5 0.5 0.5 0.5 0.5 0.7 0.7 0.7 0.7 0.7 0.4 0.4 0.4 0.4 0.4

rle(x) returns a list with items values and lengths.

rle(x)
Run Length Encoding
  lengths: int [1:6] 1 4 1 4 1 4
  values : num [1:6] 0.5 0 0.7 0 0.4 0

with opens up this list and lets us reference these entries directly.

回答4:

Here's another base R approach. Initial zeros are left as is:

v = c(0,1,2,-2.1,0,3,0,0.4,0,0)
v[v!=0] = diff(c(0, v[v!=0]))
cumsum(v)
# [1]  0.0  1.0  2.0 -2.1 -2.1  3.0  3.0  0.4  0.4  0.4

And here are some benchmarks:

roland = function(v) {v[v == 0] <- NA; na.locf(v)}
mp = function(x) {with(rle(x), rep(replace(values, values==0, values[which(values == 0)-1]), lengths))}
quant = function(dat) {not.0 <- (dat != 0); approx(which(not.0), dat[not.0], xout = seq(along.with = dat), method = "constant", rule = 2)}
eddi = function(v) {v[v!=0] = diff(c(0, v[v!=0])); cumsum(v)}

v = sample(c(-10:10, 0), 1e6, TRUE)
microbenchmark(roland(v), mp(v), quant(v), eddi(v), times = 10)
#Unit: milliseconds
#      expr      min       lq   median       uq      max neval
# roland(v) 595.1630 625.7692 638.4395 650.4758 664.9224    10
#     mp(v) 410.8224 433.6775 469.9346 496.6328 528.3218    10
#  quant(v) 646.1775 753.0684 759.9805 838.4281 883.3383    10
#   eddi(v) 265.8064 286.2922 316.7022 339.0333 354.0836    10

回答5:

Here is an alternative using approx

dat   <- c(0.5,0,0,0,0,0.7,0,0,0,0,0.4,0,0,0,0)
not.0 <- (dat != 0)
approx(which(not.0), dat[not.0], xout = seq(along.with = dat), method = "constant", yleft = 0, rule = 1:2)
# $x
# [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
#
# $y
# [1] 0.5 0.5 0.5 0.5 0.5 0.7 0.7 0.7 0.7 0.7 0.4 0.4 0.4 0.4 0.4

And here is an alternative that relies on the stated structure of the initial vector (repetitions of a non-zero value followed by 4 zeros). It adresses the speed issue but at the cost of flexibility.

dat <- c(0.5,0,0,0,0,0.7,0,0,0,0,0.4,0,0,0,0)
rep(dat[seq(1, length(dat), by = 5)], each = 5)