I'm getting a vector of numbers as output from one function, and am wanting to drop all the values higher than 2900, then pipe the remainder directly into a second function. (They'll be sorted, if that helps.) Is there a clever way to do this seemingly simple thing without having to stop and define an intermediate variable?

CodePudding user response：

Here is a way without creating a temp vector.

1. The functions f and g are simple test functions that output a sequence of integers from1 to their argument n. Function g assigns NA to half of the output vector.
2. Function h sums its input vector.
3. In the middle of the pipe, there's an anonymous function that subsets the output of f or g and pipes the resulting vector to function h.
4. In the case of the pipe from g, extra code is needed to remove NA's, if that's what the user wants.

f <- function(n) seq.int(n)
g <- function(n){
  y <- seq.int(n)
  is.na(y) <- sample(n, n/2)
  y
}
h <- function(x, na.rm = FALSE) sum(x, na.rm = na.rm)

set.seed(2022)
f(3000) |> (\(x) x[x <= 2900])() |> h()
#> [1] 4206450

set.seed(2022)
g(3000) |> (\(x) x[x <= 2900])() |> h()
#> [1] NA

set.seed(2022)
g(3000) |> (\(x) x[x <= 2900])() |> h(na.rm = TRUE)
#> [1] 2080026

set.seed(2022)
g(3000) |> (\(x) x[which(x <= 2900)])() |> h()
#> [1] 2080026

^{Created on 2022-03-12 by the reprex package (v2.0.1)}

Edit

Following Mikael Jagan's comment, the input can be piped to the first function like below.

input <- 3000
input |> f() |> (\(x) x[x <= 2900])() |> h()
#> [1] 4206450

^{Created on 2022-03-12 by the reprex package (v2.0.1)}