3

I want to select items by index from a list before applying another function to it using purrr:map. I have tried the following, but can't find a way that works.

require(dplyr)
require(purrr)

dat <- list(1:3,
     4:6,
     letters[1:3])

# I can select one item

dat[1]

# I can select two items

dat[c(1,2)]

# But how can I do this in a pipeline by index?

dat %>% map(mean)

dat %>%
  filter(c(1,2)) %>%
  map(mean)

dat %>%
  keep(1,2) %>%
  map(mean)

dat %>%
  select(1,2) %>%
  map(mean)
markus
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Wietze314
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2 Answers2

10

We can use `[` and do

dat %>% 
 .[c(1, 2)] %>%
 map(., mean)
#[[1]]
#[1] 2

#[[2]]
#[1] 5

Or define an alias in the way the magrittr package does it

extract <- `[` # literally the same as magrittr::extract
dat %>% 
  extract(c(1, 2)) %>% 
  map(., mean)

Which could also be written as

dat %>% `[`(c(1,2))

Using baseR pipe operator this would read

dat |> 
  `[`(x = _, j = c(1,2)) |> # R version >= R 4.2.0
  lapply(mean)
#[[1]]
#[1] 2
#
#[[2]]
#[1] 5
markus
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4

An option is

library(tidyverse)
keep(dat, seq_along(dat) %in% 1:2) %>% 
            map(mean)
#[[1]]
#[1] 2

#[[2]]
#[1] 5

Or map with pluck

map(1:2, ~ pluck(dat, .x) %>% 
               mean)

Or with assign_in

assign_in(dat, 3, NULL) %>% 
      map(mean)

Or another option is map_if

map_if(dat, is.numeric, mean, .else = ~ NULL) %>% 
         discard(is.null)

Or with discard

discard(dat, is.character) %>% 
        map(mean)

or with Filter and map

Filter(is.numeric, dat) %>% 
     map(mean)

NOTE: All of them gets the expected output.

akrun
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