R - 合并 SAME 列表中的两个元素。优先选择 purrr 解决方案

Question

我正在为以下问题寻找 purrr 解决方案：

说，我们有一些列表：

    list(     c("Hello",      "Well",   "You"  ),
              c("again",      "done,",  "annoy"),
              c("my friend!", "boy!",   "me!"  )      )

现在，我想合并该列表中的前两个元素。

我想要的输出是：

    list(     c("Hello",      "Well",   "You"  , "again",      
               "done,",  "annoy"),
              c("again",      "done,",  "annoy"),
              c("my friend!", "boy!",   "me!"  )      )

感谢您的帮助！谢谢

Answer 1

子集前两个 list 元素，c与 do.call 连接并将 (<-) 分配回第一个元素

lst1[[1]] <- do.call(c, lst1[1:2])

-输出

> lst1
[[1]]
[1] "Hello" "Well"  "You"   "again" "done," "annoy"

[[2]]
[1] "again" "done," "annoy"

[[3]]
[1] "my friend!" "boy!"       "me!"

有了purrr，我们可以用modify_in

library(purrr)
modify_in(lst1, 1, ~ flatten_chr(lst1[1:2]))
[[1]]
[1] "Hello" "Well"  "You"   "again" "done," "annoy"

[[2]]
[1] "again" "done," "annoy"

[[3]]
[1] "my friend!" "boy!"       "me!"

数据

lst1 <- list(c("Hello", "Well", "You"), c("again", "done,", "annoy"), 
    c("my friend!", "boy!", "me!"))

Answer 2

我认为您不需要 purrr 解决方案，但如果您出于某些工作流程原因坚持...

x <- list(c("Hello", "Well", "You"),
          c("again", "done,", "annoy"),
          c("my friend!", "boy!", "me!"))

library(purrr)
modify_at(x, 1, ~ c(., x[[2]]))

# which can simplify to...
x %>% 
  modify_at(1, c, .[[2]])

# or with more purrr!!
x %>% 
  modify_at(1, c, pluck(., 2))

但我只想做...

x[[1]] <- c(x[[1]], x[[2]])

R - 合并 SAME 列表中的两个元素。优先选择 purrr 解决方案

R - Combine two elements within the SAME list. Preferentially a purrr solution

dictionary

r

purrr

tidyverse

数据