在列表列的每一行中获取列表的第一个元素

Question

How can I get rid of the nested lists and only keep the first element of each list in ColumnB ?

ColumnA	ColumnB
first	c(1, 2, 3)
second	c(4, 5, 6)
third	c(7, 8, 9)

It should look like this:

ColumnA	ColumnB
first	1
second	4
third	7

In python , I would try it with a lambda function giving me only the first element of the list.

Answer 1

We can use map to loop over the list column and extract the first element

library(dplyr)
library(purrr)
df1 %>%
    mutate(ColumnB = map_dbl(ColumnB, first))

-output

# A tibble: 3 × 2
  ColumnA ColumnB
  <chr>     <dbl>
1 first         1
2 second        4
3 third         7

---

Or in base R use sapply to loop over the list and extract the first element

df1$ColumnB <- sapply(df1$ColumnB, `[`, 1)

data

df1 <- structure(list(ColumnA = c("first", "second", "third"), ColumnB = list(
    c(1, 2, 3), c(4, 5, 6), c(7, 8, 9))), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -3L))

Answer 2

In case your ColumnB is a real list, then we could also do:

library(tidyr)
library(dplyr)

df1 %>% 
  unnest(ColumnB) %>% 
  group_by(ColumnA) %>% 
  slice(1)

  ColumnA ColumnB
  <chr>     <dbl>
1 first         1
2 second        4
3 third         7

In case your ColumnB is a string, then we could do:

library(dplyr)
library(readr)
df %>% 
  mutate(ColumnB = parse_number(ColumnB))

  ColumnA ColumnB
1   first       1
2  second       4
3   third       7

Answer 3

Here's another method using only dplyr :

library(dplyr)

df1 %>% 
  rowwise() %>% 
  mutate(ColumnB = ColumnB[1]) %>%
  ungroup()

#> # A tibble: 3 x 2
#>   ColumnA ColumnB
#>   <chr>     <dbl>
#> 1 first         1
#> 2 second        4
#> 3 third         7