create a numeric list based on another column in tidyverse

Question

Goal: I would like to create a numeric list based on the starting value in another column (seqnbr), then unnest it.

In the data below a nested c(1:10) sequence would begin on the column seqnbr.

x <- tibble(name = c('boo', 'bah', 'bee'),
                seqnbr = c(1,2,3))

A tibble: 3 x 2
  name  seqnbr
  <chr>  <dbl>
1 boo        1
2 bah        2
3 bee        3

desired result would look like this:

# A tibble: 3 x 3
  name  seqnbr my_list
  <chr>  <dbl> <chr>  
1 boo        1 list(c(1:10) )
2 bah        2 list(c(2:10) )
3 bee        3 list(c(3:10) )

After this step, I would unnest() to make values explicit.

Tidyverse/purr solutions preferred.

Nir Graham · Accepted Answer · 2023-02-10 17:34:14Z

1

library(tidyverse)
x <- tibble(name = c('boo', 'bah', 'bee'),
            seqnbr = c(1,2,3))

(x_1 <- x |> rowwise() |> mutate(my_list= list(seq(from=seqnbr,
                               to=10))))

(x_2 <- unnest_longer(x_1,col=my_list))

answered Feb 10, 2023 at 17:34

Nir Graham

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langtang · Accepted Answer · 2023-02-10 18:03:57Z

1

You can use purrr::map, if you like:

mutate(x, my_list=map(seqnbr, ~.x:10)) %>% unnest_longer(my_list)

Output:

# A tibble: 27 × 3
   name  seqnbr my_list
   <chr>  <dbl>   <int>
 1 boo        1       1
 2 boo        1       2
 3 boo        1       3
 4 boo        1       4
 5 boo        1       5
 6 boo        1       6
 7 boo        1       7
 8 boo        1       8
 9 boo        1       9
10 boo        1      10
# … with 17 more rows

edited Feb 10, 2023 at 18:03

answered Feb 10, 2023 at 17:59

langtang

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1 Comment

Onyambu Over a year ago

The two codes give different results

Onyambu · Accepted Answer · 2023-02-10 18:06:08Z

1

You could use reframe:

library(tidyverse)
reframe(x, my_list = seqnbr:10, .by = name)

# A tibble: 27 × 2
   name  my_list
   <chr>   <int>
 1 boo         1
 2 boo         2
 3 boo         3
 4 boo         4
 5 boo         5
 6 boo         6
 7 boo         7
 8 boo         8
 9 boo         9
10 boo        10
# … with 17 more rows
# ℹ Use `print(n = ...)` to see more rows

answered Feb 10, 2023 at 18:06

Onyambu

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Comments

akrun · Accepted Answer · 2023-02-10 18:15:31Z

1

Using complete

library(dplyr)
library(tidyr)
x %>% 
   group_by(name) %>% 
   complete(seqnbr = seqnbr:10) %>%
   ungroup

answered Feb 10, 2023 at 18:15

akrun

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