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How do I add extra rows to each column across all groups in tibble using tidyverse?

November 17, 2022

How do I add extra rows for all columns (except the grouping variable) for each group id in the data frame?

Data:

> library(tidyverse)
> df <- tibble(id = c("A", "B", "C"), day = c(3, 1, 2), station = c(10, 9, 2))
> df
# A tibble: 3 × 3
  id      day station
  <chr> <dbl>   <dbl>
1 A         3      10
2 B         1       9
3 C         2       2

Expected output:

# A tibble: 6 × 3
# Groups:   id [3]
  id      day station
  <chr> <dbl>   <dbl>
1 A         2       9
2 A         3      10
3 B         0       8
4 B         1       9
5 C         1       1
6 C         2       2

I could change day using:

> df %>% group_by(id) %>% complete(day = (day - 1):day)
# A tibble: 6 × 3
# Groups:   id [3]
  id      day station
  <chr> <dbl>   <dbl>
1 A         2      NA
2 A         3      10
3 B         0      NA
4 B         1       9
5 C         1      NA
6 C         2       2

But I couldn’t use mutate appropriately to do this for station as well as since I don’t know how to refer to each column inside complete properly:

Failed attempt:

> df %>% 
+ group_by(id) %>% 
+ mutate(across(c("day", "station"), complete((.x - 1):.x)))

>Solution :

This works for your desired output:

df %>% bind_rows(df %>% mutate(across(c('day', 'station'), ~.x - 1 ))) %>% arrange(id)

# A tibble: 6 × 3
# id      day station
# <chr> <dbl>   <dbl>
# 1 A         3      10
# 2 A         2       9
# 3 B         1       9
# 4 B         0       8
# 5 C         2       2
# 6 C         1       1

Explanation:

Here I’m appending the same dataframe but with mutated columns, so:

df %>% mutate(across(c('day', 'station'), ~.x - 1 ))

Is:

# A tibble: 3 × 3
  id      day station
  <chr> <dbl>   <dbl>
1 A         2       9
2 B         0       8
3 C         1       1

Then with bind_rows I’m appending those rows to the original dataframe, which brings:

# A tibble: 6 × 3
  id      day station
  <chr> <dbl>   <dbl>
1 A         3      10
2 B         1       9
3 C         2       2
4 A         2       9
5 B         0       8
6 C         1       1

Finally I just arrange by id so the rows look like your example