Add row in each group using dplyr and add_row()

Question

If I add a new row to the iris dataset with:

iris <- as_tibble(iris)

> iris %>% 
    add_row(.before=0)

# A tibble: 151 × 5
    Sepal.Length Sepal.Width Petal.Length Petal.Width Species
          <dbl>       <dbl>        <dbl>       <dbl>   <chr>
1            NA          NA           NA          NA    <NA> <--- Good!
2           5.1         3.5          1.4         0.2  setosa
3           4.9         3.0          1.4         0.2  setosa

It works. So, why can't I add a new row on top of each "subset" with:

iris %>% 
 group_by(Species) %>% 
 add_row(.before=0)

Error: is.data.frame(df) is not TRUE

Answer 1

If you want to use a grouped operation, you need do like JasonWang described in his comment, as other functions like mutate or summarise expect a result with the same number of rows as the grouped data frame (in your case, 50) or with one row (e.g. when summarising).

As you probably know, in general do can be slow and should be a last resort if you cannot achieve your result in another way. Your task is quite simple because it only involves adding extra rows in your data frame, which can be done by simple indexing, e.g. look at the output of iris[NA, ].

What you want is essentially to create a vector

indices <- c(NA, 1:50, NA, 51:100, NA, 101:150)

(since the first group is in rows 1 to 50, the second one in 51 to 100 and the third one in 101 to 150).

The result is then iris[indices, ].

A more general way of building this vector uses group_indices.

indices <- seq(nrow(iris)) %>% 
    split(group_indices(iris, Species)) %>% 
    map(~c(NA, .x)) %>%
    unlist

(map comes from purrr which I assume you have loaded as you have tagged this with tidyverse).

Answer 2

A more recent version would be using group_modify() instead of do().

iris %>%
  as_tibble() %>%
  group_by(Species) %>% 
  group_modify(~ add_row(.x,.before=0))
#> # A tibble: 153 x 5
#> # Groups:   Species [3]
#>    Species Sepal.Length Sepal.Width Petal.Length Petal.Width
#>    <fct>          <dbl>       <dbl>        <dbl>       <dbl>
#>  1 setosa          NA          NA           NA          NA  
#>  2 setosa           5.1         3.5          1.4         0.2
#>  3 setosa           4.9         3            1.4         0.2