CODEWITHSUNDEEP
rdataframedplyrtidyr
Wasabi
Wasabi

Reputation: 3071

tidyr::complete with a vector of column names of variable length

I can use tidyr::complete to expose missing rows.

For example, with the following data.frame, I can expose the missing third quarter easily:

suppressPackageStartupMessages({
  library(dplyr)
  library(tidyr)
})

set.seed(42)

df <- data.frame(
  id = c(rep(1, 3), rep(2, 3)),
  year = rep(2020, 3),
  quarter = c(1, 2, 4),
  data = runif(3)
)

df %>% complete(nesting(id, year), quarter = 1:4)
#> # A tibble: 8 x 4
#>      id  year quarter   data
#>   <dbl> <dbl>   <dbl>  <dbl>
#> 1     1  2020       1  0.915
#> 2     1  2020       2  0.937
#> 3     1  2020       3 NA    
#> 4     1  2020       4  0.286
#> 5     2  2020       1  0.915
#> 6     2  2020       2  0.937
#> 7     2  2020       3 NA    
#> 8     2  2020       4  0.286

Created on 2020-03-02 by the reprex package (v0.3.0)

Now, I want to create a wrapper for this specific use-case: it takes in a data.frame with at least these four columns and exposes the missing quarters.

That's easy enough, just wrap the complete call in a function:

expose <- function(df) {
  complete(df, nesting(id, year), quarter = 1:4)
}

expose(df)
#> # A tibble: 8 x 4
#>      id  year quarter   data
#>   <dbl> <dbl>   <dbl>  <dbl>
#> 1     1  2020       1  0.915
#> 2     1  2020       2  0.937
#> 3     1  2020       3 NA    
#> 4     1  2020       4  0.286
#> 5     2  2020       1  0.915
#> 6     2  2020       2  0.937
#> 7     2  2020       3 NA    
#> 8     2  2020       4  0.286

However, the incoming data.frame might have other columns which are known to be constant for a given id. In this case, the function doesn't work, since it naturally sets those column to NA on the missing rows.

df <- data.frame(
  id = c(rep(1, 3), rep(2, 3)),
  name = c(rep("A", 3), rep("B", 3)),
  country = c(rep("AU", 3), rep("BR", 3)),
  year = rep(2020, 3),
  quarter = c(1, 2, 4),
  data = runif(3)
)

expose(df)
#> # A tibble: 8 x 6
#>      id  year quarter name  country   data
#>   <dbl> <dbl>   <dbl> <fct> <fct>    <dbl>
#> 1     1  2020       1 A     AU       0.830
#> 2     1  2020       2 A     AU       0.642
#> 3     1  2020       3 <NA>  <NA>    NA    
#> 4     1  2020       4 A     AU       0.519
#> 5     2  2020       1 B     BR       0.830
#> 6     2  2020       2 B     BR       0.642
#> 7     2  2020       3 <NA>  <NA>    NA    
#> 8     2  2020       4 B     BR       0.519

To avoid this, I need to add those columns to the nesting call.

If it were only one column, I could add an argument to the function for the column's name, which I'd then use with nesting(..., .data[[colname]]). However, the .data pronoun doesn't work with vectors (.data[c("name1", "name2")] doesn't work).

So, how can I add multiple variable columns to the nesting call?

Upvotes: 3

Views: 705

Answers (2)

Wasabi
Wasabi

Reputation: 3071

The rlang library includes the very powerful list2 object which allows for splicing via the "big bang" operator (!!!). This allows us to pass nesting a single object which is interpreted by the receiving function as a series of arguments.

Therefore, we can add a dots argument to the function which receives all the other columns to nest with, transform the names into symbols, and then pass that to nesting:

suppressPackageStartupMessages({
  library(dplyr)
  library(rlang)
  library(tidyr)
})

set.seed(42)

expose <- function(df, ...) {
  x <- list2(...)
  x <- lapply(x, sym)
  complete(df, nesting(id, year, !!!x), quarter = 1:4)
}

df <- data.frame(
  id = c(rep(1, 3), rep(2, 3)),
  name = c(rep("A", 3), rep("B", 3)),
  country = c(rep("AU", 3), rep("BR", 3)),
  year = rep(2020, 3),
  quarter = c(1, 2, 4),
  data = runif(3)
)

expose(df, "name", "country")
#> # A tibble: 8 x 6
#>      id  year name  country quarter   data
#>   <dbl> <dbl> <fct> <fct>     <dbl>  <dbl>
#> 1     1  2020 A     AU            1  0.915
#> 2     1  2020 A     AU            2  0.937
#> 3     1  2020 A     AU            3 NA    
#> 4     1  2020 A     AU            4  0.286
#> 5     2  2020 B     BR            1  0.915
#> 6     2  2020 B     BR            2  0.937
#> 7     2  2020 B     BR            3 NA    
#> 8     2  2020 B     BR            4  0.286

Created on 2020-03-02 by the reprex package (v0.3.0)

Upvotes: 0

asachet
asachet

Reputation: 6921

If you look at tidyr::nesting, you'll see it relies on tidyr:::dots_cols which relies on rlang to interpret the column names (in particular, rlang::enquos).

The best way to interface with tidyr::nesting is therefore to use an rlang construct. 

library(dplyr)
library(tidyr)

expose <- function(df, ...) {
  dots <- rlang::exprs(id, year, ...)
  complete(df, nesting(!!! dots), quarter = 1:4)
}

df <- data.frame(
  id = c(rep(1, 3), rep(2, 3)),
  name = c(rep("A", 3), rep("B", 3)),
  country = c(rep("AU", 3), rep("BR", 3)),
  year = rep(2020, 3),
  quarter = c(1, 2, 4),
  data = runif(3)
)

expose(df)
#> # A tibble: 8 x 6
#>      id  year quarter name  country    data
#>   <dbl> <dbl>   <dbl> <fct> <fct>     <dbl>
#> 1     1  2020       1 A     AU       0.0417
#> 2     1  2020       2 A     AU       0.365 
#> 3     1  2020       3 <NA>  <NA>    NA     
#> 4     1  2020       4 A     AU       0.690 
#> 5     2  2020       1 B     BR       0.0417
#> 6     2  2020       2 B     BR       0.365 
#> 7     2  2020       3 <NA>  <NA>    NA     
#> 8     2  2020       4 B     BR       0.690
expose(df, name)
#> # A tibble: 8 x 6
#>      id  year name  quarter country    data
#>   <dbl> <dbl> <fct>   <dbl> <fct>     <dbl>
#> 1     1  2020 A           1 AU       0.0417
#> 2     1  2020 A           2 AU       0.365 
#> 3     1  2020 A           3 <NA>    NA     
#> 4     1  2020 A           4 AU       0.690 
#> 5     2  2020 B           1 BR       0.0417
#> 6     2  2020 B           2 BR       0.365 
#> 7     2  2020 B           3 <NA>    NA     
#> 8     2  2020 B           4 BR       0.690
expose(df, name, country)
#> # A tibble: 8 x 6
#>      id  year name  country quarter    data
#>   <dbl> <dbl> <fct> <fct>     <dbl>   <dbl>
#> 1     1  2020 A     AU            1  0.0417
#> 2     1  2020 A     AU            2  0.365 
#> 3     1  2020 A     AU            3 NA     
#> 4     1  2020 A     AU            4  0.690 
#> 5     2  2020 B     BR            1  0.0417
#> 6     2  2020 B     BR            2  0.365 
#> 7     2  2020 B     BR            3 NA     
#> 8     2  2020 B     BR            4  0.690

Created on 2020-03-02 by the reprex package (v0.3.0)

Upvotes: 2

Related Questions