dplyr case_when 具有动态案例数

Question

Wanting to use dplyr and case_when to collapse a series of indicator columns into a single column. The challenge is I want to be able to collapse over an unspecified/dynamic number of columns.

Consider the following dataset, gear has been split into a series of indicator columns.

library(dplyr)
data(mtcars)
mtcars = mtcars %>%
  mutate(g2 = ifelse(gear == 2, 1, 0),
         g3 = ifelse(gear == 3, 1, 0),
         g4 = ifelse(gear == 4, 1, 0)) %>%
  select(g2, g3, g4)

I am trying to write a function that does the reverse.

When I know how many cases this can be done as follows:

combine_indices = function(db, cols, vals){
  db %>% mutate(new_col = case_when(!!sym(cols[1]) == 1 ~ vals[1],
                                    !!sym(cols[2]) == 1 ~ vals[2],
                                    !!sym(cols[3]) == 1 ~ vals[3]))
}

cols = c("g2", "g3", "g4")
vals = c(2,3,4)
combine_indices(mtcars, cols, vals)

However, I would like the combine_indices function to handle any number of index columns (right now it works for exactly three).

According to the documentation ( ?case_when ), "if your patterns are stored in a list, you can splice that in with !!! ". But I can not get this working:

patterns = list(sym(cols[1] == 1 ~ vals[1],
                sym(cols[2] == 1 ~ vals[2],
                sym(cols[3] == 1 ~ vals[3])

mtcars %>% mutate(new_col = case_when(!!!patterns))

Only produces a new column filled with NAs.

If !!!patterns worked, then it would be straightforward to take the lists cols and vals and generate patterns . However, I can not get the quosures correct. Hoping someone more familiar with quosures knows how.

Note - some similar questions here of SO were solved using joins or other functions. However, I am restricted to using case_when because of how it translates to sql when using dbplyr.

Answer 1

We can create a string of conditions, use parse_exprs and splice it ( !!! ).

library(dplyr)
library(rlang)

combine_indices = function(db, cols, vals){
   db %>% mutate(new_col = case_when(!!!parse_exprs(paste(cols, '== 1 ~', vals))))
}


cols = c("g2", "g3", "g4")
vals = c(2,3,4)
combine_indices(mtcars, cols, vals)

which returns:

#   g2 g3 g4 new_col
#1   0  0  1       4
#2   0  0  1       4
#3   0  0  1       4
#4   0  1  0       3
#5   0  1  0       3
#6   0  1  0       3
#....

where paste generates the conditions for case_when dynamically.

paste(cols, '== 1 ~', vals)
#[1] "g2 == 1 ~ 2" "g3 == 1 ~ 3" "g4 == 1 ~ 4"

Answer 2

This solution should create a column for any value in the gear column:

data <- mtcars %>% 
  mutate(mygear = gear) %>%
  pivot_wider(values_from = gear, names_from = gear, names_prefix = "g") %>% 
  mutate_at(vars(starts_with('g')), function(x) x/.$mygear) %>%
  mutate_if(is.numeric , replace_na, replace = 0) %>%
  rename(gear = mygear)

I do need to create a temporary column mygear as pivot_wider does not retain the pivot column.

> data
# A tibble: 32 x 14
     mpg   cyl  disp    hp  drat    wt  qsec    vs    am  carb  gear    g4    g3    g5
   <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
 1  21       6  160    110  3.9   2.62  16.5     0     1     4     4     1     0     0
 2  21       6  160    110  3.9   2.88  17.0     0     1     4     4     1     0     0
 3  22.8     4  108     93  3.85  2.32  18.6     1     1     1     4     1     0     0
 4  21.4     6  258    110  3.08  3.22  19.4     1     0     1     3     0     1     0
 5  18.7     8  360    175  3.15  3.44  17.0     0     0     2     3     0     1     0
 6  18.1     6  225    105  2.76  3.46  20.2     1     0     1     3     0     1     0
 7  14.3     8  360    245  3.21  3.57  15.8     0     0     4     3     0     1     0
 8  24.4     4  147.    62  3.69  3.19  20       1     0     2     4     1     0     0
 9  22.8     4  141.    95  3.92  3.15  22.9     1     0     2     4     1     0     0
10  19.2     6  168.   123  3.92  3.44  18.3     1     0     4     4     1     0     0
# … with 22 more rows

Answer 3

You need to make object in list unevaluated expression by expr() to evaluate by case_when .
To be honest I didn't understand it completely, but it is work.

patterns <- list(expr(!!sym(cols[1]) == 1 ~ vals[1]),
                 expr(!!sym(cols[2]) == 1 ~ vals[2]),
                 expr(!!sym(cols[3]) == 1 ~ vals[3]))

OR more simply

patterns <- exprs(!!sym(cols[1]) == 1 ~ vals[1],
                  !!sym(cols[2]) == 1 ~ vals[2],
                  !!sym(cols[3]) == 1 ~ vals[3])

mtcars %>% mutate(new_col = case_when(!!!patterns))

Answer 4

For the sake of completeness, for this particular use case only the result can be obtained using matrix multiplication:

library(dplyr)
combine_indices = function(db, cols, vals){
  db %>% mutate(new_col = as.matrix(db[, cols]) %*% vals)
}

cols = c("g2", "g3", "g4")
vals = c(2, 3, 4)
combine_indices(mtcars, cols, vals)

 g2 g3 g4 new_col 1 0 0 1 4 2 0 0 1 4 3 0 0 1 4 4 0 1 0 3 5 0 1 0 3 6 0 1 0 3 7 0 1 0 3 8 0 0 1 4 9 0 0 1 4 10 0 0 1 4 11 0 0 1 4 12 0 1 0 3 13 0 1 0 3 14 0 1 0 3 15 0 1 0 3 16 0 1 0 3 17 0 1 0 3 18 0 0 1 4 19 0 0 1 4 20 0 0 1 4 21 0 1 0 3 22 0 1 0 3 23 0 1 0 3 24 0 1 0 3 25 0 1 0 3 26 0 0 1 4 27 0 0 0 0 28 0 0 0 0 29 0 0 0 0 30 0 0 0 0 31 0 0 0 0 32 0 0 1 4

Explanation

For row 1, we get

0 * 2 + 0 * 3 + 1 * 4 = 4

Answer 5

Perhaps I'm looking at it wrong, but I think this can be done more efficiently with a join:

cols <- tibble(g2 = c(1, 0, 0), g3 = c(0, 1, 0), g4 = c(0, 0, 1), val = c(2, 3, 4))
cols
# # A tibble: 3 x 4
#      g2    g3    g4   val
#   <dbl> <dbl> <dbl> <dbl>
# 1     1     0     0     2
# 2     0     1     0     3
# 3     0     0     1     4

# using your mtcars
left_join(mtcars, cols, by = c("g2", "g3", "g4"))
#    g2 g3 g4 val
# 1   0  0  1   4
# 2   0  0  1   4
# 3   0  0  1   4
# 4   0  1  0   3
# 5   0  1  0   3
# 6   0  1  0   3
# 7   0  1  0   3
# 8   0  0  1   4
# 9   0  0  1   4
# 10  0  0  1   4
# 11  0  0  1   4
# 12  0  1  0   3
# 13  0  1  0   3
# 14  0  1  0   3
# 15  0  1  0   3
# 16  0  1  0   3
# 17  0  1  0   3
# 18  0  0  1   4
# 19  0  0  1   4
# 20  0  0  1   4
# 21  0  1  0   3
# 22  0  1  0   3
# 23  0  1  0   3
# 24  0  1  0   3
# 25  0  1  0   3
# 26  0  0  1   4
# 27  0  0  0  NA
# 28  0  0  0  NA
# 29  0  0  0  NA
# 30  0  0  0  NA
# 31  0  0  0  NA
# 32  0  0  1   4