pmap_ variants operating on data.frames as lists

Question

I have a recollection that purrr::pmap_* can treat a data.frame as a list but the syntax eludes me.

Imagine we wanted to fit a separate lm object for each value of mtcars$vs and mtcars$am

library(tidyverse)
library(broom)

d1 <- mtcars %>% 
  group_by(
    vs, am
  ) %>% 
  nest %>% 
  mutate(
    coef = data %>% 
      map(
        ~lm(mpg ~ wt, data =.) %>% 
          tidy
      )
  )

If I wanted to extract the coefficient estimates as an un-nested data.frame, and append the values of am and vs, I might try

d1[, -3] %>% 
  pmap_dfr(
    function(i, j, k)
      k %>% 
      mutate(
        vs = i,
        am = j
      )
  )

But this results in an error. More explicitly declaring these variables as separate lists has the desired effect

list(
  d1$vs,
  d1$am,
  d1$coef
  ) %>% 
  pmap_dfr(
    function(i, j, k)
      k %>% 
      mutate(
        vs = i,
        am = j
      )
  )

Is there a succinct way for pmap_* to treat a data.frame as a list?

Answer 1

This is because the second list has no names attribute. If you unname d1 it works. The fact that you used the list function in the second example doesn't make a difference (except that it removed the names), because both objects are lists (data frames are lists).

d1[, -3] %>% 
  unname %>% 
  pmap_dfr(
    function(i, j, k)
      k %>% 
      mutate(
        vs = i,
        am = j
      )
  )


# # A tibble: 8 x 7
#   term        estimate std.error statistic   p.value    vs    am
#   <chr>          <dbl>     <dbl>     <dbl>     <dbl> <dbl> <dbl>
# 1 (Intercept)    42.4      3.30      12.8  0.000213      0     1
# 2 wt             -7.91     1.14      -6.93 0.00227       0     1
# 3 (Intercept)    44.1      6.96       6.34 0.00144       1     1
# 4 wt             -7.77     3.36      -2.31 0.0689        1     1
# 5 (Intercept)    31.5      8.98       3.51 0.0171        1     0
# 6 wt             -3.38     2.80      -1.21 0.281         1     0
# 7 (Intercept)    25.1      3.51       7.14 0.0000315     0     0
# 8 wt             -2.44     0.842     -2.90 0.0159        0     0

You can also name the arguments in your first code block's function to match (or use ..1 etc) for the same result

d1[, -3] %>% 
  pmap_dfr(
    function(vs, am, coef)
      coef %>% 
      mutate(
        vs = vs,
        am = am
      )
  )


# # A tibble: 8 x 7
#   term        estimate std.error statistic   p.value    vs    am
#   <chr>          <dbl>     <dbl>     <dbl>     <dbl> <dbl> <dbl>
# 1 (Intercept)    42.4      3.30      12.8  0.000213      0     1
# 2 wt             -7.91     1.14      -6.93 0.00227       0     1
# 3 (Intercept)    44.1      6.96       6.34 0.00144       1     1
# 4 wt             -7.77     3.36      -2.31 0.0689        1     1
# 5 (Intercept)    31.5      8.98       3.51 0.0171        1     0
# 6 wt             -3.38     2.80      -1.21 0.281         1     0
# 7 (Intercept)    25.1      3.51       7.14 0.0000315     0     0
# 8 wt             -2.44     0.842     -2.90 0.0159        0     0

You could also use wap from the experimental rap package

library(rap)

d1[, -3] %>% 
  wap( ~ coef %>% 
          mutate(
            vs = vs,
            am = am)) %>% 
  bind_rows
# # A tibble: 8 x 7
#   term        estimate std.error statistic   p.value    vs    am
#   <chr>          <dbl>     <dbl>     <dbl>     <dbl> <dbl> <dbl>
# 1 (Intercept)    42.4      3.30      12.8  0.000213      0     1
# 2 wt             -7.91     1.14      -6.93 0.00227       0     1
# 3 (Intercept)    44.1      6.96       6.34 0.00144       1     1
# 4 wt             -7.77     3.36      -2.31 0.0689        1     1
# 5 (Intercept)    31.5      8.98       3.51 0.0171        1     0
# 6 wt             -3.38     2.80      -1.21 0.281         1     0
# 7 (Intercept)    25.1      3.51       7.14 0.0000315     0     0
# 8 wt             -2.44     0.842     -2.90 0.0159        0     0

Answer 2

We can use the standard option to extract the components (..1, ..2, etc)

d1[, -3]  %>% 
    pmap_dfr(~ ..3 %>%
                  mutate(vs = ..1, am = ..2))
# A tibble: 8 x 7
#  term        estimate std.error statistic   p.value    vs    am
#  <chr>          <dbl>     <dbl>     <dbl>     <dbl> <dbl> <dbl>
#1 (Intercept)    42.4      3.30      12.8  0.000213      0     1
#2 wt             -7.91     1.14      -6.93 0.00227       0     1
#3 (Intercept)    44.1      6.96       6.34 0.00144       1     1
#4 wt             -7.77     3.36      -2.31 0.0689        1     1
#5 (Intercept)    31.5      8.98       3.51 0.0171        1     0
#6 wt             -3.38     2.80      -1.21 0.281         1     0
#7 (Intercept)    25.1      3.51       7.14 0.0000315     0     0
#8 wt             -2.44     0.842     -2.90 0.0159        0     0