dplyr - mutate using function that uses other column data as argument? - r

Question

I have a list with 3 regression models, called logregs. My data has a column called type that only has integers 1, 2, and 3, which are used to decide which regression model from logregs should be used, and a column called adstock which is the only independent variable used in the regression models.
I'm trying to do something like:
dataframe %>% mutate(probability = predict(logregs[[type]], type = "prediction", newdata = adstock) )
Sample data frame:
structure(list(type = c(3L, 3L, 3L, 3L, 3L, 3L), adstock = c(1.7984,
1.7984, 2.7984, 6.7984, 6.5968, 4.992)), row.names = c(NA, 6L
), class = "data.frame")
(unfortunately, the logregs models are too large to dput here)
How is this achievable using dplyr?

Answer 1

Yes, but you need to take some more care on subsetting logregs, and use data.frame on your newdata=.
I'll generate a quick set of models based on mtcars.
library(dplyr)
models <- mtcars %>%
group_by(cyl = as.character(cyl)) %>%
nest() %>%
mutate(mdl = map(data, ~ lm(mpg ~ disp, data = .x))) %>%
arrange(cyl) %>%
select(cyl, mdl) %>%
deframe()
models
# $`4`
# Call:
# lm(formula = mpg ~ disp, data = .x)
# Coefficients:
# (Intercept) disp
# 40.8720 -0.1351
# $`6`
# Call:
# lm(formula = mpg ~ disp, data = .x)
# Coefficients:
# (Intercept) disp
# 19.081987 0.003605
# $`8`
# Call:
# lm(formula = mpg ~ disp, data = .x)
# Coefficients:
# (Intercept) disp
# 22.03280 -0.01963
Note that they are indexed on the character of the number of cylinders, since otherwise numeric indexing can be confusing.
Let's modify the mtcars$disp a little and to use it again:
set.seed(42)
mtcars %>%
mutate(disp = disp + sample(20, size=n(), replace = TRUE) - 10) %>%
group_by(cyl) %>%
sample_n(2)
# # A tibble: 6 x 11
# # Groups: cyl [3]
# mpg cyl disp hp drat wt qsec vs am gear carb
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 30.4 4 75.7 52 4.93 1.62 18.5 1 1 4 2
# 2 21.5 4 129. 97 3.7 2.46 20.0 1 0 3 1
# 3 21 6 169 110 3.9 2.62 16.5 0 1 4 4
# 4 19.2 6 173. 123 3.92 3.44 18.3 1 0 4 4
# 5 18.7 8 363 175 3.15 3.44 17.0 0 0 3 2
# 6 16.4 8 281. 180 3.07 4.07 17.4 0 0 3 3
The [[ indexing on your logregs expects a single type, but you're actually passing a vector. Since my data here is still grouped, I can go with the first of the group variable (cyl) and do just a single call to predict per group:
set.seed(42)
mtcars %>%
mutate(disp = disp + sample(20, size=n(), replace = TRUE) - 10) %>%
group_by(cyl) %>%
sample_n(2) %>%
mutate(mpg2 = predict(models[[as.character(cyl)[1]]], newdata = data.frame(disp)))
# # A tibble: 6 x 12
# # Groups: cyl [3]
# mpg cyl disp hp drat wt qsec vs am gear carb mpg2
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 30.4 4 75.7 52 4.93 1.62 18.5 1 1 4 2 30.6
# 2 21.5 4 129. 97 3.7 2.46 20.0 1 0 3 1 23.4
# 3 21 6 169 110 3.9 2.62 16.5 0 1 4 4 19.7
# 4 19.2 6 173. 123 3.92 3.44 18.3 1 0 4 4 19.7
# 5 18.7 8 363 175 3.15 3.44 17.0 0 0 3 2 14.9
# 6 16.4 8 281. 180 3.07 4.07 17.4 0 0 3 3 16.5
If you don't want to (or cannot) group, then you need to run one prediction per row. This is expensive in that it runs predict with a single newdata= argument, but ... it still works. To do this, we'll map it:
library(purrr) # map* functions
set.seed(42)
mtcars %>%
mutate(disp = disp + sample(20, size=n(), replace = TRUE) - 10) %>%
group_by(cyl) %>%
sample_n(2) %>%
ungroup() %>%
mutate(mpg2 = map2_dbl(cyl, disp, ~ predict(models[[as.character(.x)]], newdata = data.frame(disp=.y))))
# # A tibble: 6 x 12
# mpg cyl disp hp drat wt qsec vs am gear carb mpg2
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 30.4 4 75.7 52 4.93 1.62 18.5 1 1 4 2 30.6
# 2 21.5 4 129. 97 3.7 2.46 20.0 1 0 3 1 23.4
# 3 21 6 169 110 3.9 2.62 16.5 0 1 4 4 19.7
# 4 19.2 6 173. 123 3.92 3.44 18.3 1 0 4 4 19.7
# 5 18.7 8 363 175 3.15 3.44 17.0 0 0 3 2 14.9
# 6 16.4 8 281. 180 3.07 4.07 17.4 0 0 3 3 16.5
Note that I had to name the column of newdata=data.frame(disp=.y): when we did it before, data.frame(disp) names it the name of the import variable. In this case, .y is not known to the model, so we have to explicitly name it.