CODEWITHSUNDEEP
rtidyverse
lowndrul
lowndrul

Reputation: 3815

Mutate columns of dataframe based on a character array of expressions?

How can I mutate columns of a dataframe based on a character array of expressions? E.g.,

I have:

library(tidyverse)
dat <- data_frame(id = 0:4, 
                  brand = c(NA, 'coke', 'pepsi', 'other', 'pepsi'), 
                  price = as.character(c(NA, 1, 1.10, 1.25, .99)))

model_feature_definitions_tmp <-
  data_frame(feature_id = 0:3,
             feature_name = c("intercept", "brand_coke", "brand_pepsi", "price"),
             feature_definition = c("as.numeric(id != 0)", "as.numeric(brand == 'coke')",
                                    "as.numeric(brand == 'pepsi')", "as.numeric(price)"))

I want:

# # A tibble: 5 x 4
# intercept brand_coke brand_pepsi price
# <dbl>      <dbl>       <dbl> <dbl>
# 1         0         NA          NA    NA
# 2         1          1           0  1.00
# 3         1          0           1  1.10
# 4         1          0           0  1.25
# 5         1          0           1  0.99

The following works:

library(tidyverse)
res_list <- list()
n <- nrow(model_feature_definitions_tmp)

for (i in 1:n) {
  mfd_i <- slice(model_feature_definitions_tmp, i)
  dat %>%
    transmute(eval(parse(text=mfd_i$feature_definition))) ->
    res_list[[i]]
}

res_list %>%
  bind_cols() %>% 
  setNames(model_feature_definitions_tmp$feature_name) ->
  model_feature_space

But I doubt this is the best approach. I imagine there's a better approach that doesn't involve for-loops or *apply functions. Maybe the purrr package could be used here?

tidyverse solutions are ideal, but not necessary.

Upvotes: 1

Views: 75

Answers (1)

Eric
Eric

Reputation: 3473

Unquote splicing (rlang's !!!) works well for this task.

library(tidyverse)

dat <-
  data_frame(
    id = 0:4, 
    brand = c(NA, 'coke', 'pepsi', 'other', 'pepsi'), 
    price = as.character(c(NA, 1, 1.10, 1.25, .99))
  )

defs <-
  data_frame(
    feature_name = c("intercept", "brand_coke", "brand_pepsi", "price"),
    feature_definition = 
      c("as.numeric(id != 0)", "as.numeric(brand == 'coke')",
        "as.numeric(brand == 'pepsi')", "as.numeric(price)")
  )

Essentially you're trying to do the following (I think?):

dat %>%
  transmute(
    intercept   = as.numeric(id != 0),
    brand_coke  = as.numeric(brand == 'coke'),
    brand_pepsi = as.numeric(brand == 'pepsi'),
    price       = as.numeric(price)
  )

Which is equivalent to capturing the quoted expressions first and then splicing them into the ... of dplyr::transmute:

quosures1 <- 
  quos(
    intercept   = as.numeric(id != 0),
    brand_coke  = as.numeric(brand == 'coke'),
    brand_pepsi = as.numeric(brand == 'pepsi'),
    price       = as.numeric(price)
  )

transmute(dat, !!! quosures1)

But, you have your expressions stored as strings, so they must be parsed into expressions that can then be quoted. Here I map over the strings to generate a list of expressions that I splice into quos to make a list of quosures. I name the elements of the list so that they are used as LHS names in transmute:

quosures2 <- 
  quos(!!! map(defs$feature_definition, rlang::parse_expr)) %>%
  set_names(defs$feature_name)

transmute(dat, !!! quosures2)

Of course, I think the first version (without the quoting and splicing) will be easier for future you to read, but if you want to reduce code duplication I could see the argument for the second example (quosures1). I tend to avoid storing expressions as strings for this reason.

Upvotes: 1

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