Parse and Evaluate Column of String Expressions in R?

Question

How can I parse and evaluate a column of string expressions in R as part of a pipeline?

In the example below, I produce my desired column, evaluated. But I know this isn't the right approach. I tried taking a tidyverse approach. But I'm just very confused.

library(tidyverse)
df <- tibble(name = LETTERS[1:3], 
             to_evaluate = c("1-1+1", "iter+iter", "4*iter-1"), 
             evaluated = NA)
iter = 1
for (i in 1:nrow(df)) {
  df[i,"evaluated"] <- eval(parse(text=df$to_evaluate[[i]]))
}
print(df)
# # A tibble: 3 x 3
# name  to_evaluate evaluated
# <chr> <chr>           <dbl>
# 1 A     1-1+1               1
# 2 B     iter+iter           2
# 3 C     4*iter-1            3

As part of a pipeline, I tried:

df %>% mutate(evaluated = eval(parse(text=to_evaluate)))
df %>% mutate(evaluated = !!parse_exprs(to_evaluate))
df %>% mutate(evaluated = parse_exprs(to_evaluate))
df %>% mutate(evaluated = eval(parse_expr(to_evaluate)))
df %>% mutate(evaluated = parse_exprs(to_evaluate))
df %>% mutate(evaluated = eval(parse_exprs(to_evaluate)))
df %>% mutate(evaluated = eval_tidy(parse_exprs(to_evaluate)))

None of these work.

Possible duplicate https://stackoverflow.com/questions/55807468/evaluate-different-logical-conditions-from-string-for-each-row/ — Ronak Shah, Aug 05 '19 at 01:27

tmfmnk · Answer 1 · 2019-08-04T18:50:58.857

4

You can try:

df %>%
 rowwise() %>%
 mutate(iter = 1,
        evaluated = eval(parse(text = to_evaluate))) %>%
 select(-iter)

  name  to_evaluate evaluated
  <chr> <chr>           <dbl>
1 A     1-1+1               1
2 B     iter+iter           2
3 C     4*iter-1            3

Following this logic, also other possibilities could work. Using rlang::parse_expr():

df %>%
 rowwise() %>%
 mutate(iter = 1,
        evaluated = eval(rlang::parse_expr(to_evaluate))) %>%
 select(-iter)

On the other hand, I think it is important to quote @Martin Mächler:

The (possibly) only connection is via parse(text = ....) and all good R programmers should know that this is rarely an efficient or safe means to construct expressions (or calls). Rather learn more about substitute(), quote(), and possibly the power of using do.call(substitute, ......).

edited Aug 04 '19 at 18:50

answered Aug 04 '19 at 18:44

tmfmnk

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1

You can do all of this with mutate and pmap and well. `df %>% mutate(evaluated = pmap_dbl(., function(name, to_evaluate, evaluated) eval(parse(text=to_evaluate))))` – rpolicastro Aug 04 '19 at 18:59
@rpolicastro I think it is worth of a separate post, please consider posting it :) – tmfmnk Aug 04 '19 at 19:16
1

I'll post another answer for it, thanks. Wasn't sure if it was different enough from yours to be worth it. – rpolicastro Aug 04 '19 at 19:42
This simplifies your second solution a bit: `iter <- 1; df %>% rowwise() %>% mutate(evaluated = eval(rlang::parse_expr(to_evaluate)))`. No need to define `iter` in the pipe. – lowndrul Aug 05 '19 at 14:20

score 2 · Answer 2 · answered Aug 04 '19 at 19:42

Here's a slightly different way that does everything within mutate.

df %>% mutate(
    evaluated = pmap_dbl(., function(name, to_evaluate, evaluated) 
                         eval(parse(text=to_evaluate)))
)

# A tibble: 3 x 3
  name  to_evaluate evaluated
  <chr> <chr>           <dbl>
1 A     1-1+1               1
2 B     iter+iter           2
3 C     4*iter-1            3

score 1 · Answer 3 · answered Aug 12 '19 at 17:28

Note that values of additional variables (such as iter=1 in your case) can be passed directly to eval():

df %>% 
  mutate( evaluated = map_dbl(to_evaluate, ~eval(parse(text=.x), list(iter=1))) )

One advantage is that it automatically restricts the scope of the variable, keeping its value right next to where it is used.

Parse and Evaluate Column of String Expressions in R?

3 Answers3

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