How do I filter using regex across multiple columns?

Question

I want to filter a data frame to include rows where the value of any column containing the string "bean" starts with "black" or contains "vanilla."

My code looks like this.

  library(dplyr)
  df2 <- df1 %>%
    filter(
      if_any(
        .cols = contains('bean'),
        grepl(
          pattern = "^black*|*vanilla*",
          ignore.case = T,
          x = .
        )
      )
    )

But that gives me this error:

Error: Problem with `filter()` input `..1`.
x Problem with `across()` input `.fns`.
ℹ Input `.fns` must be NULL, a function, a formula, or a list of functions/formulas.
ℹ Input `..1` is `if_all(...)`.
Run `rlang::last_error()` to see where the error occurred.

Running last_error() gives me this:

<error/dplyr_error>
Problem with `filter()` input `..1`.
x Problem with `across()` input `.fns`.
ℹ Input `.fns` must be NULL, a function, a formula, or a list of functions/formulas.
ℹ Input `..1` is `if_any(...)`.
Backtrace:
Run `rlang::last_trace()` to see the full context.

Running last_trace() gives me this:

<error/dplyr_error>
Problem with `filter()` input `..1`.
x Problem with `across()` input `.fns`.
ℹ Input `.fns` must be NULL, a function, a formula, or a list of functions/formulas.
ℹ Input `..1` is `if_any(...)`.
Backtrace:
     █
  1. ├─`%>%`(...)
  2. ├─dplyr::filter(...)
  3. ├─dplyr:::filter.data.frame(...)
  4. │ └─dplyr:::filter_rows(.data, ...)
  5. │   ├─base::withCallingHandlers(...)
  6. │   └─mask$eval_all_filter(dots, env_filter)
  7. ├─dplyr::if_any(...)
  8. │ └─dplyr::across(...)
  9. │   └─dplyr:::across_setup(...)
 10. │     └─dplyr:::across_setup_impl(...)
 11. │       └─rlang::abort(c("Problem with `across()` input `.fns`.", i = "Input `.fns` must be NULL, a function, a formula, or a list of functions/formulas."))
 12. │         └─rlang:::signal_abort(cnd)
 13. │           └─base::signalCondition(cnd)
 14. └─(function (e) ...

akrun · Accepted Answer · 2021-03-11 20:13:41Z

2

We need a lambda function (~) as the x is .. The use of * in the pattern implies zero or more characters. Not sure if the OP meant it that way

library(dplyr) # version 1.0.4
df1 %>%
filter(
  if_any(
    .cols = contains('bean'),
    ~ grepl(
      pattern = "^black*|*vanilla*",
      ignore.case = TRUE,
      x = .
    )
  )
)

-output

#   bean_1 bean_2 col3
#1   black velvet    1
#2 vanilla    red    4

Or if we don't specify the x, OP's current code would work because it already specified the pattern and ignore.case as named arguments

df1 %>%
 filter(
  if_any(
    .cols = contains('bean'),
    grepl,
      pattern = "^black*|*vanilla*",
      ignore.case = TRUE
     
    
  )
)

-output

#  bean_1 bean_2 col3
#1   black velvet    1
#2 vanilla    red    4

data

df1 <- data.frame(bean_1 = c('black', 'velvet', 'orange', 'vanilla'),
       bean_2 = c('velvet', 'red', 'green', 'red'), col3 = 1:4)

edited Mar 11, 2021 at 20:13

answered Mar 11, 2021 at 20:07

akrun

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2 Comments

Username Over a year ago

What does the tilda mean?

akrun Over a year ago

@Username sorry, was away. The ~ is a short form in tidyverse for function(x). it can be otherwise written as function(x) grepl( pattern = "^black*|*vanilla*", ignore.case = TRUE, x = x) which is a lamdba/anonymous function

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