CODEWITHSUNDEEP
rstringr
ibm
ibm

Reputation: 864

str_c to build a quotation-wrapped comma separated vector

iris %>% select(where(is.numeric)) %>% colnames() %>% str_c(., collapse = '","')

Output: "Sepal.Length","Sepal.Width","Petal.Length","Petal.Width" It gets me close to what I want (I don't really care about the space missing after the comma) but I can't remove the backslash afterwards with str_remove_all because it doesn't recognize the escaped backslash

Ideal: "Sepal.Length","Sepal.Width","Petal.Length","Petal.Width"

iris %>% select(where(is.numeric)) %>% colnames() %>% str_c(., collapse = '",') str_remove_all(., "\\")

output: Error in stri_replace_all_regex(string, pattern, fix_replacement(replacement), : Unrecognized backslash escape sequence in pattern. (U_REGEX_BAD_ESCAPE_SEQUENCE)

Upvotes: 1

Views: 269

Answers (1)

Ronak Shah
Ronak Shah

Reputation: 388817

You can use :

library(dplyr)

iris %>% 
      select(where(is.numeric)) %>% 
      colnames() %>% 
      sprintf('"%s"', .) %>%
      toString()

#[1] "\"Sepal.Length\", \"Sepal.Width\", \"Petal.Length\", \"Petal.Width\""

The backslashes are not part of actual string, that is how R displays double quotes. To see the actual string you can use cat :

iris %>% 
      select(where(is.numeric)) %>% 
      colnames() %>% 
      sprintf('"%s"', .) %>%
      toString() %>% 
      cat

#"Sepal.Length", "Sepal.Width", "Petal.Length", "Petal.Width"

Upvotes: 1

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