Variables with brackets

[ksingh19]

Hi,

I have several variables which include brackets and texts within it. However, I would like to remove all these contents along with brackets. Is there a way to do this with tidyverse? Examples of variable names:

2020 - 1 (N = 1211)
2019 - 2 (N = 1191)
2019 - 1 (N = 1234)

Basically, I would like to get rid of everything starting from "("

Thank you!

[FJCC]

Do you really want to leave the trailing space character that was just before the (?

library(stringr)
TEXT <- c("2020 - 1 (N = 1211)",
"2019 - 2 (N = 1191)",
"2019 - 1 (N = 1234)")
NewTEXT <- str_replace(TEXT, pattern = "\\([^\\)]+\\)", replacement = "")
NewTEXT
#> [1] "2020 - 1 " "2019 - 2 " "2019 - 1 "

^{Created on 2020-09-09 by the reprex package (v0.3.0)}

[ksingh19]

Thanks you! Yes, I don't want the trailing space character before bracket.
Also, can you please help me read this pattern as to how it works.

[FJCC]

Here is a modified version that removes the trailing space and has been simplified a little.

NewTEXT <- str_replace(TEXT, pattern = " \\([^)]+\\)", replacement = "")

The pattern argument is a regular expression, which allows defining patterns in text. It looks complicated at first but can be understood by explaining each piece.

In a regular expression, placing text within [^ ] means you want to match any character that is not included within the brackets. So [^)] represents any character that is not a ).

Placing a + after a character means "one or more of the preceding character". So [^)]+ represents one or more characters that are not ).

What we want to search for is a space, followed by (, followed by one or more characters that are not ), followed by ). You might think that would look like
" ([^)]+)"
However parentheses are special characters within a regular expression. They are used to make groups of text. To prevent the parentheses from being treated as a special character, they must be preceded by two back slashes. That makes the final regular expression
" \\([^)]+\\)"

(Outside of R, it is sufficient to precede special characters in a regular expression by one back slash.)

Learning regular expressions is very helpful in many programming situations.

[ksingh19]

Perfect! Thank you so much!

[ksingh19]

Hi,

Because I have several variables, so I kept all those variables in cols as below and it doesn't work the way it should. What am I doing wrong here while including all the variables.

cols <- c(2:10)
df[cols] <- str_replace(cols, pattern = " \\([^)]+\\)", replacement = "")

Thank you!

[FJCC]

The problem is that you are asking str_replace to act on the cols vector, not on any of the columns of df. Try running this code that acts on columns 2:4 of the data frame. I used the mutate function combined with across() to pick which columns are affected.

library(dplyr)
library(stringr)
TEXT <- c("2020 - 1 (N = 1211)",
          "2019 - 2 (N = 1191)",
          "2019 - 1 (N = 1234)")
DF <- data.frame(Name = c("A", "B", "C"), 
                 C1 = TEXT,
                 C2 = TEXT,
                 C3 = TEXT)
DF
DF <- DF %>% mutate(across(.cols = 2:4, 
                     .fns = ~ str_replace(., pattern = "\\([^\\)]+\\)", replacement = "")))
DF

[ksingh19]

Sorry, I think I was not clear earlier.
I tried to create a reprex for this as follows with only few variables:

df <- data.frame(
       stringsAsFactors = FALSE,
            check.names = FALSE,
                  Brand = c("A", "B", "C"),
                               `Total (N = 9719)` = c("0.67400000000000004",
                                                      "0.56999999999999995",
                                                      "0.50700000000000001"),
                            `2020 - 1 (N = 1211)` = c("0.63200000000000001",
                                                      "0.57699999999999996",
                                                      "0.46500000000000002"),
                            `2019 - 2 (N = 1191)` = c("0.67900000000000005",
                                                      "0.56799999999999995",
                                                      "0.48")
                          )

It seems like your example is working on values per variable where C1, C2, and C3 are the variable names. I am trying to remove the bracket with text in the variable names itself. So, the name of C2 is actually 2020 - 1 (N=1211) in my case and the name of C3 is 2019 - 2 (N = 1191) and so on.

Can we still remove the brackets and text within it from the variable names itself?

Thanks for all your help!

[FJCC]

Here is an example using the data frame you provided.

library(stringr)
df <- data.frame(
  stringsAsFactors = FALSE,
  check.names = FALSE,
  Brand = c("A", "B", "C"),
  `Total (N = 9719)` = c("0.67400000000000004",
                         "0.56999999999999995",
                         "0.50700000000000001"),
  `2020 - 1 (N = 1211)` = c("0.63200000000000001",
                            "0.57699999999999996",
                            "0.46500000000000002"),
  `2019 - 2 (N = 1191)` = c("0.67900000000000005",
                            "0.56799999999999995",
                            "0.48")
)

Nms <- colnames(df)
colnames(df) <- str_replace(Nms, pattern = "\\([^\\)]+\\)", replacement = "")
colnames(df)

[ksingh19]

Yes, it works! Thank you so much!