Remove/Replace/Extract Percentages

Usage

rm_percent(text.var, trim = !extract, clean = TRUE, pattern = "@rm_percent", replacement = "", extract = FALSE, dictionary = getOption("regex.library"), ...)

Arguments

text.var
The text variable.
trim
logical. If TRUE removes leading and trailing white spaces.
clean
trim logical. If TRUE extra white spaces and escaped character will be removed.
pattern
A character string containing a regular expression (or character string for fixed = TRUE) to be matched in the given character vector. Default, @rm_percent uses the rm_percent regex from the regular expression dictionary from the dictionary argument.
replacement
Replacement for matched pattern.
extract
logical. If TRUE the percentages are extracted into a list of vectors.
dictionary
A dictionary of canned regular expressions to search within if pattern begins with "@rm_".
...
Other arguments passed to gsub.

Value

Returns a character string with percentages removed.

Description

Remove/replace/extract percentages from a string.

Examples

x <- c("There is $5.50 for me.", "that's 45.6% of the pizza", "14% is $26 or $25.99") rm_percent(x)
[1] "There is $5.50 for me." "that's of the pizza" "is $26 or $25.99"
rm_percent(x, extract=TRUE)
[[1]] [1] NA [[2]] [1] "45.6%" [[3]] [1] "14%"