Hi,

stringr::str_replace() treats the 2nd argument ('pattern') as a regular expression and some characters have a special meaning when they are used in a regular expression. For example the dot plays the role of a wildcard (i.e. it means "any character"):

  > str_replace("aaXcc", "a.c", "ZZ")
  [1] "aZZc"

If you want to treat a special character literally, you need to escape it with a double backslahe '\\':

  > str_replace(c("aaXcc", "aa.cc"), "a.c", "ZZ")
  [1] "aZZc" "aZZc"

  > str_replace(c("aaXcc", "aa.cc"), "a\\.c", "ZZ")
  [1] "aaXcc" "aZZc"

Turns out that parenthesis are also special characters so you also need to escape them:

  > str_replace("aa(X)cc", "a(X)c", "ZZ")
  [1] "aa(X)cc"

  > str_replace("aa(X)cc", "a\\(X\\)c", "ZZ")
  [1] "aZZc"

There are plenty of example in the man page for str_replace() (see '?str_replace') including examples showing the use of parenthesis in the pattern.

Hope this helps,

H.


On 3/16/21 5:34 PM, p...@philipsmith.ca wrote:
I have a problem with the str_replace() function in the stringr package. Please refer to my reprex below.

I start with a vector of strings, called x. Some of the strings contain apostrophes and brackets. I make a simple replacement as with x1, and there is no problem. I make another simple replacement, x2, where the pattern string has an apostrophe. Again no problem. Then I make a third replacement, x3, where the pattern has opening and closing brackets and the function still works fine. Finally I make a replacement where the pattern has both an apostrophe and opening and closing brackets and the replacement does not work. I tried to solve this by putting backslashes before the apostrophe and/or the brackets, but that accomplished nothing. I am stumped.

# Reprex for str_replace problem

library(stringr)

x <- c(
   "Clothing and footwear",
   "Women's clothing",
   "Women's footwear (excluding athletic)",
   "Clothing accessories (belts and so on)",
   "Clothing and footwear",
   "Women's clothing",
   "Women's footwear (excluding athletic)",
   "Clothing accessories (belts and so on)"
)
x
x1 <- str_replace(x,
   "Clothing and footwear",
   "Clothing and shoes"
)
x1
x2 <- str_replace(x,
   "Women's clothing",
   "Women's clothing goods"
)
x2
x3 <- str_replace(x,
   "Clothing accessories (belts and so on)",
   "Clothing accessories")
x3
x4 <- str_replace(x,
   "Women's footwear (excluding athletic)",
   "Women's footwear")
x4

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--
Hervé Pagès

Bioconductor Core Team
hpages.on.git...@gmail.com

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