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Replace or remove backslashes in a string in R

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Escape ’til infinity — removing or replacing backslashes in a string can be burdensome in any programming language, if you’re not familiar what regex library is below the hood. In this post, we’ll remove a backslash from a string in R.

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all credits to xckd

Using gsub, there’s two paradigms to choose from. The first is by selecting it via a regex pattern. Believe it or not, but you’ll have to provide four (!) backslashes. This is because all characters are first parsed by the R parser, before being sent to the command line — i.e. you have to escape the escape.

my_str <- 'I am a \ backslash'
my_str <- gsub(
  pattern = ('\\\\'), 
  replacement = '', 
  x = my_str
  )

However, there’s a less confusing way. Instead of using regex patterns, you can simply match literal strings by using gsub‘s fixed parameter.

my_str <- gsub(
  pattern = ('\\'), 
  replacement = '', 
  x = my_str,
  fixed = T
)

You can also do it using stringr:

str_replace(my_str, fixed("\\"), "backslash")

Let me know in the comments below if you figured this out.

By the way, if you’re having trouble understanding some of the code and concepts, I can highly recommend “An Introduction to Statistical Learning: with Applications in R”, which is the must-have data science bible. If you simply need an introduction into R, and less into the Data Science part, I can absolutely recommend this book by Richard Cotton. Hope it helps!

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Technologies get updated, syntax changes and honestly… I make mistakes too. If something is incorrect, incomplete or doesn’t work, let me know in the comments below and help thousands of visitors.

3 thoughts on “Replace or remove backslashes in a string in R”

  1. Thank you for the suggestions. It is still not working for me. I cannot convert a path with backslashes to a string in R. I get a message that either \ or \_ is an unrecognized escape character. I still have to replace \ with / by hand in R.
    Robert Cline

  2. How can I do the reverse? Replace one backslash with two. Somehow my syntax is incorrect if I simply apply the same logic the the replace argument.

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