Stringr str_replace_all misses repeated terms

user_2658344 0
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I'm having an issue with the stringr::str_replace_all function. I'm trying to replace all instances of iv with insuredvehicle, but the function only seems to catch the first term.

temp_data <- data.table(text = 'the driver of the 1st vehicle hit the iv iv at a stop')
temp_data[, new_text := stringr::str_replace_all(pattern = ' iv ', replacement = ' insuredvehicle ', string = text)]

The outcome looks like the following, which missed the 2nd iv term:

1: the driver of the 1st vehicle hit the insuredvehicle iv at a stop

I believe the issue is that the 2 instances share a space, which is part of the search pattern. I did that because I want to replace the iv term, and not iv within driver.

I DON'T want to simply consolidate the repeated terms to 1. I'd like the result to look like:

1: the driver of the 1st vehicle hit the insuredvehicle insuredvehicle at a stop

I'd appreciate any help getting this to work!

stringr str-replace r regex
4 Answers

Maybe if you include a word boundary in your regex, than remove the white spaces from the replacement? It is ideal when you want just a full word matching the pattern, but not parts of words, while staying away from these blank space issues. \\bseems to do the trick

temp_data[, new_text := stringr::str_replace_all(pattern = '\\biv\\b', replacement = 'insuredvehicle', string = text)]

new_text

1: the driver of the 1st vehicle hit the insuredvehicle insuredvehicle at a stop
user_139723330

You can use lookarounds:

temp_data[, new_text := stringr::str_replace_all(pattern = '(?<= )iv(?= )', replacement = 'insuredvehicle', string = text)]

Output:

"the driver of the 1st vehicle hit the insuredvehicle insuredvehicle at a stop"
user_58056700

Use gsub:

gsub("\\biv\\b", "insuredvehicle", temp_data$text)
[1] "the driver of the 1st vehicle hit the uninsuredvehicle uninsuredvehicle at a stop"
user_80399780

Use space boundaries:

temp_data <- data.table(text = 'the driver of the 1st vehicle hit the iv iv at a stop')
temp_data[, new_text := stringr::str_replace_all(pattern = '(?<!\\S)iv(?!\\S)', replacement = 'insuredvehicle', string = text)]

See regex proof.

EXPLANATION

--------------------------------------------------------------------------------
  (?<!                     look behind to see if there is not:
--------------------------------------------------------------------------------
    \S                       non-whitespace (all but \n, \r, \t, \f,
                             and " ")
--------------------------------------------------------------------------------
  )                        end of look-behind
--------------------------------------------------------------------------------
  iv                       'iv'
--------------------------------------------------------------------------------
  (?!                      look ahead to see if there is not:
--------------------------------------------------------------------------------
    \S                       non-whitespace (all but \n, \r, \t, \f,
                             and " ")
--------------------------------------------------------------------------------
  )                        end of look-ahead
user_113298900
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