backreference

I have strings in the form \d+_\d+ and I want to add 1 to the second number. Since my explanation is so very clear, let me give you a few examples: 1234567_2 should become 1234567_3 1234_10 should become 1234_11 Here is my first attempt: $new = preg_replace("/(\d+)_(\d+)/", "$1_".((int)$2)+1, $old); This results in a syntax error: Parse error: syntax error, unexpected T_LNUMBER, expecting T_VARIABLE or '$' in [...] on line 201 Here is my second attempt $new = preg_replace("/(\d+)_(\d+)/", "$1_".("$2"+1), $old); This transforms $old = 1234567_2 into $new = 1234567_1, which is not the desired

While running some tests for this answer , I noticed the following unexpected behavior. This will remove all occurrences of <tag> after the first: var input = "<text><text>extra<words><text><words><something>"; Regex.Replace(input, @"(<[^>]+>)(?<=\1.*\1)", ""); // <text>extra<words><something> But this will not: Regex.Replace(input, @"(?<=\1.*)(<[^>]+>)", ""); // <text><text>extra<words><text><words><something> Similarly, this will remove all occurences of <tag> before the last: Regex.Replace(input, @"(<[^>]+>)(?=.*\1)", ""); // extra<text><words><something> But this will not: Regex.Replace

In R I can use \\1 to reference to a capturing group. However, when using the stringi package, this doesn't work as expected. library(stringi) fileName <- "hello-you.lst" (fileName <- stri_replace_first_regex(fileName, "(.*)\\.lst$", "\\1")) [1] "1" Expected output: hello-you . In the documentation I couldn't find anything concerning this problem. You need to use $1 instead of \\1 in the replacement string: library(stringi) fileName <- "hello-you.lst" fileName <- stri_replace_first_regex(fileName, "(.*)\\.lst$", "$1") [1] "hello-you" From the doc , stri_*_regex uses ICU's regular expressions

In a regex replacement pattern, a backreference looks like \1 . If you want to include a digit after that backreference, this will fail because the digit is considered to be part of the backreference number: # replace all twin digits by zeroes, but retain white space in between re.sub(r"\d(\s*)\d", r"0\10", "0 1") >>> sre_constants.error: invalid group reference Substitution pattern r"0\1 0" would work fine but in the failing example back-reference \1 is interpreted as \10 . How can the digit '0' be separated from the back-reference \1 that precedes it? You can use \g<1> , as mentioned in the

Say I have strings like the following: old_string = "I love the number 3 so much" I would like to spot the integer numbers (in the example above, there is only one number, 3 ), and replace them with a value larger by 1, i.e., the desired result should be new_string = "I love the number 4 so much" In Python, I can use: r = re.compile(r'([0-9])+') new_string = r.sub(r'\19', s) to append a 9 at the end of the integer numbers matched. However, I would like to apply something more general on \1 . If I define a function: def f(i): return i + 1 How do I apply f() on \1 , so that I can replace the

In Perl regexes, expressions like \1 , \2 , etc. are usually interpreted as "backreferences" to previously captured groups, but not so when the \1 , \2 , etc. appear within a character class. In the latter case, the \ is treated as an escape character (and therefore \1 is just 1 , etc.). Therefore, if (for example) one wanted to match a string (of length greater than 1) whose first character matches its last character, but does not appear anywhere else in the string, the following regex will not do: /\A # match beginning of string; (.) # match and capture first character (referred to