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I have the following string
string <- c('a - b - c - d',
'z - c - b',
'y',
'u - z')
I would like to subset it such that everything after the second occurrence of ' - ' is thrown away.
The result would be this:
> string
[1] "a - b" "z - c" "y" "u - z"
I used substr(x = string, 1, regexpr(string, pattern = '[^ - ]*$') - 4), but it excludes the last occurrence of ' - ', which is not what I want .
349
If you need to remove everything after the last occurrence of a specific character, use the lastIndexOf method. We have two sets of brackets in the string and we only want to remove everything after the last set of brackets.
What is most elegant and simple method to achieve this; sed, awk or just unix commands like cut? Everything after the 2nd - should be stripped out. The regex should also match zero occurrences of the pattern, so zero or one occurrence should be ignored and from the 2nd occurrence everything should be removed.
Use the String.slice () method to remove everything after a specific character, e.g. const removed = str.slice (0, str.indexOf (' ['));. The slice method will return the part of the string before the specified character. We passed the following arguments to the String.slice method:
To force it to return only the part after the last occurrence of the letter a, we need to add an end-of-string character ($) at the end: The final method makes use of a positive look-behind, which might not be supported in all regex engines:
Note that you cannot use a negated character class to negate a sequence of characters. [^ - ]*$ matches any 0+ chars other than a space (yes, it matches -, too, because the - created a range between a space and a space) followed by the end of the string marker ($).
You may use a sub function with the following regex:
^(.*? - .*?) - .*
to replace with \1. See the regex demo.
R code:
> string <- c('a - b - c - d', 'z - c - b', 'y', 'u - z')
> sub("^(.*? - .*?) - .*", "\\1", string)
[1] "a - b" "z - c" "y" "u - z"
Details:
^ - start of a string(.*? - .*?) - Group 1 (referred to with the \1 backreference in the replacement pattern) capturing any 0+ chars lazily up to the first space, hyphen, space and then again any 0+ chars up to the next leftmost occurrence of space, hyphen, space - - a space, hyphen and a space.* - any zero or more chars up to the end of the string.
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