I'm reading the regular expressions reference and I'm thinking about ? and ?? characters. Could you explain me with some examples their usefulness? I don't understand them enough.
thank you
This is an excellent question, and it took me a while to see the point of the lazy ?? quantifier myself.
? - Optional (greedy) quantifier
The usefulness of ? is easy enough to understand. If you wanted to find both http and https, you could use a pattern like this:
https?
This pattern will match both inputs, because it makes the s optional.
?? - Optional (lazy) quantifier
?? is more subtle. It usually does the same thing ? does. It doesn't change the true/false result when you ask: "Does this input satisfy this regex?" Instead, it's relevant to the question: "Which part of this input matches this regex, and which parts belong in which groups?" If an input could satisfy the pattern in more than one way, the engine will decide how to group it based on ? vs. ?? (or * vs. *?, or + vs. +?).
Say you have a set of inputs that you want to validate and parse. Here's an (admittedly silly) example:
Input:
http123
https456
httpsomething
Expected result:
Pass/Fail Group 1 Group 2
Pass http 123
Pass https 456
Pass http something
You try the first thing that comes to mind, which is this:
^(http)([a-z\d]+)$
Pass/Fail Group 1 Group 2 Grouped correctly?
Pass http 123 Yes
Pass http s456 No
Pass http something Yes
They all pass, but you can't use the second set of results because you only wanted 456 in Group 2.
Fine, let's try again. Let's say Group 2 can be letters or numbers, but not both:
(https?)([a-z]+|\d+)
Pass/Fail Group 1 Group 2 Grouped correctly?
Pass http 123 Yes
Pass https 456 Yes
Pass https omething No
Now the second input is fine, but the third one is grouped wrong because ? is greedy by default (the + is too, but the ? came first). When deciding whether the s is part of https? or [a-z]+|\d+, if the result is a pass either way, the regex engine will always pick the one on the left. So Group 2 loses s because Group 1 sucked it up.
To fix this, you make one tiny change:
(https??)([a-z]+|\d+)$
Pass/Fail Group 1 Group 2 Grouped correctly?
Pass http 123 Yes
Pass https 456 Yes
Pass http something Yes
Essentially, this means: "Match https if you have to, but see if this still passes when Group 1 is just http." The engine realizes that the s could work as part of [a-z]+|\d+, so it prefers to put it into Group 2.
The key difference between ? and ?? concerns their laziness. ?? is lazy, ? is not.
Let's say you want to search for the word "car" in a body of text, but you don't want to be restricted to just the singular "car"; you also want to match against the plural "cars".
Here's an example sentence:
I own three cars.
Now, if I wanted to match the word "car" and I only wanted to get the string "car" in return, I would use the lazy ?? like so:
cars??
This says, "look for the word car or cars; if you find either, return car and nothing more".
Now, if I wanted to match against the same words ("car" or "cars") and I wanted to get the whole match in return, I'd use the non-lazy ? like so:
cars?
This says, "look for the word car or cars, and return either car or cars, whatever you find".
In the world of computer programming, lazy generally means "evaluating only as much as is needed". So the lazy ?? only returns as much as is needed to make a match; since the "s" in "cars" is optional, don't return it. On the flip side, non-lazy (sometimes called greedy) operations evaluate as much as possible, hence the ? returns all of the match, including the optional "s".
Personally, I find myself using ? as a way of making other regular expression operators lazy (like the * and + operators) more often than I use it for simple character optionality, but YMMV.
See it in Code
Here's the above implemented in Clojure as an example:
(re-find #"cars??" "I own three cars.")
;=> "car"
(re-find #"cars?" "I own three cars.")
;=> "cars"
The item re-find is a function that takes its first argument as a regular expression #"cars??" and returns the first match it finds in the second argument "I own three cars."
Some Other Uses of Question marks in regular expressions
Apart from what's explained in other answers, there are still 3 more uses of Question Marks in regular expressions.
Negative Lookahead
Negative lookaheads are used if you want to
match something not followed by something else. The negative
lookahead construct is the pair of parentheses, with the opening
parenthesis followed by a question mark and an exclamation point. x(?!x2)
example
Consider a word There
Now, by default, the RegEx e will find the third letter e in word There.
There
^
However if you don't want the e which is immediately followed by r, then you can use RegEx e(?!r). Now the result would be:
There
^
Positive Lookahead
Positive lookahead works just the same. q(?=u) matches a q that
is immediately followed by a u, without making the u part of the
match. The positive lookahead construct is a pair of parentheses,
with the opening parenthesis followed by a question mark and an
equals sign.
example
Consider a word getting
Now, by default, the RegEx t will find the third letter t in word getting.
getting
^
However if you want the t which is immediately followed by i, then you can use RegEx t(?=i). Now the result would be:
getting
^
Non-Capturing Groups
Whenever you place a Regular Expression in parenthesis(), they
create a numbered capturing group. It stores the part of the string
matched by the part of the regular expression inside the
parentheses.
If you do not need the group to capture its match, you can optimize
this regular expression into
(?:Value)
See also this and this.
? simply makes the previous item (character, character class, group) optional:
colou?r
matches "color" and "colour"
(swimming )?pool
matches "a pool" and "the swimming pool"
?? is the same, but it's also lazy, so the item will be excluded if at all possible. As those docs note, ?? is rare in practice. I have never used it.
Running the test harness from Oracle documentation with the reluctant quantifier of the "once or not at all" match X?? shows that it works as a guaranteed always-empty match.
$ java RegexTestHarness
Enter your regex: x?
Enter input string to search: xx
I found the text "x" starting at index 0 and ending at index 1.
I found the text "x" starting at index 1 and ending at index 2.
I found the text "" starting at index 2 and ending at index 2.
Enter your regex: x??
Enter input string to search: xx
I found the text "" starting at index 0 and ending at index 0.
I found the text "" starting at index 1 and ending at index 1.
I found the text "" starting at index 2 and ending at index 2.
https://docs.oracle.com/javase/tutorial/essential/regex/quant.html
It seems identical to the empty matcher.
Enter your regex:
Enter input string to search: xx
I found the text "" starting at index 0 and ending at index 0.
I found the text "" starting at index 1 and ending at index 1.
I found the text "" starting at index 2 and ending at index 2.
Enter your regex:
Enter input string to search:
I found the text "" starting at index 0 and ending at index 0.
Enter your regex: x??
Enter input string to search:
I found the text "" starting at index 0 and ending at index 0.
Related
I'm trying to use a regex replace each character after a given position (say, 3) with a placeholder character, for an arbitrary-length string (the output length should be the same as that of the input). I think a lookahead (lookbehind?) can do it, but I can't get it to work.
What I have right now is:
regex: /.(?=.{0,2}$)/
input string: 'hello there'
replace string: '_'
current output: 'hello th___' (last 3 substituted)
The output I'm looking for would be 'hel________' (everything but the first 3 substituted).
I'm doing this in Typescript, to replace some old javascript that is using ugly split/concatenate logic. However, I know how to make the regex calls, so the answer should be pretty language agnostic.
If you know the string is longer than given position n, the start-part can be optionally captured
(^.{3})?.
and replaced with e.g. $1_ (capture of first group and _). Won't work if string length is <= n.
See this demo at regex101
Another option is to use a lookehind as far as supported to check if preceded by n characters.
(?<=.{3}).
See other demo at regex101 (replace just with underscore) - String length does not matter here.
To mention in PHP/PCRE the start-part could simply be skipped like this: ^.{1,3}(*SKIP)(*F)|.
My input is of this format: (xxx)yyyy(zz)(eee)fff where {x,y,z,e,f} are all numbers. But fff is optional though.
Input: x = (123)4567(89)(660)
Expected output: Only the eeepart i.e. the number inside 3rd "()" i.e. 660 in my example.
I am able to achieve this so far:
re.search("\((\d*)\)", x).group()
Output: (123)
Expected: (660)
I am surely missing something fundamental. Please advise.
Edit 1: Just added fff to the input data format.
You could find all those matches that have round braces (), and print the third match with findall
import re
n = "(123)4567(89)(660)999"
r = re.findall("\(\d*\)", n)
print(r[2])
Output:
(660)
The (eee) part is identical to the (xxx) part in your regex. If you don't provide an anchor, or some sequencing requirement, then an unanchored search will match the first thing it finds, which is (xxx) in your case.
If you know the (eee) always appears at the end of the string, you could append an "at-end" anchor ($) to force the match at the end. Or perhaps you could append a following character, like a space or comma or something.
Otherwise, you might do well to match the other parts of the pattern and not capture them:
pattern = r'[0-9()]{13}\((\d{3})\)'
If you want to get the third group of numbers in brackets, you need to skip the first two groups which you can do with a repeating non-capturing group which looks for a set of digits enclosed in () followed by some number of non ( characters:
x = '(123)4567(89)(660)'
print(re.search("(?:\(\d+\)[^(]*){2}(\(\d+\))", x).group(1))
Output:
(660)
Demo on rextester
Example 1:
THE COMPANIES ACT
(Cap 486)
IT IS notified
Example 2:
THE COMPANIES ACT
(Cap. 486)
Incorporations
IT IS notified
My current regex: THE COMPANIES ACT\n\(((?:Cap.|Cap) .*?)\)(?:\nIncorporations|\nincorporations)\nIT IS notifiedonly matches Example 2.
I would like it to match both examples.
You should make (?:\nIncorporations|\nincorporations) optional by appending ? (0 or 1 match) after it. Otherwise, the first example doesn't match as you have specified that you want to match (?:\nIncorporations|\nincorporations) in any case.
As ncorporations is common in both *ncorporations, you could consider (?:\n[Ii]ncorporations)? instead of (?:\nIncorporations|\nincorporations)? and (?:Cap\.?) instead of (?:Cap.|Cap), to shorten it a bit and also to escape the dot (since . means any character).
So I've bought a book on R and automated data collection, and one of the first examples are leaving me baffled.
I have a table with a date-column consisting of numbers looking like this "2001-". According to the tutorial, the line below will remove the "-" from the dates by singling out the first four digits:
yend_clean <- unlist(str_extract_all(danger_table$yend, "[[:digit:]]4$"))
When I run this command, "yend_clean" is simply set to "character (empty)".
If I remove the ”4$", I get all of the dates split into atoms so that the list that originally looked like this "1992", "2003" now looks like this "1", "9" etc.
So I suspect that something around the "4$" is the problem. I can't find any documentation on this that helps me figure out the correct solution.
Was hoping someone in here could point me in the right direction.
This is a regular expression question. Your regular expression is wrong. Use:
unlist(str_extract_all("2003-", "^[[:digit:]]{4}"))
or equivalently
sub("^(\\d{4}).*", "\\1", "2003-")
of if really all you want is to remove the "-"
sub("-", "", "2003-")
Repetition in regular expressions is controlled by the {} parameter. You were missing that. Additionally $ means match the end of the string, so your expression translates as:
match any single digit, followed by a 4, followed by the end of the string
When you remove the "4", then the pattern becomes "match any single digit", which is exactly what happens (i.e. you get each digit matched separately).
The pattern I propose says instead:
match the beginning of the string (^), followed by a digit repeated four times.
The sub variation is a very common technique where we create a pattern that matches what we want to keep in parentheses, and then everything else outside of the parentheses (.* matches anything, any number of times). We then replace the entire match with just the piece in the parens (\\1 means the first sub-expression in parentheses). \\d is equivalent to [[:digit:]].
A good website to learn about regex
A visualization tool to see how specific regular expressions match strings
If you mean the book Automated Data Collection with R, the code could be like this:
yend_clean <- unlist(str_extract_all(danger_table$yend, "[[:digit:]]{4}[-]$"))
yend_clean <- unlist(str_extract_all(yend_clean, "^[[:digit:]]{4}"))
Assumes that you have a string, "1993–2007, 2010-", and you want to get the last given year, which is "2010". The first line, which means four digits and a dash and end, return "2010-", and the second line return "2010".
I'm reading the regular expressions reference and I'm thinking about ? and ?? characters. Could you explain me with some examples their usefulness? I don't understand them enough.
thank you
This is an excellent question, and it took me a while to see the point of the lazy ?? quantifier myself.
? - Optional (greedy) quantifier
The usefulness of ? is easy enough to understand. If you wanted to find both http and https, you could use a pattern like this:
https?
This pattern will match both inputs, because it makes the s optional.
?? - Optional (lazy) quantifier
?? is more subtle. It usually does the same thing ? does. It doesn't change the true/false result when you ask: "Does this input satisfy this regex?" Instead, it's relevant to the question: "Which part of this input matches this regex, and which parts belong in which groups?" If an input could satisfy the pattern in more than one way, the engine will decide how to group it based on ? vs. ?? (or * vs. *?, or + vs. +?).
Say you have a set of inputs that you want to validate and parse. Here's an (admittedly silly) example:
Input:
http123
https456
httpsomething
Expected result:
Pass/Fail Group 1 Group 2
Pass http 123
Pass https 456
Pass http something
You try the first thing that comes to mind, which is this:
^(http)([a-z\d]+)$
Pass/Fail Group 1 Group 2 Grouped correctly?
Pass http 123 Yes
Pass http s456 No
Pass http something Yes
They all pass, but you can't use the second set of results because you only wanted 456 in Group 2.
Fine, let's try again. Let's say Group 2 can be letters or numbers, but not both:
(https?)([a-z]+|\d+)
Pass/Fail Group 1 Group 2 Grouped correctly?
Pass http 123 Yes
Pass https 456 Yes
Pass https omething No
Now the second input is fine, but the third one is grouped wrong because ? is greedy by default (the + is too, but the ? came first). When deciding whether the s is part of https? or [a-z]+|\d+, if the result is a pass either way, the regex engine will always pick the one on the left. So Group 2 loses s because Group 1 sucked it up.
To fix this, you make one tiny change:
(https??)([a-z]+|\d+)$
Pass/Fail Group 1 Group 2 Grouped correctly?
Pass http 123 Yes
Pass https 456 Yes
Pass http something Yes
Essentially, this means: "Match https if you have to, but see if this still passes when Group 1 is just http." The engine realizes that the s could work as part of [a-z]+|\d+, so it prefers to put it into Group 2.
The key difference between ? and ?? concerns their laziness. ?? is lazy, ? is not.
Let's say you want to search for the word "car" in a body of text, but you don't want to be restricted to just the singular "car"; you also want to match against the plural "cars".
Here's an example sentence:
I own three cars.
Now, if I wanted to match the word "car" and I only wanted to get the string "car" in return, I would use the lazy ?? like so:
cars??
This says, "look for the word car or cars; if you find either, return car and nothing more".
Now, if I wanted to match against the same words ("car" or "cars") and I wanted to get the whole match in return, I'd use the non-lazy ? like so:
cars?
This says, "look for the word car or cars, and return either car or cars, whatever you find".
In the world of computer programming, lazy generally means "evaluating only as much as is needed". So the lazy ?? only returns as much as is needed to make a match; since the "s" in "cars" is optional, don't return it. On the flip side, non-lazy (sometimes called greedy) operations evaluate as much as possible, hence the ? returns all of the match, including the optional "s".
Personally, I find myself using ? as a way of making other regular expression operators lazy (like the * and + operators) more often than I use it for simple character optionality, but YMMV.
See it in Code
Here's the above implemented in Clojure as an example:
(re-find #"cars??" "I own three cars.")
;=> "car"
(re-find #"cars?" "I own three cars.")
;=> "cars"
The item re-find is a function that takes its first argument as a regular expression #"cars??" and returns the first match it finds in the second argument "I own three cars."
Some Other Uses of Question marks in regular expressions
Apart from what's explained in other answers, there are still 3 more uses of Question Marks in regular expressions.
Negative Lookahead
Negative lookaheads are used if you want to
match something not followed by something else. The negative
lookahead construct is the pair of parentheses, with the opening
parenthesis followed by a question mark and an exclamation point. x(?!x2)
example
Consider a word There
Now, by default, the RegEx e will find the third letter e in word There.
There
^
However if you don't want the e which is immediately followed by r, then you can use RegEx e(?!r). Now the result would be:
There
^
Positive Lookahead
Positive lookahead works just the same. q(?=u) matches a q that
is immediately followed by a u, without making the u part of the
match. The positive lookahead construct is a pair of parentheses,
with the opening parenthesis followed by a question mark and an
equals sign.
example
Consider a word getting
Now, by default, the RegEx t will find the third letter t in word getting.
getting
^
However if you want the t which is immediately followed by i, then you can use RegEx t(?=i). Now the result would be:
getting
^
Non-Capturing Groups
Whenever you place a Regular Expression in parenthesis(), they
create a numbered capturing group. It stores the part of the string
matched by the part of the regular expression inside the
parentheses.
If you do not need the group to capture its match, you can optimize
this regular expression into
(?:Value)
See also this and this.
? simply makes the previous item (character, character class, group) optional:
colou?r
matches "color" and "colour"
(swimming )?pool
matches "a pool" and "the swimming pool"
?? is the same, but it's also lazy, so the item will be excluded if at all possible. As those docs note, ?? is rare in practice. I have never used it.
Running the test harness from Oracle documentation with the reluctant quantifier of the "once or not at all" match X?? shows that it works as a guaranteed always-empty match.
$ java RegexTestHarness
Enter your regex: x?
Enter input string to search: xx
I found the text "x" starting at index 0 and ending at index 1.
I found the text "x" starting at index 1 and ending at index 2.
I found the text "" starting at index 2 and ending at index 2.
Enter your regex: x??
Enter input string to search: xx
I found the text "" starting at index 0 and ending at index 0.
I found the text "" starting at index 1 and ending at index 1.
I found the text "" starting at index 2 and ending at index 2.
https://docs.oracle.com/javase/tutorial/essential/regex/quant.html
It seems identical to the empty matcher.
Enter your regex:
Enter input string to search: xx
I found the text "" starting at index 0 and ending at index 0.
I found the text "" starting at index 1 and ending at index 1.
I found the text "" starting at index 2 and ending at index 2.
Enter your regex:
Enter input string to search:
I found the text "" starting at index 0 and ending at index 0.
Enter your regex: x??
Enter input string to search:
I found the text "" starting at index 0 and ending at index 0.