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I need to capture multiple groups of the same pattern. Suppose, I have the following string:
HELLO,THERE,WORLD
And I've written the following pattern
^(?:([A-Z]+),?)+$
What I want it to do is to capture every single word, so that Group 1 is : "HELLO", Group 2 is "THERE" and Group 3 is "WORLD". What my regex is actually capturing is only the last one, which is "WORLD".
I'm testing my regular expression here and I want to use it with Swift (maybe there's a way in Swift to get intermediate results somehow, so that I can use them?)
UPDATE: I don't want to use split. I just need to now how to capture all the groups that match the pattern, not only the last one.
With one group in the pattern, you can only get one exact result in that group. If your capture group gets repeated by the pattern (you used the + quantifier on the surrounding non-capturing group), only the last value that matches it gets stored.
You have to use your language's regex implementation functions to find all matches of a pattern, then you would have to remove the anchors and the quantifier of the non-capturing group (and you could omit the non-capturing group itself as well).
Alternatively, expand your regex and let the pattern contain one capturing group per group you want to get in the result:
^([A-Z]+),([A-Z]+),([A-Z]+)$
The key distinction is repeating a captured group instead of capturing a repeated group.
As you have already found out, the difference is that repeating a captured group captures only the last iteration. Capturing a repeated group captures all iterations.
In PCRE (PHP):
((?:\w+)+),?
Match 1, Group 1. 0-5 HELLO
Match 2, Group 1. 6-11 THERE
Match 3, Group 1. 12-20 BRUTALLY
Match 4, Group 1. 21-26 CRUEL
Match 5, Group 1. 27-32 WORLD
Since all captures are in Group 1, you only need $1 for substitution.
I used the following general form of this regular expression:
((?:{{RE}})+)
Example at regex101
I think you need something like this....
b="HELLO,THERE,WORLD"
re.findall('[\w]+',b)
Which in Python3 will return
['HELLO', 'THERE', 'WORLD']
After reading Byte Commander's answer, I want to introduce a tiny possible improvement:
You can generate a regexp that will match either n words, as long as your n is predetermined. For instance, if I want to match between 1 and 3 words, the regexp:
^([A-Z]+)(?:,([A-Z]+))?(?:,([A-Z]+))?$
will match the next sentences, with one, two or three capturing groups.
HELLO,LITTLE,WORLD
HELLO,WORLD
HELLO
You can see a fully detailed explanation about this regular expression on Regex101.
As I said, it is pretty easy to generate this regexp for any groups you want using your favorite language. Since I'm not much of a swift guy, here's a ruby example:
def make_regexp(group_regexp, count: 3, delimiter: ",")
regexp_str = "^(#{group_regexp})"
(count - 1).times.each do
regexp_str += "(?:#{delimiter}(#{group_regexp}))?"
end
regexp_str += "$"
return regexp_str
end
puts make_regexp("[A-Z]+")
That being said, I'd suggest not using regular expression in that case, there are many other great tools from a simple split to some tokenization patterns depending on your needs. IMHO, a regular expression is not one of them. For instance in ruby I'd use something like str.split(",") or str.scan(/[A-Z]+/)
Just to provide additional example of paragraph 2 in the answer. I'm not sure how critical it is for you to get three groups in one match rather than three matches using one group. E.g., in groovy:
def subject = "HELLO,THERE,WORLD"
def pat = "([A-Z]+)"
def m = (subject =~ pat)
m.eachWithIndex{ g,i ->
println "Match #$i: ${g[1]}"
}
Match #0: HELLO
Match #1: THERE
Match #2: WORLD
The problem with the attempted code, as discussed, is that there is one capture group matching repeatedly so in the end only the last match can be kept.
Instead, instruct the regex to match (and capture) all pattern instances in the string, what can be done in any regex implementation (language). So come up with the regex pattern for this.
The defining property of the shown sample data is that the patterns of interest are separated by commas so we can match anything-but-a-comma, using a negated character class
[^,]+
and match (capture) globally, to get all matches in the string.
If your pattern need be more restrictive then adjust the exclusion list. For example, to capture words separated by any of the listed punctuation
[^,.!-]+
This extracts all words from hi,there-again!, without the punctuation. (The - itself should be given first or last in a character class, unless it's used in a range like a-z or 0-9.)
In Python
import re
string = "HELLO,THERE,WORLD"
pattern = r"([^,]+)"
matches = re.findall(pattern,string)
print(matches)
In Perl (and many other compatible systems)
use warnings;
use strict;
use feature 'say';
my $string = 'HELLO,THERE,WORLD';
my #matches = $string =~ /([^,]+)/g;
say "#matches";
(In this specific example the capturing () in fact aren't needed since we collect everything that is matched. But they don't hurt and in general they are needed.)
The approach above works as it stands for other patterns as well, including the one attempted in the question (as long as you remove the anchors which make it too specific). The most common one is to capture all words (usually meaning [a-zA-Z0-9_]), with the pattern \w+. Or, as in the question, get only the substrings of upper-case ascii letters[A-Z]+.
I know that my answer came late but it happens to me today and I solved it with the following approach:
^(([A-Z]+),)+([A-Z]+)$
So the first group (([A-Z]+),)+ will match all the repeated patterns except the final one ([A-Z]+) that will match the final one. and this will be dynamic no matter how many repeated groups in the string.
You actually have one capture group that will match multiple times. Not multiple capture groups.
javascript (js) solution:
let string = "HI,THERE,TOM";
let myRegexp = /([A-Z]+),?/g; // modify as you like
let match = myRegexp.exec(string); // js function, output described below
while (match != null) { // loops through matches
console.log(match[1]); // do whatever you want with each match
match = myRegexp.exec(string); // find next match
}
Syntax:
// matched text: match[0]
// match start: match.index
// capturing group n: match[n]
As you can see, this will work for any number of matches.
Sorry, not Swift, just a proof of concept in the closest language at hand.
// JavaScript POC. Output:
// Matches: ["GOODBYE","CRUEL","WORLD","IM","LEAVING","U","TODAY"]
let str = `GOODBYE,CRUEL,WORLD,IM,LEAVING,U,TODAY`
let matches = [];
function recurse(str, matches) {
let regex = /^((,?([A-Z]+))+)$/gm
let m
while ((m = regex.exec(str)) !== null) {
matches.unshift(m[3])
return str.replace(m[2], '')
}
return "bzzt!"
}
while ((str = recurse(str, matches)) != "bzzt!") ;
console.log("Matches: ", JSON.stringify(matches))
Note: If you were really going to use this, you would use the position of the match as given by the regex match function, not a string replace.
Design a regex that matches each particular element of the list rather then a list as a whole. Apply it with /g
Iterate throught the matches, cleaning them from any garbage such as list separators that got mixed in. You may require another regex, or you can get by with simple replace substring method.
The sample code is in JS, sorry :) The idea must be clear enough.
const string = 'HELLO,THERE,WORLD';
// First use following regex matches each of the list items separately:
const captureListElement = /^[^,]+|,\w+/g;
const matches = string.match(captureListElement);
// Some of the matches may include the separator, so we have to clean them:
const cleanMatches = matches.map(match => match.replace(',',''));
console.log(cleanMatches);
repeat the A-Z pattern in the group for the regular expression.
data="HELLO,THERE,WORLD"
pattern=r"([a-zA-Z]+)"
matches=re.findall(pattern,data)
print(matches)
output
['HELLO', 'THERE', 'WORLD']
In my code I Want answer [('22', '254', '15', '36')] but got [('15', '36')]. My regex (?:([0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)\.){3} is not run for 3 time may be!
import re
def fun(st):
print(re.findall("(?:([0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)\.){3}([0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)",st))
ip="22.254.15.36"
print(fun(ip))
Overview
As I mentioned in the comments below your question, most regex engines only capture the last match. So when you do (...){3}, only the last match is captured: E.g. (.){3} used against abc will only return c.
Also, note that changing your regex to (2[0-4]\d|25[0-5]|[01]?\d{1,2}) performs much better and catches full numbers (currently you'll grab 25 instead of 255 on the last octet for example - unless you anchor it to the end).
To give you a fully functional regex for capturing each octet of the IP:
(2[0-4]\d|25[0-5]|[01]?\d{1,2})\.(2[0-4]\d|25[0-5]|[01]?\d{1,2})\.(2[0-4]\d|25[0-5]|[01]?\d{1,2})\.(2[0-4]\d|25[0-5]|[01]?\d{1,2})
Personally, however, I'd separate the logic from the validation. The code below first validates the format of the string and then checks whether or not the logic (no octets greater than 255) passes while splitting the string on ..
Code
See code in use here
import re
ip='22.254.15.36'
if re.match(r"(?:\d{1,3}\.){3}\d{1,3}$", ip):
print([octet for octet in ip.split('.') if int(octet) < 256])
Result: ['22', '254', '15', '36']
If you're using this method to extract IPs from an arbitrary string, you can replace re.match() with re.search() or re.findall(). In that case you may want to remove $ and add some logic to ensure you're not matching special cases like 11.11.11.11.11: (?<!\d\.)\b(?:\d{1,3}\.){3}\d{1,3}\b(?!\.\d)
You only have two capturing groups in your regex:
(?: # non-capturing group
( # group 1
[0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?
)\.
){3}
( # group 2
[0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?
)
That the first group can be repeated 3 times doesn't make it capture 3 times. The regex engine will only ever return 2 groups, and the last match in a given group will fill that group.
If you want to capture each of the parts of an IP address into separate groups, you'll have to explicitly define groups for each:
pattern = (
r'([0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)\.'
r'([0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)\.'
r'([0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)\.'
r'([0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)')
def fun(st, p=re.compile(pattern)):
return p.findall(st)
You could avoid that much repetition with a little string and list manipulation:
octet = r'([0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)'
pattern = r'\.'.join([octet] * 4)
Next, the pattern will just as happily match the 25 portion of 255. Better to put matching of the 200-255 range at the start over matching smaller numbers:
octet = r'(2(?:5[0-5]|[0-4]\d)|[01]?[0-9]{1,2})'
pattern = r'\.'.join([octet] * 4)
This still allows leading 0 digits, by the way, but is
If all you are doing is passing in single IP addresses, then re.findall() is overkill, just use p.match() (matching only at the string start) or p.search(), and return the .groups() result if there is a match;)
def fun(st, p=re.compile(pattern + '$')):
match = p.match(st)
return match and match.groups()
Note that no validation is done on the surrounding data, so if you are trying to extract IP addresses from a larger body of text you can't use re.match(), and can't add the $ anchor and the match could be from a larger number of octets (e.g. 22.22.22.22.22.22). You'd have to add some look-around operators for that:
# only match an IP address if there is no indication that it is part of a larger
# set of octets; no leading or trailing dot or digits
pattern = r'(?<![\.\d])' + pattern + r'(?![\.\d])'
I encountered a very similar issue.
I found two solutions, using the official documentation.
The answer of #ctwheels above did mention the cause of the problem, and I really appreciate it, but it did not provide a solution.
Even when trying the lookbehind and the lookahead, it did not work.
First solution:
re.finditer
re.finditer iterates over match objects !!
You can use each one's 'group' method !
>>> def fun(st):
pr=re.finditer("(?:([0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)\.){3}([0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)",st)
for p in pr:
print(p.group(),end="")
>>> fun(ip)
22.254.15.36
Or !!!
Another solution haha : You can still use findall, but you'll have to make every group a non-capturing group ! (Since the main problem is not with findall, but with the group function that is used by findall (which, we all know, only returns the last match):
"re.findall:
...If one or more groups are present in the pattern, return a list of groups"
(Python 3.8 Manuals)
So:
>>> def fun(st):
print(re.findall("(?:(?:[0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)\.){3}(?:[0-1]?[0-9]{0,2}|2?[0-4]?[0-9]|25[0-5]?)",st))
>>> fun(ip)
['22.254.15.36']
Have fun !
Suppose I want to match a string like this:
123(432)123(342)2348(34)
I can match digits like 123 with [\d]* and (432) with \([\d]+\).
How can match the whole string by repeating either of the 2 patterns?
I tried [[\d]* | \([\d]+\)]+, but this is incorrect.
I am using python re module.
I think you need this regex:
"^(\d+|\(\d+\))+$"
and to avoid catastrophic backtracking you need to change it to a regex like this:
"^(\d|\(\d+\))+$"
You can use a character class to match the whole of string :
[\d()]+
But if you want to match the separate parts in separate groups you can use re.findall with a spacial regex based on your need, for example :
>>> import re
>>> s="123(432)123(342)2348(34)"
>>> re.findall(r'\d+\(\d+\)',s)
['123(432)', '123(342)', '2348(34)']
>>>
Or :
>>> re.findall(r'(\d+)\((\d+)\)',s)
[('123', '432'), ('123', '342'), ('2348', '34')]
Or you can just use \d+ to get all the numbers :
>>> re.findall(r'\d+',s)
['123', '432', '123', '342', '2348', '34']
If you want to match the patter \d+\(\d+\) repeatedly you can use following regex :
(?:\d+\(\d+\))+
You can achieve it with this pattern:
^(?=.)\d*(?:\(\d+\)\d*)*$
demo
(?=.) ensures there is at least one character (if you want to allow empty strings, remove it).
\d*(?:\(\d+\)\d*)* is an unrolled sub-pattern. Explanation: With a bactracking regex engine, when you have a sub-pattern like (A|B)* where A and B are mutually exclusive (or at least when the end of A or B doesn't match respectively the beginning of B or A), you can rewrite the sub-pattern like this: A*(BA*)* or B*(AB*)*. For your example, it replaces (?:\d+|\(\d+\))*
This new form is more efficient: it reduces the steps needed to obtain a match, it avoids a great part of the eventual bactracking.
Note that you can improve it more, if you emulate an atomic group (?>....) with this trick (?=(....))\1 that uses the fact that a lookahead is naturally atomic:
^(?=.)(?=(\d*(?:\(\d+\)\d*)*))\1$
demo (compare the number of steps needed with the previous version and check the debugger to see what happens)
Note: if you don't want two consecutive numbers enclosed in parenthesis, you only need to change the quantifier * with + inside the non-capturing group and to add (?:\(\d+\))? at the end of the pattern, before the anchor $:
^(?=.)\d*(?:\(\d+\)\d+)*(?:\(\d+\))?$
or
^(?=.)(?=(\d*(?:\(\d+\)\d+)*(?:\(\d+\))?))\1$
I'd like to match three-character sequences of letters (only letters 'a', 'b', 'c' are allowed) separated by comma (last group is not ended with comma).
Examples:
abc,bca,cbb
ccc,abc,aab,baa
bcb
I have written following regular expression:
re.match('([abc][abc][abc],)+', "abc,defx,df")
However it doesn't work correctly, because for above example:
>>> print bool(re.match('([abc][abc][abc],)+', "abc,defx,df")) # defx in second group
True
>>> print bool(re.match('([abc][abc][abc],)+', "axc,defx,df")) # 'x' in first group
False
It seems only to check first group of three letters but it ignores the rest. How to write this regular expression correctly?
Try following regex:
^[abc]{3}(,[abc]{3})*$
^...$ from the start till the end of the string
[...] one of the given character
...{3} three time of the phrase before
(...)* 0 till n times of the characters in the brackets
What you're asking it to find with your regex is "at least one triple of letters a, b, c" - that's what "+" gives you. Whatever follows after that doesn't really matter to the regex. You might want to include "$", which means "end of the line", to be sure that the line must all consist of allowed triples. However in the current form your regex would also demand that the last triple ends in a comma, so you should explicitly code that it's not so.
Try this:
re.match('([abc][abc][abc],)*([abc][abc][abc])$'
This finds any number of allowed triples followed by a comma (maybe zero), then a triple without a comma, then the end of the line.
Edit: including the "^" (start of string) symbol is not necessary, because the match method already checks for a match only at the beginning of the string.
The obligatory "you don't need a regex" solution:
all(letter in 'abc,' for letter in data) and all(len(item) == 3 for item in data.split(','))
You need to iterate over sequence of found values.
data_string = "abc,bca,df"
imatch = re.finditer(r'(?P<value>[abc]{3})(,|$)', data_string)
for match in imatch:
print match.group('value')
So the regex to check if the string matches pattern will be
data_string = "abc,bca,df"
match = re.match(r'^([abc]{3}(,|$))+', data_string)
if match:
print "data string is correct"
Your result is not surprising since the regular expression
([abc][abc][abc],)+
tries to match a string containing three characters of [abc] followed by a comma one ore more times anywhere in the string. So the most important part is to make sure that there is nothing more in the string - as scessor suggests with adding ^ (start of string) and $ (end of string) to the regular expression.
An alternative without using regex (albeit a brute force way):
>>> def matcher(x):
total = ["".join(p) for p in itertools.product(('a','b','c'),repeat=3)]
for i in x.split(','):
if i not in total:
return False
return True
>>> matcher("abc,bca,aaa")
True
>>> matcher("abc,bca,xyz")
False
>>> matcher("abc,aaa,bb")
False
If your aim is to validate a string as being composed of triplet of letters a,b,and c:
for ss in ("abc,bbc,abb,baa,bbb",
"acc",
"abc,bbc,abb,bXa,bbb",
"abc,bbc,ab,baa,bbb"):
print ss,' ',bool(re.match('([abc]{3},?)+\Z',ss))
result
abc,bbc,abb,baa,bbb True
acc True
abc,bbc,abb,bXa,bbb False
abc,bbc,ab,baa,bbb False
\Z means: the end of the string. Its presence obliges the match to be until the very end of the string
By the way, I like the form of Sonya too, in a way it is clearer:
bool(re.match('([abc]{3},)*[abc]{3}\Z',ss))
To just repeat a sequence of patterns, you need to use a non-capturing group, a (?:...) like contruct, and apply a quantifier right after the closing parenthesis. The question mark and the colon after the opening parenthesis are the syntax that creates a non-capturing group (SO post).
For example:
(?:abc)+ matches strings like abc, abcabc, abcabcabc, etc.
(?:\d+\.){3} matches strings like 1.12.2., 000.00000.0., etc.
Here, you can use
^[abc]{3}(?:,[abc]{3})*$
^^
Note that using a capturing group is fraught with unwelcome effects in a lot of Python regex methods. See a classical issue described at re.findall behaves weird post, for example, where re.findall and all other regex methods using this function behind the scenes only return captured substrings if there is a capturing group in the pattern.
In Pandas, it is also important to use non-capturing groups when you just need to group a pattern sequence: Series.str.contains will complain that this pattern has match groups. To actually get the groups, use str.extract. and
the Series.str.extract, Series.str.extractall and Series.str.findall will behave as re.findall.
Is there a way using a regex to match a repeating set of characters? For example:
ABCABCABCABCABC
ABC{5}
I know that's wrong. But is there anything to match that effect?
Update:
Can you use nested capture groups? So Something like (?<cap>(ABC){5}) ?
Enclose the regex you want to repeat in parentheses. For instance, if you want 5 repetitions of ABC:
(ABC){5}
Or if you want any number of repetitions (0 or more):
(ABC)*
Or one or more repetitions:
(ABC)+
edit to respond to update
Parentheses in regular expressions do two things; they group together a sequence of items in a regular expression, so that you can apply an operator to an entire sequence instead of just the last item, and they capture the contents of that group so you can extract the substring that was matched by that subexpression in the regex.
You can nest parentheses; they are counted from the first opening paren. For instance:
>>> re.search('[0-9]* (ABC(...))', '123 ABCDEF 456').group(0)
'123 ABCDEF'
>>> re.search('[0-9]* (ABC(...))', '123 ABCDEF 456').group(1)
'ABCDEF'
>>> re.search('[0-9]* (ABC(...))', '123 ABCDEF 456').group(2)
'DEF'
If you would like to avoid capturing when you are grouping, you can use (?:. This can be helpful if you don't want parentheses that you're just using to group together a sequence for the purpose of applying an operator to change the numbering of your matches. It is also faster.
>>> re.search('[0-9]* (?:ABC(...))', '123 ABCDEF 456').group(1)
'DEF'
So to answer your update, yes, you can use nested capture groups, or even avoid capturing with the inner group at all:
>>> re.search('((?:ABC){5})(DEF)', 'ABCABCABCABCABCDEF').group(1)
'ABCABCABCABCABC'
>>> re.search('((?:ABC){5})(DEF)', 'ABCABCABCABCABCDEF').group(2)
'DEF'
ABC{5} matches ABCCCCC. To match 5 ABC's, you should use (ABC){5}. Parentheses are used to group a set of characters. You can also set an interval for occurrences like (ABC){3,5} which matches ABCABCABC, ABCABCABCABC, and ABCABCABCABCABC.
(ABC){1,} means 1 or more repetition which is exactly the same as (ABC)+.
(ABC){0,} means 0 or more repetition which is exactly the same as (ABC)*.
(ABC){5} Should work for you
Parentheses "()" are used to group characters and expressions within larger, more complex regular expressions. Quantifiers that immediately follow the group apply to the whole group.
(ABC){5}
As to the update to the question-
You can nest capture groups. The capture group index is incremented per open paren.
(((ABC)*)(DEF)*)
Feeding that regex ABCABCABCDEFDEFDEF, capture group 0 matches the whole thing, 1 is also the whole thing, 2 is ABCABCABC, 3 is ABC, and 4 is DEF (because the star is outside of the capture group).
If you have variation inside a capture group and a repeat just outside, then things can get a little wonky if you're not expecting it...
(a[bc]*c)*
when fed abbbcccabbc will return the last match as capture group 1, in this example just the abbc, since the capture group gets reset with the repeat operator.