I have a string like codecodecodecodecode...... I need to find a repeated word in that string.
I found a way but the regular expression always returns half of the repeated part I want.
^(.*)\1+$
at the group(1) I want to see just "code"
If it is greedy, it will first match till the end of the line, and will then backtrack until it can repeat 1 or more times till the end of the string, and for an evenly divided part like this of 4 words, you can capture 2 words and match the same 2 words with the backreference \1
If you have 5 words like codecodecodecodecode as in your example there will be a single group, as the only repetition it can do until the end of the string is 5 repetitions.
The quantifier should be non greedy (and repeat 1+ times to not match an empty string) to match as least as possible characters that can be repeated to the right till the end of the string.
^(.+?)\1+$
regex demo
Related
I'm trying to match some sort of amount, here are all possibilities:
$5.6 million
$4,1 million
$8,1M
$6.3M
$333,333
$2 million
$5 million
I have already this regex:
\$\d{1,3}(?:,\d{3})*(?:\s+(?:thousand|[mb]illion|[MB]illion)|[M])?
See online demo.
But I'm not able to match those ones:
$5.6 million
$4,1 million
$8,1M
$6.3M
Any help would be appreciated.
Let's look at your regular expression:
\$\d{1,3}(?:,\d{3})*(?:\s+(?:thousand|[mb]illion|[MB]illion)|[M])?
\$\d{1,3} is fine. What follows? One way to answer that is to consider the following three possibilities.
The string to be matched ends ' million'
This string (which begins with a space, in case you missed that) is preceded by an empty string or a single digit preceded by a comma or period:
(?:[,.]\d)? million
Evidently, "million" can be "thousand" or "billion", and the first in last might be capitalized, so we change the expression to
(?:[,.]\d)? (?:[MmBb]illion|thousand)
One potential problem is that this matches '$5.6 millionaire'. We can avoid that problem by tacking on a word boundary preventing the match to be followed by a word character:
(?:[,.]\d)? (?:[MmBb]illion|thousand)\b
The string ends 'M'
In this case the 'M' must be preceded by a single digit preceded by a comma or period:
[,.]\dM\b
You could accept 'B' as well by changing M to [MB].
The string ends with three digits preceded by a comma
Here we need
,\d{3}\b
Here the word boundary avoids matching, for example, $333,3333'. It will not match, however, '$333,333,333' or '$333,333,333,333'. If we want to match those we could change the expression to
(?:,\d{3})+\b
or to match '$333' as well, change it to
(?:,\d{3})*\b
Construct the alternation
We therefore can use the following regular expression.
\$\d{1,3}(?:(?:[,.]\d)? (?:[MmBb]illion|thousand)\b|[,.]\dMb|,\d{3}b)
Factoring out the end-of-string anchor we obtain
\$\d{1,3}(?:(?:[,.]\d)? (?:[MmBb]illion|thousand)|[,.]\dM|,\d{3})b
Demo
You can use
(?i)\$\d+(?:[.,]\d+)*(?:\s+(?:thousand|[mb]illion)|m)?
If you need to make sure you do not match m that is part of another word:
(?i)\$\d+(?:[.,]\d+)*(?:\s+(?:thousand|[mb]illion)|m)?\b
See the regex demo. Details:
(?i) - case insensitive option
\$ - a $ char
\d+ - one or more digits
(?:[.,]\d+)* - zero or more repetitions of . or , and then one or more digits
(?:\s+(?:thousand|[mb]illion)|m)? - an optional occurrence of
\s+(?:thousand|[mb]illion) - one or more whitespaces and then thousand, million or billion
| - or
m - an m char
\b - a word boundary.
I want to match strings in which every second character is same.
for example 'abababababab'
I have tried this : '''(([a-z])[^/2])*'''
The output should return the complete string as it is like 'abababababab'
This is actually impossible to do in a real regular expression with an amount of states polynomial to the alphabet size, because the expression is not a Chomsky level-0 grammar.
However, Python's regexes are not actually regular expressions, and can handle much more complex grammars than that. In particular, you could put your grammar as the following.
(..)\1*
(..) is a sequence of 2 characters. \1* matches the exact pair of characters an arbitrary (possibly null) number of times.
I interpreted your question as wanting every other character to be equal (ababab works, but abcbdb fails). If you needed only the 2nd, 4th, ... characters to be equal you can use a similar one.
.(.)(.\1)*
You could match the first [a-z] followed by capturing ([a-z]) in a group. Then repeat 0+ times matching again a-z and a backreference to group 1 to keep every second character the same.
^[a-z]([a-z])(?:[a-z]\1)*$
Explanation
^ Start of the string
[a-z]([a-z]) Match a-z and capture in group 1 matching a-z
)(?:[a-z]\1)* Repeat 0+ times matching a-z followed by a backreference to group 1
$ End of string
Regex demo
Though not a regex answer, you could do something like this:
def all_same(string):
return all(c == string[1] for c in string[1::2])
string = 'abababababab'
print('All the same {}'.format(all_same(string)))
string = 'ababacababab'
print('All the same {}'.format(all_same(string)))
the string[1::2] says start at the 2nd character (1) and then pull out every second character (the 2 part).
This returns:
All the same True
All the same False
This is a bit complicated expression, maybe we would start with:
^(?=^[a-z]([a-z]))([a-z]\1)+$
if I understand the problem right.
Demo
Following regex matches both 59-59-59 and 59-59-59-59 and outputs only 59
The intent is to match four and only numbers followed by - with the max number being 59. Numbers less than 10 are represented as 00-09.
print(re.match(r'(\b[0-5][0-9]-{1,4}\b)','59-59-59').groups())
--> output ('59-',)
I need a pattern match that matches exactly 59-59-59-59
and does not match 59--59-59or 59-59-59-59-59
Try using the following pattern, if using re.match:
[0-5][0-9](?:-[0-5][0-9]){3}$
This is phrased to match an initial number starting with 0 through 5, followed by any second digit. Then, this is followed by a dash and a number with the same rules, this quantity three times exactly. Note that re.match anchor at the beginning by default, so we only need an ending anchor $.
Code:
print(re.match(r'([0-5][0-9](?:-[0-5][0-9]){3})$', '59-59-59-59').groups())
('59-59-59-59',)
If you intend to actually match the same number four times in a row, then see the answer by #Thefourthbird.
If you want to find such a string in a larger text, then consider using re.search. In that case, use this pattern:
(?:^|(?<=\s))[0-5][0-9](?:-[0-5][0-9]){3}(?=\s|$)
Note that instead of using word boundaries \b I used lookarounds to enforce the end of the "word" here. This means that the above pattern will not match something like 59-59-59-59-59.
In your pattern, this part -{1,4} matches 1-4 times a hyphen so 59-- will match.
If all the matches should be the same as 59, you could use a backreference to the first capturing group and repeat that 3 times with a prepended hyphen.
\b([0-5][0-9])(?:-\1){3}\b
Your code might look like:
import re
res = re.match(r'\b([0-5][0-9])(?:-\1){3}\b', '59-59-59-59')
if res:
print(res.group())
If there should not be partial matches, you could use an anchors to assert the ^ start and the end $ of the string:
^([0-5][0-9])(?:-\1){3}$
I have such list (it's only a part);
not match me
norme
16/02574/REMMAJ
20160721
17/00016/FULM
OUT/2017/1071
SMD/2017/0391
17/01090/FULM
2017/30597
17/03940/MAO
18/00076/FULM
CH/17/323
18/00840/OUTMEI
17/00902/EIAM
PL/2017/02671/MINFOT
I need to find general rule to match them all but not this first rows (simple words) or any of \d nor \w if not mixed with each other and slash. Numbers like \d{8} are allowed.
I don't know how to use something like MUST clause applied for each of these 3 groups together - neither can be miss.
These patterns either match not fully or match words. Need as simple regex as possible if possible.
\d{8}|(\w+|/+|\d+)
\d{8}|[\w/\d]+
EDIT
It's funny, but some not provided examples doesn't match for proposed expressions. For example:
7/2018/4127
NWB/18CM032
but I know why and this is outside the scope. However, adding functionality for mixed numbers and letters in one group, like NWB/18CM032 would be great and wouldn't break previous idea I think.
You could match either 1 or more times an uppercase char or 1-8 digits and repeat that zero or more times with a forward slash prepended:
^(?:[a-z0-9]+(?:/[a-z0-9]+)+|\d{8})$
That will match
^ Start of string
(?: Non capturing group
[a-z0-9]+ Match a char a-z or a digit 1+ times
(?:/[a-z0-9]+)+ Match a / followed by a char or digit 1+ times and repeat 1+ times.
| Or
\d{8} Match 8 digits
) Close group
$ End of string
See it on regex101
If I have a text (e.g. This is g56875f562f624g64a4b54a4g51bb3) how can I match the substrings of it that are made up of [a,b,0-9], are of length 5, contain at least one letter (a or b) and don't start or end with a space (so 51bb3 shouldn't be matched since it's at the end of the string)?
The matches in the example would be 64a4b, 4a4b5, a4b54, 4b54a and b54a4.
I want to use Python.
Start by matching exactly 5 occurences of [a,b,0-9]:
[ab0-9]{5}
Then wrap it in a lookahead so that it can produce overlapping matches:
(?=([ab0-9]{5}))
Then add another lookahead that asserts that there's an a or a b somewhere within the next 5 characters:
(?=.{,4}[ab])(?=([ab0-9]{5}))
And finally add lookarounds that assert the absence of whitespace:
(?<!\s)(?<!^)(?=.{,4}[ab])(?=([ab0-9]{5})(?!\s|$))
See also the online demo.