I have a bunch of 5-letter strings. For each string, I would like to match only if the string contains 3 instances of the same letter, i.e.:
Case 1: 'aabbc' -> no match
Case 2: 'bbbcc' -> match 'bbb'
Case 3: 'ddcdc' -> match 'ddd'
My best regex attempt is:
(.){1}(?!\1)*\1{1}(?!\1)*\1{1}
This works for case 1 (where there is no match) and case 2 (where the 3 instances are adjacent), but not for case 3 (where the 3 instances are separated by at least one other letter).
Is there a regex that will work for case 3? Ideally I would like to also extract the locations of the 3 matching instances from the string.
The below pattern catches what you need and should capture the edge cases. The first capturing group can be modified to just be the subset of characters you need to search for if there is a limited list of expected values. Putting the \1s in capturing groups means that you should be able to extract the index of the capturing groups from the match via .start() (getting the starting index of the capturing group), meeting your bonus goal.
>>> pattern = r"(.).*(\1).*(\1)"
>>> x = re.search(pattern, "ababb")
>>> x.groups()
('b', 'b', 'b')
>>> x.start(1)
1
>>> x.start(2)
3
>>> x.start(3)
4
I think the pattern ([a-z]).*?\1.*?\1 does what you want, although there are likely to be edge cases that would complicate it.
The pattern looks for a lowercase letter three times, with 0 or more characters between them.
You could then extract just the capturing groups to get your match locations.
At the moment, the pattern only looks for any lowercase letter repeated three times, but you could change the initial capturing group - ([a-z]) - if you wanted to capture something else.
Demo
You can use the following regex to determine if one character appears at least three times.
^.*(.).*\1.*\1
Demo
This does not check that the characters are letters but it does work with any characters. To restrict to letters change each . to [a-z] or [a-zA-Z], as appropriate.
To see if one character appears exactly 3 times, change the regex to:
^(?!.*(.)(?:.*\1){3,}).*(.).*\2.*\2
Demo
^ # match beginning of line
(?! # begin negative lookahead
.* # match 0+ chars
(.) # match a char in cap grp 2
(?:.*\1) # match 0+ chars followed by content of cap grp 1
# in a non-cap grp
{3,} # execute non-cap grp 3+ times
) # end negative lookahead
.* # match 0+ chars
(.) # match char in cap grp 2
.* # match 0+ chars
\2 # match content of cap grp 2
.* # match 0+ chars
\2 # match content of cap grp 2
Related
I was trying out to solve a problem on regex:
There is an input sentence which is of one of these forms: Number1,2,3 or Number1/2/3 or Number1-2-3 these are the 3 delimiters: , / -
The expected output is: Number1,Number2,Number3
Pattern I've tried so far:
(?\<=,)\[^,\]+(?=,)
but this misses out on the edge cases i.e. 1st element and last element. I am also not able to generate for '/'.
You could separate out the key from values, then use a list comprehension to build the output you want.
inp = "Number1,2,3"
matches = re.search(r'(\D+)(.*)', inp)
output = [matches[1] + x for x in re.split(r'[,/]', matches[2])]
print(output) # ['Number1', 'Number2', 'Number3']
You can do it in several steps: 1) validate the string to match your pattern, and once validated 2) add the first non-digit chunk to the numbers while replacing - and / separator chars with commas:
import re
texts = ['Number1,2,3', 'Number1/2/3', 'Number1-2-3']
for text in texts:
m = re.search(r'^(\D+)(\d+(?=([,/-]))(?:\3\d+)*)$', text)
if m:
print( re.sub(r'(?<=,)(?=\d)', m.group(1).replace('\\', '\\\\'), text.replace('/',',').replace('-',',')) )
else:
print(f"NO MATCH in '{text}'")
See this Python demo.
Output:
Number1,Number2,Number3
Number1,Number2,Number3
Number1,Number2,Number3
The ^(\D+)(\d+(?=([,/-]))(?:\3\d+)*)$ regex validates your three types of input:
^ - start of string
(\D+) - Group 1: one or more non-digits
(\d+(?=([,/-]))(?:\3\d+)*) - Group 2: one or more digits, and then zero or more repetitions of ,, / or - and one or more digits (and the separator chars should be consistent due to the capture used in the positive lookahead and the \3 backreference to that value used in the non-capturing group)
$ - end of string.
The re.sub pattern, (?<=,)(?=\d), matches a location between a comma and a digit, the Group 1 value is placed there (note the .replace('\\', '\\\\') is necessary since the replacement is dynamic).
import re
for text in ("Number1,2,3", "Number1-2-3", "Number1/2/3"):
print(re.sub(r"(\D+)(\d+)[/,-](\d+)[/,-](\d+)", r"\1\2,\1\3,\1\4", text))
\D+ matches "Number" or any other non-number text
\d+ matches a number (or more than one)
[/,-] matches any of /, ,, -
The rest is copy paste 3 times.
The substitution consists of backreferences to the matched "Number" string (\1) and then each group of the (\d+)s.
This works if you're sure that it's always three numbers divided by that separator. This does not ensure that it's the same separator between each number. But it's short.
Output:
Number1,Number2,Number3
Number1,Number2,Number3
Number1,Number2,Number3
If you can make use of the pypi regex module you can use the captures collection with a named capture group.
([^\d\s,/]+)(?<num>\d+)([,/-])(?<num>\d+)(?:\3(?<num>\d+))*(?!\S)
([^\d\s,/]+) Capture group 1, match 1+ chars other than the listed
(?<num>\d+) Named capture group num matching 1+ digits
([,/-]) Capture either , / - in group 3
(?<num>\d+) Named capture group num matching 1+ digits
(?:\3(?<num>\d+))* Optionally repeat a backreference to group 3 to keep the separators the same and match 1+ digits in group num
(?!\S) Assert a whitspace boundary to the right to prevent a partial match
Regex demo | Python demo
import regex as re
pattern = r"([^\d\s,/]+)(?<num>\d+)([,/-])(?<num>\d+)(?:\3(?<num>\d+))*(?!\S)"
s = "Number1,2,3 or Number4/5/6 but not Number7/8,9"
matches = re.finditer(pattern, s)
for _, m in enumerate(matches, start=1):
print(','.join([m.group(1) + c for c in m.captures("num")]))
Output
Number1,Number2,Number3
Number4,Number5,Number6
I am trying to extract first 5 character+digit from last hyphen.
Here is the example
String -- X008-TGa19-ER751QF7
Output -- X008-TGa19-ER751
String -- X002-KF13-ER782cPU80
Output -- X002-KF13-ER782
My attempt -- I could manage to take element from the last -- (\w+)[^-.]*$
But now how to take first 5, then return my the entire value as the output as shown in the example.
You can optionally repeat a - and 1+ word chars from the start of the string. Then match the last - and match 5 word chars.
^\w+(?:-\w+)*-\w{5}
^ Start of string
\w+ Math 1+ word chars
(?:-\w+)* Optionally repeat - and 1+ word chars
-\w{5} Match - and 5 word chars
Regex demo
import re
regex = r"^\w+(?:-\w+)*-\w{5}"
s = ("X008-TGa19-ER751QF7\n"
"X002-KF13-ER782cPU80")
print(re.findall(regex, s, re.MULTILINE))
Output
['X008-TGa19-ER751', 'X002-KF13-ER782']
Note that \w can also match _.
If there can also be other character in the string, to get the first 5 digits or characters except _ after the last hyphen, you can match word characters without an underscore using a negated character class [^\W_]{5}
Repeat that 5 times while asserting no more underscore at the right.
^.*-[^\W_]{5}(?=[^-]*$)
Regex demo
(\w+-\w+-\w{5}) seems to capture what you're asking for.
Example:
https://regex101.com/r/PcPSim/1
If you are open for non-regex solution, you can use this which is based on splitting, slicing and joining the strings:
>>> my_str = "X008-TGa19-ER751QF7"
>>> '-'.join(s[:5] for s in my_str.split('-'))
'X008-TGa19-ER751'
Here I am splitting the string based on hyphen -, slicing the string to get at max five chars per sub-string, and joining it back using str.join() to get the string in your desired format.
^(.*-[^-]{5})[^-]*$
Capture group 1 is what you need
https://regex101.com/r/SYz9i5/1
Explanation
^(.*-[^-]{5})[^-]*$
^ Start of line
( Capture group 1 start
.* Any number of any character
- hyphen
[^-]{5} 5 non-hyphen character
) Capture group 1 end
[^-]* Any number of non-hyphen character
$ End of line
Another simpler one is
^(.*-.{5}).*$
This should be quite straight-forward.
This is making use of behaviour greedy match of first .*, which will try to match as much as possible, so the - will be the last one with at least 5 character following it.
https://regex101.com/r/CFqgeF/1/
I'm a beginner to regex and I am trying to make an expression to find if there are two of the same digits next to each other, and the digit behind and in front of the pair is different.
For example,
123456678 should match as there is a double 6,
1234566678 should not match as there is no double with different surrounding numbers.
12334566 should match because there are two 3s.
So far i have this which works only with 1, and as long as the double is not at the start or end of the string, however I can deal with that by adding a letter at the start and end.
^.*([^1]11[^1]).*$
I know i can use [0-9] instead of the 1s but the problem is having them all be the same digit.
Thank you!
I have divided my answer into four sections.
The first section contains my solution to the problem. Readers interested in nothing else may skip the other sections.
The remaining three sections are concerned with identifying the pairs of equal digits that are preceded by a different digit and are followed by a different digit. The first of the three sections matches them; the other two capture them in a group.
I've included the last section because I wanted to share The Greatest Regex Trick Ever with those unfamiliar with it, because I find it so very cool and clever, yet simple. It is documented here. Be forewarned that, to build suspense, the author at that link has included a lengthy preamble before the drum-roll reveal.
Determine if a string contains two consecutive equal digits that are preceded by a different digit and are followed by a different digit
You can test the string as follows:
import re
r = r'(\d)(?!\1)(\d)\2(?!\2)\d'
arr = ["123456678", "1123455a666788"]
for s in arr:
print(s, bool(re.search(r, s)) )
displays
123456678 True
1123455a666788 False
Run Python code | Start your engine!1
The regex engine performs the following operations.
(\d) : match a digit and save to capture group 1 (preceding digit)
(?!\1) : next character cannot equal content of capture group 1
(\d) : match a digit in capture group 2 (first digit of pair)
\2 : match content of capture group 2 (second digit of pair)
(?!\2) : next character cannot equal content of capture group 2
\d : match a digit
(?!\1) and (?!\2) are negative lookaheads.
Use Python's regex module to match pairs of consecutive digits that have the desired property
You can use the following regular expression with Python’s regex module to obtain the matching pairs of digits.
r'(\d)(?!\1)\K(\d)\2(?=\d)(?!\2)'
Regex Engine
The regex engine performs the following operations.
(\d) : match a digit and save to capture group 1 (preceding digit)
(?!\1) : next character cannot equal content of capture group 1
\K : forget everything matched so far and reset start of match
(\d) : match a digit in capture group 2 (first digit of pair)
\2 : match content of capture group 2 (second digit of pair)
(?=\d) : next character must be a digit
(?!\2) : next character cannot equal content of capture group 2
(?=\d) is a positive lookahead. (?=\d)(?!\2) could be replaced with (?!\2|$|\D).
Save pairs of consecutive digits that have the desired property to a capture group
Another way to obtain the matching pairs of digits, which does not require the regex module, is to extract the contents of capture group 2 from matches of the following regular expression.
r'(\d)(?!\1)((\d)\3)(?!\3)(?=\d)'
Re engine
The following operations are performed.
(\d) : match a digit in capture group 1
(?!\1) : next character does not equal last character
( : begin capture group 2
(\d) : match a digit in capture group 3
\3 : match the content of capture group 3
) : end capture group 2
(?!\3) : next character does not equal last character
(?=\d) : next character is a digit
Use The Greatest Regex Trick Ever to identify pairs of consecutive digits that have the desired property
We use the following regular expression to match the string.
r'(\d)(?=\1)|\d(?=(\d)(?!\2))|\d(?=\d(\d)\3)|\d(?=(\d{2})\d)'
When there is a match, we pay no attention to which character was matched, but examine the content of capture group 4 ((\d{2})), as I will explain below.
The Trick in action
The first three components of the alternation correspond to the ways that a string of four digits can fail to have the property that the second and third digits are equal, the first and second are unequal and the third and fourth are equal. They are:
(\d)(?=\1) : assert first and second digits are equal
\d(?=(\d)(?!\2)) : assert second and third digits are not equal
\d(?=\d(\d)\3) : assert third and fourth digits are equal
It follows that if there is a match of a digit and the first three parts of the alternation fail the last part (\d(?=(\d{2})\d)) must succeed, and the capture group it contains (#4) must contain the two equal digits that have the required properties. (The final \d is needed to assert that the pair of digits of interest is followed by a digit.)
If there is a match how do we determine if the last part of the alternation is the one that is matched?
When this regex matches a digit we have no interest in what digit that was. Instead, we look to capture group 4 ((\d{2})). If that group is empty we conclude that one of the first three components of the alternation matched the digit, meaning that the two digits following the matched digit do not have the properties that they are equal and are unequal to the digits that precede and follow them.
If, however, capture group 4 is not empty, it means that none of the first three parts of the alternation matched the digit, so the last part of the alternation must have matched and the two digits following the matched digit, which are held in capture group 4, have the desired properties.
1. Move the cursor around for detailed explanations.
With regex, it is much more convenient to use a PyPi regex module with the (*SKIP)(*FAIL) based pattern:
import regex
rx = r'(\d)\1{2,}(*SKIP)(*F)|(\d)\2'
l = ["123456678", "1234566678"]
for s in l:
print(s, bool(regex.search(rx, s)) )
See the Python demo. Output:
123456678 True
1234566678 False
Regex details
(\d)\1{2,}(*SKIP)(*F) - a digit and then two or more occurrences of the same digit
| - or
(\d)\2 - a digit and then the same digit.
The point is to match all chunks of identical 3 or more digits and skip them, and then match a chunk of two identical digits.
See the regex demo.
Inspired by the answer or Wiktor Stribiżew, another variation of using an alternation with re is to check for the existence of the capturing group which contains a positive match for 2 of the same digits not surrounded by the same digit.
In this case, check for group 3.
((\d)\2{2,})|\d(\d)\3(?!\3)\d
Regex demo | Python demo
( Capture group 1
(\d)\2{2,} Capture group 2, match 1 digit and repeat that same digit 2+ times
) Close group
| Or
\d(\d) Match a digit, capture a digit in group 3
\3(?!\3)\d Match the same digit as in group 3. Match the 4th digit, but is should not be the same as the group 3 digit
For example
import re
pattern = r"((\d)\2{2,})|\d(\d)\3(?!\3)\d"
strings = ["123456678", "12334566", "12345654554888", "1221", "1234566678", "1222", "2221", "66", "122", "221", "111"]
for s in strings:
match = re.search(pattern, s)
if match and match.group(3):
print ("Match: " + match.string)
else:
print ("No match: " + s)
Output
Match: 123456678
Match: 12334566
Match: 12345654554888
Match: 1221
No match: 1234566678
No match: 1222
No match: 2221
No match: 66
No match: 122
No match: 221
No match: 111
If for example 2 or 3 digits only is also ok to match, you could check for group 2
(\d)\1{2,}|(\d)\2
Python demo
You can also use a simple way .
import re
l=["123456678",
"1234566678",
"12334566 "]
for i in l:
matches = re.findall(r"((.)\2+)", i)
if any(len(x[0])!=2 for x in matches):
print "{}-->{}".format(i, False)
else:
print "{}-->{}".format(i, True)
You can customize this based on you rules.
Output:
123456678-->True
1234566678-->False
12334566 -->True
I have a filename having numerals like test_20200331_2020041612345678.csv.
So I just want to read only first 8 characters from the number between last underscore and .csv using a regex.
For e.g: From the file name test_20200331_2020041612345678.csv --> i want to read only 20200416 using regex.
Regex tried: (?<=_)(\d+)(?=\.)
But it is returning the full number between underscore and period i.e 2020041612345678
Also, when tried quantifier like (?<=_)(\d{8})(?=\.) its not matching with any string
The (?<=_)(\d{8})(?=\.) does not work because the (?=\.) positive lookahead requires the presence of a . char immediately to the right of the current location, i.e. right after the eigth digit, but there are more digits in between.
You may add \d* before \. to match any amount of digits after the required 8 digits, use
(?<=_)\d{8}(?=\d*\.)
Or, with a capturing group, you do not even need lookarounds (just make sure you access Group 1 when a match is obtained):
_(\d{8})\d*\.
See the regex demo
Python demo:
import re
s = "test_20200331_2020041612345678.csv"
m = re.search(r"(?<=_)\d{8}(?=\d*\.)", s)
# m = re.search(r"_(\d{8})\d*\.", s) # capturing group approach
if m:
print(m.group()) # => 20200416
# print(m.group(1)) # capturing group approach
I'm trying to implement some kind of markdown like behavior for a Python log formatter.
Let's take this string as example:
**This is a warning**: Virus manager __failed__
A few regexes later the string has lost the markdown like syntax and been turned into bash code:
\033[33m\033[1mThis is a warning\033[0m: Virus manager \033[4mfailed\033[0m\033[0m
But that should be compressed to
\033[33;1mThis is a warning\033[0m: Virus manager \033[4mfailed\033[0m
I tried these, beside many other non working solutions:
(\\033\[([\d]+)m){2,} => Capture: \033[33m\033[1m with g1 '\033[1m' and g2 '1' and \033[0m\033[0mwith g1 '\033[0m' and g2 '0'
(\\033\[([\d]+)m)+ many results, not ok
(?:(\\033\[([\d]+)m)+) many results, although this is the recommended way for repeated patterns if I understood correctly, not ok
and others..
My goal is to have as results:
Input
\033[33m\033[1mThis is a warning\033[0m: Virus manager \033[4mfailed\033[0m\033[0m
Output
Match 1
033[33m\033[1m
Group1: 33
Group2: 1
Match 2
033[0m\033[0m
Group1: 0
Group2: 0
In other words, capture the ones that are "duplicated" and not the ones alone, so I can fuse them with a regex sub.
You want to match consectuively repeating \033[\d+m chunks of text and join the numbers after [ with a semi-colon.
You may use
re.sub(r'(?:\\033\[\d+m){2,}', lambda m: r'\033['+";".join(set(re.findall(r"\[(\d+)", m.group())))+'m', text)
See the Python demo online
The (?:\\033\[\d+m){2,} pattern will match two or more sequences of \033[ + one or more digits + m chunks of texts and then, the match will be passed to the lambda expression, where the output will be: 1) \033[, 2) all the numbers after [ extracted with re.findall(r"\[(\d+)", m.group()) and deduplicated with the set, and then 3) m.
The patterns in the string to be modified have not been made clear from the question. For example, is 033 fixed or might it be 025 or even 25? I've made certain assumptions in using the regex
r" ^(\\0(\d+)\[\2)[a-z]\\0\2\[(\d[a-z].+)
to obtain two capture groups that are to be combined, separated by a semi-colon. I've attempted to make clear my assumptions below, in part to help the OP modify this regex to satisfy alternative requirements.
Demo
The regex performs the following operations:
^ # match beginning of line
( # begin cap grp 1
\\0 # match '\0'
(\d+) # match 1+ digits in cap grp 2
\[ # match '['
\2 # match contents of cap grp 2
) # end cap grp 1
[a-z] # match a lc letter
\\0 # match '\0'
\2 # match contents of cap grp 2
\[ # match '['
(\d[a-z].+) # match a digit, then lc letter then 1+ chars to the
# end of the line in cap grp 3
As you see, the portion of the string captured in group 1 is
\033[33
I've assumed that the part of this string that is now 033 must be two or more digits beginning with a zero, and the second appearance of a string of digits consists of the same digits after the zero. This is done by capturing the digits following '0' (33) in capture group 2 and then using a back-reference \2.
The next part of the string is to be replaced and therefore is not captured:
m\\033[
I've assumed that m must be one lower case letter (or should it be a literal m?), the backslash and zero and required and the following digits must again match the content of capture group 2.
The remainder of the string,
1mThis is a warning\033[0m: Virus manager \033[4mfailed\033[0m\033[0m
is captured in capture group 3. Here I've assumed it begins with one digit (perhaps it should be \d+) followed by one lower case letter that needn't be the same as the lower case letter matched earlier (though that could be enforced with another capture group). At that point I match the remainder of the line with .+, having given up matching patterns in that part of the string.
One may alternatively have just two capture groups, the capture group that is now #2, becoming #1, and #2 being the part of the string that is to be replaced with a semicolon.
This is pretty straightforward for the cases you desribe here; simply write out from left to right what you want to match and capture. Repeating capturing blocks won't help you here, because only the most recently captured values would be returned as a result.
\\033\[(\d+)m\\033\[(\d+)m