I am having trouble matching a pattern of this format: p#.g.com where # is not a 1 or a 2. For instance if the pattern is p1.g.com, I don't need to match. If it it p2.g.com, I don't need to match.
But if it is any other number, such as p3.g.com or p29.g.com, then I need to match.
My current pattern is r"(?P<url>p([^1,2])\.g\.com)", but this fails if the pattern is p##.g.com, basically any two digit number it fails on. There is no upper limit on the #, so it could be a 3 or 999 or anything in between.
I also tried r"(?P<url>p([^1,2])\d+\.g\.com)" but that does not match any number beginning with a 1 or a 2. For instance 11 or 23 are not matched, which I do want matched.
Try this regex:
p(?:[03-9]|\d{2,})\.g\.com
Demo
Explanation:
Matches character p
Start of non-capturing group
Match one of:
The digits 0 or 3-9
Any double digit number like 10 or higher
Matches character .g.com
Related
I want to test a number consisting of 9 fixed digits.
The number consists of 7 consecutive numbers in the middle. I want to ignore the first and last character. The pattern is 5YYYYYYYX
I am testing my regex using the below sample
577777773
I was able to write a regex that catches the middle 7 numbers. But i want to exclude the first and last character.
(?<!^)([0-9])\1{7}(?!$)
Any advice on how to do this
You could write the pattern as:
(?<=^\d)(\d)\1{6}(?=\d$)
Explanation
(?<=^\d) Assert a digit at the start of the string to the left
(\d) Capture a digit in group 1
\1{6} Repeat the captured value in group 1 six times
(?=\d$) Assert a digit at the end of the string to the right
See a regex demo.
Or a capture group variant instead of lookarounds:
^\d((\d)\2{6})\d$
See another regex demo.
If the patterns should not be bounded to the start and the end of the string, you can use word boundaries \b on the left and right instead of ^ and $
To match 7 consecutive digits in the middle, and the first and last char can not be the same as the consecutive ones:
^(?!(\d)\1)\d((\d)\3{6})(?!\3)\d$
Explanation
^ Start of string
(?!(\d)\1) Negative lookahead, assert not 2 of the same numbers at the start by capturing a single digit in group 1 and matching the same digit directly after it
\d Match a single digit (the first one)
( Capture group 2
(\d)\3{6} Capture a digit in group 3, and repeat that 6 times after it
) Close group 2
(?!\3)\d Match the last digit when it is not the same as the 7 preceding digits
$ End of string
See a regex demo.
The value of the 7 consecutive digits are in group 2
You may use this alternative solution using \B (not a word boundary):
\B(\d)\1{6}\B
RegEx Demo
RegEx Breakup:
\B: Inverse of word boundary
(\d): Match a digit and capture in group #1
\1{6}: Match 6 more occurrences of same digit captured in group #1
\B: Inverse of word boundary
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'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
I want to write regex pattern by def function lets name it is_number(string), which check if string is integer from range -49 to 49. Also number should no contain insignificant zeros.
So i want to pass test:
self.assertTrue(is_number("50"))
self.assertTrue(is_number("-50"))
self.assertTrue(is_number("-9"))
self.assertFalse(is_number("7"))
self.assertFalse(is_number("-200"))
self.assertTrue(is_number("-21"))
self.assertTrue(is_number("18"))
self.assertTrue(is_number("0"))
self.assertTrue(is_number("49"))
self.assertFalse(is_number("100"))
self.assertTrue(is_number("-49"))
I tried something like, but it doesnt work:
def is_number(string):
pattern = r'[-]?\d[1,4]{1,2}*'
return re.search(pattern, string)
You might use
^-?(?:[0-9]|[1-4][0-9])$
That will match
^ Start of string
-? Optional -
(?: Non capturing group
[0-9] Match a digit 0-9
| Or
[1-4][0-9] Match a digit 1-4 and a digit 0-9 to match a range 10 - 49
) Close group
$ End of string
Regex demo
If you also want to match 50 and -50 and 7 should not match you could add 50 to the alternation and match digits 0-6, 8 and 9 using
^-?(?:[0-689]|[1-4][0-9]|50)$
Regex demo
The pattern matches either
double digit numbers with leading 1, 2, 3 or 4 (positive and negative)
or any single digit number (positive and negative)
Regex:
^(-?[1-4]\d|-?\d)$
To fulfill the range of 49 to negative 49 your tests should actually look like this:
self.assertFalse(is_number("50")) # 50 must be assertFalse
self.assertFalse(is_number("-50")) # -50 must be assertFalse
self.assertTrue(is_number("-9"))
self.assertTrue(is_number("7")) # 7 must be assertTrue
self.assertFalse(is_number("-200"))
self.assertTrue(is_number("-21"))
self.assertTrue(is_number("18"))
self.assertTrue(is_number("0"))
self.assertTrue(is_number("49"))
self.assertFalse(is_number("100"))
self.assertTrue(is_number("-49"))
Try this regex pattern,
^[-]?[0-4]?\d$
The first digit must be within the four digits 0,1,2,3 and 4 and the last digit can be any.
pattern = r'^-?[0-4]?\d$'
Try out this regex pattern
I am attempting to write a regex to match numbers in a given string, the below manages to retrieve the first number within the string, however it stops there, I would like it to match all numbers within the file,
thanks in advance
regular expression :
([^\s+\w+\n\r]*(\d))+
string :
hi there this is 1
yes this is 2
actual match : 1
ideal match : 1,2
On site regex101.com/#python type g in the right box near your expression. This box is called modifier. And as others mention in comments use re.findall(pattern, your_string) in python. Notice also that you are actually looking for two substrings - you have two pairs of braces in your regexp.
"([\d]+)"g
Sample
test 13231 test 123123
123 asdfasdf
1a2a3 a
will match
MATCH 1
1. [5-10] `13231`
MATCH 2
1. [16-22] `123123`
MATCH 3
1. [23-26] `123`
MATCH 4
1. [37-38] `1`
MATCH 5
1. [39-40] `2`
MATCH 6
1. [41-42] `3`
and the explaination
"([\d]+)"g
1st Capturing group ([\d]+)
[\d]+ match a single character present in the list below
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
\d match a digit [0-9]
g modifier: global. All matches (don't return on first match)
Why dont you use \d+ simply?
see demo on regex101.com/#python