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Regex match 7 Consecutive Numbers and Ignore first and last Characters

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

Python: find a string between 2 strings in text

I have a text like this
s = """
...
(1) Literature
1. a.
2. b.
3. c.
...
"""
I want to cut Literature section but I have some problem with detection.
I use here
re.search("(1) Literature\n\n(.*).\n\n", s).group(1)
but search return None.
Desire output is
(1) Literature
1. a.
2. b.
3. c.
What did I do wrong?

You could match (1) Literature and 2 newlines, and then capture all lines that start with digits followed by a dot.
\(1\) Literature\n\n((?:\d+\..*(?:\n|$))+)
The pattern matches:
\(1\) Literature\n\n Match (1) Literature and 2 newlines
( Capture group 1
(?: Non capture group
\d+\..*(?:\n|$) Match 1+ digits and a dot followed by either a newline or end of string
)+ Close non capture group and repeat it 1 or more times to match all the lines
) Close group 1
Regex demo
Another option is to capture all following lines that do not start with ( digits ) using a negative lookahead, and then trim the leading and trailing whitespaces.
\(1\) Literature((?:\n(?!\(\d+\)).*)*)
Regex demo

Parentheses have a special meaning in regex. They are used to group matches.
(1) - Capture 1 as the first capturing group.
Since the string has parentheses in it, the match is not successful. And .* capturing end with line end.
Check Demo
Based on your regex, I assumed you wanted to capture the line with the word Literature, 5 lines below it. Here is a regex to do so.
\(1\) Literature(.*\n){5}
Regex Demo
Note the scape characters used on parentheses around 1.
EDIT
Based on zr0gravity7's comment, I came up with this regex to capture the middle section on the string.
\(1\)\sLiterature\n+((.*\n){3})
This regex will capture the below string in capturing group 1.
1. a.
2. b.
3. c.
Regex Demo

You may use this regex with a capture group:
r'\(1\)\s+Literature\s+((?:.+\n)+)'
RegEx Demo
Explanation:
\(1\): Match (1) text
\s+: Match 1+ whitespaces
Literature:
\s+:
(: Start capture group #1
(?:.+\n)+: Match a line with 1+ character followed by newline. Repeat this 1 or more times to allow it to match multiple such lines
): End capture group #1

Regex for capturing the generic question with that structure:
\(\d+\)\s+(\w+)\s+((?:\d+\.\s.+\n)+)
It will capture the title "Literature", then the choices in another group (for a total of 2 groups).
It is not possible to capture repeating groups, so in order to get each of your "1. a." in a separate group you would have to match the second group from above again, with this pattern:
((\d+\.\s+.+)\n)+) then globally match to get all groups.

regex to find a pair of adjacent digits with different digits around them

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

Searching multiple repeating patterns of text using regular exressions

I am trying to search for texts from a document, which have repeating portions and occur multiple times in the document. However, using the regex.match, it shows only the first match from the document and not others.
The patterns which I want to search looks like:
clauses 5.3, 12 & 15
clause 10 C, 10 CA & 10 CC
The following line shows the regular expression which I am using.
regex_crossref_multiple_1=r'(clause|Clause|clauses|Clauses)\s*\d+[.]?\d*\s*[a-zA-Z]*((,|&|and)\s*\d+[.]?\d*\s*[A-Z]*)+'
The code used for matching and the results are shown below:
cross=regex.search(regex_crossref_multiple_1,des)
(des is string containing text)
For printing the results, I am using print(cross.group()).
Result:
clauses 5.3, 12 & 15
However, there are other patterns as well in des which I am not getting in the result.
Please let me know what can be the problem.
The input string(des) is can be found from following link.
https://docs.google.com/document/d/1LPmYaD6VE724OYoXDGPfInvx8WTu5JfrTqTOIv8zAlg/edit?usp=sharing
In case, the contractor completes the work ahead of stipulated date of
completion or justified extended date of completion as determined
under clauses 5.3, 12 & 15, a bonus # 0.5 % (zero point five per cent) of
the tendered value per month computed on per day basis, shall be
payable to the contractor, subject to a maximum limit of 2 % (two
percent) of the tendered value. Provided that justified time for extra
work shall be calculated on pro-rata basis as cost of extra work excluding
amount payable/ paid under clause 10 C, 10 CA & 10 CC X stipulated
period /tendered value. The amount of bonus, if payable, shall be paid
along with final bill after completion of work. Provided always that
provision of the Clause 2A shall be applicable only when so provided in
‘Schedule F’

You could match clauses followed by an optional digits part and optional chars A-Z and then use a repeating pattern to match the optional following comma and the digits.
For the last part of the pattern you can optionally match either a ,, & or and followed by a digit and optional chars A-Z.
\b[Cc]lauses?\s+\d+(?:\.\d+)?(?:\s*[A-Z]+)?(?:,\s+\d+(?:\.\d+)?(?:\s*[A-Z]+)?)*(?:\s+(?:[,&]|and)\s+\d+(?:\.\d+)?(?:\s*[A-Z]+)?)?\b
Explanation
\b Word boundary
[Cc]lauses?\s+\d+(?:\.\d+)? Match clauses followed by digits and optional decimal part
(?:\s*[A-Z]+)? Optionally match whitespace chars and 1+ chars A-Z
(?: Non capture group
,\s+\d+(?:\.\d+)? Match a comma, digits and optional decimal part
(?:\s*[A-Z]+)? Optionally match whitespace chars and 1+ chars A-Z
)* Close group and repeat 0+ times
(?: Non capture group
\s+(?:[,&]|and) Match 1+ whitespace char and either ,, & or and
\s+\d+(?:\.\d+)? Match 1+ whitespace chars, 1+ digits with an optional decimal part
(?:\s*[A-Z]+)? Match optional whitespace chars and 1+ chars A-Z
)? Close group and make optional
\b Word boundary
Regex demo

Python Regular expression to match if any number appears more than a certain amount of times

I'm in need of a regular expression for python that is able to match all strings where any number appears a certain amount of times (4 times in a 5 digit number is my desired result in this example).
For example, consider this list:
["11211", "23424", "22323", "99991", "88988", "11122"]
I would like a regEx that returns
["11211", "99991", "88988"]
because in these three cases, there is a digit that appears more than 4 times in the number.
I am not even sure if this is easily doable with just one single regEx, apart from hardcoding the digits from 0-9, which does not seem to be an elegant solution.
Here is a regEx that matches four 1's in a list of 5 number strings:
four1 = re.compile(".*1.*1.*1.*1.*")
But is there a more elegant solution than these two to not only search for four 1's, but four of any kind, as long as they are four times the same number?
four1 = re.compile("(.*1.*1.*1.*1.*")|(.*2.*2.*2.*2.*")| ...
or
four1 = re.compile(".*1.*1.*1.*1.*")
four2 = re.compile(".*2.*2.*2.*2.*")
...
Thank you for your help.

You may use this regex with a capture group and a back-reference:
(\d)(?:\d*?\1){3}
RegEx Demo
RegEx Description:
(\d): Match a single digit and capture in group #1
(?:: Start non-capture group
\d*?: Match 0 or more digits
\1: Back-reference to capture group #1 to make sure we match repeating digits of capture group #1
): End non-capture group
{3}: Match 3 instances of above non-capture group
Code:
import re
arr = ["11211", "23424", "22323", "99991", "88988", "11122"]
reg = re.compile(r'(\d)(?:\d*?\1){3}')
for s in arr:
if reg.search(s):
print s
output:
11211
99991
88988