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Remove all numbers except for the ones combined to string using python regex

Trying to use a regex function to remove a word, whitespaces, special characters and numbers but not the one combined with to a word/string. E.g.
ORIGIN
1 malwmrllp1 lallalwgpd paaafvnghl cgshlvealy lvcgergffy tpktrreaed
61 lqvgqvelgg gpgagslqpl alegslqkrg iveqcctsic slyqlenycn
//
The \W+ removes all numbers including 1 in malwmrll1
import re
text_file = open('mytext.txt').read()
new_txt = re.sub('[\\b\\d+\\b\s*$+\sORIGIN$\W+]', '', text_file)
print(new_txt, len(new_txt))
My output is:
malwmrllplallalwgpdpaaafvnghlcgshlvealylvcgergffytpktrreaedlqvgqvelgggpgagslqplalegslqkrgiveqcctsicslyqlenycn 109
The desired output should be: malwmrll1plallalwgpdpaaafvnghlcgshlvealylvcgergffytpktrreaedlqvgqvelgggpgagslqplalegslqkrgiveqcctsicslyqlenycn 110

Right, depending on your desired result showing underscores at all or not, try to use re.findall and raw-string notation. You currently use a character class that makes no sense:
\b(?!(?:ORIGIN|[_\d]+)\b)\w+
See an online demo
\b - Word-boundary;
(?!(?:ORIGIN|[_\d]+)\b) - Negative lookahead with nested non-capture group to match either ORIGIN or 1+ underscore/digit combinations before a trailing word-boundary;
\w+ - 1+ word-characters.
import re
text_file = """ORIGIN
1 malwmrllp1 lallalwgpd paaafvnghl cgshlvealy lvcgergffy tpktrreaed
61 lqvgqvelgg gpgagslqpl alegslqkrg iveqcctsic slyqlenycn
//"""
new_txt=''.join(re.findall(r'\b(?!(?:ORIGIN|[_\d]+)\b)\w+', text_file))
print(new_txt, len(new_txt))
Prints:
malwmrllp1lallalwgpdpaaafvnghlcgshlvealylvcgergffytpktrreaedlqvgqvelgggpgagslqplalegslqkrgiveqcctsicslyqlenycn 110

Using RE for this is an interesting academic exercise but extending the functionality is fraught with danger unless one is very familiar with the technique.
This answer may look long-winded but you should be able to see how easy it would be to extend it so that other tokens/patterns can be excluded or included. It's also readily maintainable because anyone else having to modify the code isn't going to get a migraine while trying to figure out how the RE works.
FILENAME = 'mytext.txt'
def keep(t):
if t.isdigit() or t == 'ORIGIN' or t == '//':
return False
return True
with open(FILENAME) as f:
new_txt = ''.join(filter(keep, f.read().split()))
print(new_txt, len(new_txt))
Output:
malwmrllp1lallalwgpdpaaafvnghlcgshlvealylvcgergffytpktrreaedlqvgqvelgggpgagslqplalegslqkrgiveqcctsicslyqlenycn 110

Another idea:
new_txt = re.sub('[\\W_]+|\\b(?:\\d+|ORIGIN)\\b', '', text_file)
Strip out all non word characters + underscore OR digits / "ORIGIN" between word boundaries.
See this demo at tio.run (the regex is very basic, explanation at regex101)

Extract Only Digits from Dollar Figures

What I'm trying to do is extract only the digits from dollar figures.
Format of Input
...
$1,289,868
$62,000
$421
...
Desired Output
...
1289868
62000
421
...
The regular expression that I was using to extract only the digits and commas is:
r'\d+(,\d+){0,}'
which of course outputs...
...
1,289,868
62,000
421
...
What I'd like to do is convert the output to an integer (int(...)), but obviously this won't work with the commas. I'm sure I could figure this out on my own, but I'm running really short on time right now.
I know I can simply use r'\d+', but this obviously separates each chunk into separate matches...

You can't match discontinuous texts within one match operation. You can't put a regex into re.findall against 1,345,456 to receive 1345456. You will need to first match the strings you need, and then post-process them within code.
A regex you may use to extract the numbers themselves
re.findall(r'\$(\d{1,3}(?:,\d{3})*)', s)
See this regex demo.
Alternatively, you may use a bit more general regex to be used with re.findall:
r'\$(\d+(?:,\d+)*)'
See this regex demo.
Note that re.findall will only return the captured part of the string (the one matched with the (...) part in the regex).
Details
\$ - a dollar sign
(\d{1,3}(?:,\d{3})*) - Capturing group 1:
\d{1,3} - 1 to 3 digits (if \d+ is used, 1 or more digits)
(?:,\d{3})* - 0 or more sequences of
, - a comma
\d{3} - 3 digits (or if \d+ is used, 1 or more digits).
Python code sample (with removing commas):
import re
s = """$1,289,868
$62,000
$421"""
result = [x.replace(",", "") for x in re.findall(r'\$(\d{1,3}(?:,\d{3})*)', s)]
print(result) # => ['1289868', '62000', '421']

Using re.sub
Ex:
import re
s = """$1,289,868
$62,000
$421"""
print([int(i) for i in re.sub(r'[^0-9\s]', "", s).splitlines()])
Output:
[1289868, 62000, 421]

You don't need regex for this.
int(''.join(filter(str.isdigit, "$1,000,000")))
works just fine.
If you did want to use regex for some reason:
int(''.join(re.findall(r"\d", "$1,000,000")))

If you know how to extract the numbers with comma groupings, the easiest thing to do is just transform that into something int can handle:
for match in matches:
i = int(match.replace(',', ''))
For example, if match is '1,289,868', then match.replace(',', '') is '1289868', and obviously int(<that>) is 1289868.

You dont need regex for this. Just string operations should be enough
>>> string = '$1,289,868\n$62,000\n$421'
>>> [w.lstrip('$').replace(',', '') for w in string.splitlines()]
['1289868', '62000', '421']
Or alternatively, you can use locale.atoi to convert string of digits with commas to int
>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF8')
>>> list(map(lambda x: locale.atoi(x.lstrip('$')), string.splitlines()))
[1289868, 62000, 421]

repeated pattern in regex

I am trying to catch a repeated pattern in my string. The subpattern starts with the beginning of word or ":" and ends with ":" or end of word. I tried findall and search in combination of multiple matching ((subpattern)__(subpattern))+ but was not able what is wrong:
cc = "GT__abc23_1231:TF__XYZ451"
import regex
ma = regex.match("(\b|\:)([a-zA-Z]*)__(.*)(:|\b)", cc)
Expected output:
GT, abc23_1231, TF, XYZ451
I saw a bunch of questions like this, but it did not help.

It seems you can use
(?:[^_:]|(?<!_)_(?!_))+
See the regex demo
Pattern details:
(?:[^_:]|(?<!_)_(?!_))+ - 1 or more sequences of:
[^_:] - any character but _ and :
(?<!_)_(?!_) - a single _ not enclosed with other _s
Python demo with re based solution:
import re
p = re.compile(r'(?:[^_:]|(?<!_)_(?!_))+')
s = "GT__abc23_1231:TF__XYZ451"
print(p.findall(s))
# => ['GT', 'abc23_1231', 'TF', 'XYZ451']
If the first character is always not a : and _, you may use an unrolled regex like:
r'[^_:]+(?:_(?!_)[^_:]*)*'
It won't match the values that start with single _ though (so, an unrolled regex is safer).

Use the smallest common denominator in "starts and ends with a : or a word-boundary", that is the word-boundary (your substrings are composed with word characters):
>>> import re
>>> cc = "GT__abc23_1231:TF__XYZ451"
>>> re.findall(r'\b([A-Za-z]+)__(\w+)', cc)
[['GT', 'abc23_1231'], ['TF', 'XYZ451']]
Testing if there are : around is useless.
(Note: no need to add a \b after \w+, since the quantifier is greedy, the word-boundary becomes implicit.)
[EDIT]
According to your comment: "I want to first split on ":", then split on double underscore.", perhaps you dont need regex at all:
>>> [x.split('__') for x in cc.split(':')]
[['GT', 'abc23_1231'], ['TF', 'XYZ451']]

regEx: To match two groups of chars

I want a regEx to match some text that contains both alpha and numeric chars. But I do NOT want it to match only alpha or numbers.
E.g. in python:
s = '[mytaskid: 3fee46d2]: STARTED at processing job number 10022001'
# ^^^^^^^^ <- I want something that'll only match this part.
import re
rr = re.compile('([0-9a-z]{8})')
print 'sub=', rr.sub('########', s)
print 'findall=', rr.findall(s)
generates following output:
sub= [########: ########]: STARTED at ########ng job number ########
findall= ['mytaskid', '3fee46d2', 'processi', '10022001']
I want it to be:
sub= [mytaskid: ########]: STARTED at processing job number 10022001
findall= ['3fee46d2']
Any ideas... ??
In this case it's exactly 8 chars always, it would be even more wonderful to have a regEx that doesn't have {8} in it, i.e. it can match even if there are more or less than 8 chars.
-- edit --
Question is more to understand if there is a way to write a regEx such that I can combine 2 patterns (in this case [0-9] and [a-z]) and ensure the matched string matches both patterns, but number of chars matched from each set is variable. E.g. s could also be
s = 'mytaskid 3fee46d2 STARTED processing job number 10022001'
-- answer --
Thanks to all for the answers, all them give me what I want, so everyone gets a +1 and the first one to answer gets the accepted answer. Although jerry explains it the best. :)
If anyone is a stickler for performance, there is nothing to choose from, they're all the same.
s = '[mytaskid: 3fee46d2]: STARTED at processing job number 10022001'
# ^^^^^^^^ <- I want something that'll only match this part.
def testIt(regEx):
from timeit import timeit
s = '[mytaskid: 3333fe46d2]: STARTED at processing job number 10022001'
assert (re.sub('\\b(?=[a-z0-9]*[0-9])[a-z0-9]*[a-z][a-z0-9]*\\b', '########', s) ==
'[mytaskid: ########]: STARTED at processing job number 10022001'), '"%s" does not work.' % regEx
print 'sub() with \'', regEx, '\': ', timeit('rr.sub(\'########\', s)', number=500000, setup='''
import re
s = '%s'
rr = re.compile('%s')
''' % (s, regEx)
)
print 'findall() with \'', regEx, '\': ', timeit('rr.findall(s)', setup='''
import re
s = '%s'
rr = re.compile('%s')
''' % (s, regEx)
)
testIt('\\b[0-9a-z]*(?:[a-z][0-9]|[0-9][a-z])[0-9a-z]*\\b')
testIt('\\b[a-z\d]*(?:\d[a-z]|[a-z]\d)[a-z\d]*\\b')
testIt('\\b(?=[a-z0-9]*[0-9])[a-z0-9]*[a-z][a-z0-9]*\\b')
testIt('\\b(?=[0-9]*[a-z])(?=[a-z]*[0-9])[a-z0-9]+\\b')
produced:
sub() with ' \b[0-9a-z]*(?:[a-z][0-9]|[0-9][a-z])[0-9a-z]*\b ': 0.328042736387
findall() with ' \b[0-9a-z]*(?:[a-z][0-9]|[0-9][a-z])[0-9a-z]*\b ': 0.350668751542
sub() with ' \b[a-z\d]*(?:\d[a-z]|[a-z]\d)[a-z\d]*\b ': 0.314759661193
findall() with ' \b[a-z\d]*(?:\d[a-z]|[a-z]\d)[a-z\d]*\b ': 0.35618526928
sub() with ' \b(?=[a-z0-9]*[0-9])[a-z0-9]*[a-z][a-z0-9]*\b ': 0.322802906619
findall() with ' \b(?=[a-z0-9]*[0-9])[a-z0-9]*[a-z][a-z0-9]*\b ': 0.35330467656
sub() with ' \b(?=[0-9]*[a-z])(?=[a-z]*[0-9])[a-z0-9]+\b ': 0.320779061371
findall() with ' \b(?=[0-9]*[a-z])(?=[a-z]*[0-9])[a-z0-9]+\b ': 0.347522144274

Try following regex:
\b[0-9a-z]*(?:[a-z][0-9]|[0-9][a-z])[0-9a-z]*\b
This will match a word containing a digit followed an alphabet or vice versa.
Hence it will cover a complete set of those words which contain at-least one digit and one alphabet.
Note: Although it is not the case with python, I have observed that not all varieties of tools support lookahead and lookbehind. So I prefer to avoid them if possible.

You need to use the look ahead (?=...).
This one matches all words with at least one out of [123] and [abc].
>>> re.findall('\\b(?=[abc321]*[321])[abc321]*[abc][abc321]*\\b', ' 123abc 123 abc')
['123abc']
This way you can do AND for constraints to the same string.
>>> help(re)
(?=...) Matches if ... matches next, but doesn't consume the string.
An other way is to ground it and to say: with one of [abc] and one of [123] means there is at least a [123][abc] or a [abc][123] in the string resulting in
>>> re.findall('\\b[abc321]*(?:[abc][123]|[123][abc])[abc321]*\\b', ' 123abc 123 abc')
['123abc']

Not the most beautiful regular expression, but it works:
\b[a-z\d]*(?:\d[a-z]|[a-z]\d)[a-z\d]*\b

If the format is the same each time, that is:
[########: ########]: STARTED at ########ng job number ########
You can use:
([^\]\s]+)\]
With re.findall, or re.search and getting .group(1) if you use re.search.
[^\]\s]+ is a negated class and will match any character except space (and family) or closing square bracket.
The regex basically looks for characters (except ] or spaces) up until a closing square bracket.
If you want to match any string containing both alpha and numeric characters, you will need a lookahead:
\b(?=[0-9]*[a-z])(?=[a-z]*[0-9])[a-z0-9]+\b
Used like so:
result = re.search(r'\b(?=[0-9]*[a-z])(?=[a-z]*[0-9])[a-z0-9]+\b', text, re.I)
re.I is for ignorecase.
\b is a word boundary and will match only between a 'word' character and a 'non-word' character (or start/end of string).
(?=[0-9]*[a-z]) is a positive lookahead and makes sure there's at least 1 alpha in the part to be matched.
(?=[a-z]*[0-9]) is a similar lookahead but checks for digits.

You can use more specific regular expression and skip the findall.
import re
s = '[mytaskid: 3fee46d2]: STARTED at processing job number 10022001'
mo = re.search(':\s+(\w+)', s)
print mo.group(1)

python regex: get end digits from a string

I am quite new to python and regex (regex newbie here), and I have the following simple string:
s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716"""
I would like to extract only the last digits in the above string i.e 767980716 and I was wondering how I could achieve this using python regex.
I wanted to do something similar along the lines of:
re.compile(r"""-(.*?)""").search(str(s)).group(1)
indicating that I want to find the stuff in between (.*?) which starts with a "-" and ends at the end of string - but this returns nothing..
I was wondering if anyone could point me in the right direction..
Thanks.

You can use re.match to find only the characters:
>>> import re
>>> s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716"""
>>> re.match('.*?([0-9]+)$', s).group(1)
'767980716'
Alternatively, re.finditer works just as well:
>>> next(re.finditer(r'\d+$', s)).group(0)
'767980716'
Explanation of all regexp components:
.*? is a non-greedy match and consumes only as much as possible (a greedy match would consume everything except for the last digit).
[0-9] and \d are two different ways of capturing digits. Note that the latter also matches digits in other writing schemes, like ୪ or ൨.
Parentheses (()) make the content of the expression a group, which can be retrieved with group(1) (or 2 for the second group, 0 for the whole match).
+ means multiple entries (at least one number at the end).
$ matches only the end of the input.

Nice and simple with findall:
import re
s=r"""99-my-name-is-John-Smith-6376827-%^-1-2-767980716"""
print re.findall('^.*-([0-9]+)$',s)
>>> ['767980716']
Regex Explanation:
^ # Match the start of the string
.* # Followed by anthing
- # Upto the last hyphen
([0-9]+) # Capture the digits after the hyphen
$ # Upto the end of the string
Or more simply just match the digits followed at the end of the string '([0-9]+)$'

Your Regex should be (\d+)$.
\d+ is used to match digit (one or more)
$ is used to match at the end of string.
So, your code should be: -
>>> s = "99-my-name-is-John-Smith-6376827-%^-1-2-767980716"
>>> import re
>>> re.compile(r'(\d+)$').search(s).group(1)
'767980716'
And you don't need to use str function here, as s is already a string.

Use the below regex
\d+$
$ depicts the end of string..
\d is a digit
+ matches the preceding character 1 to many times

Save the regular expressions for something that requires more heavy lifting.
>>> def parse_last_digits(line): return line.split('-')[-1]
>>> s = parse_last_digits(r"99-my-name-is-John-Smith-6376827-%^-1-2-767980716")
>>> s
'767980716'

I have been playing around with several of these solutions, but many seem to fail if there are no numeric digits at the end of the string. The following code should work.
import re
W = input("Enter a string:")
if re.match('.*?([0-9]+)$', W)== None:
last_digits = "None"
else:
last_digits = re.match('.*?([0-9]+)$', W).group(1)
print("Last digits of "+W+" are "+last_digits)

Try using \d+$ instead. That matches one or more numeric characters followed by the end of the string.