Develop Reference

How to empty or optional match group in Python RegEx= [duplicate]

This is an example string:
123456#p654321
Currently, I am using this match to capture 123456 and 654321 in to two different groups:
([0-9].*)#p([0-9].*)
But on occasions, the #p654321 part of the string will not be there, so I will only want to capture the first group. I tried to make the second group "optional" by appending ? to it, which works, but only as long as there is a #p at the end of the remaining string.
What would be the best way to solve this problem?

You have the #p outside of the capturing group, which makes it a required piece of the result. You are also using the dot character (.) improperly. Dot (in most reg-ex variants) will match any character. Change it to:
([0-9]*)(?:#p([0-9]*))?
The (?:) syntax is how you get a non-capturing group. We then capture just the digits that you're interested in. Finally, we make the whole thing optional.
Also, most reg-ex variants have a \d character class for digits. So you could simplify even further:
(\d*)(?:#p(\d*))?
As another person has pointed out, the * operator could potentially match zero digits. To prevent this, use the + operator instead:
(\d+)(?:#p(\d+))?

Your regex will actually match no digits, because you've used * instead of +.
This is what (I think) you want:
(\d+)(?:#p(\d+))?

How to match substring or whole string

In Python regex, how would I match only the facebook.com...777 substrings given either string? I don't want the ?sfnsn=mo at the end.
I have (?<=https://m\.)([^\s]+) to match everything after the https://m.. I also have (?=\?sfnsn) to match every thing in front of ?sfnsn.
How do I combine the regex to only return the facebook.com...777 part for either string.
have: https://m.facebook.com/story.php?story_fbid=123456789&id=7777777777?sfnsn=mo
want: facebook.com/story.php?story_fbid=123456789&id=7777777777
have: https://m.facebook.com/story.php?story_fbid=123456789&id=7777777777
want: facebook.com/story.php?story_fbid=123456789&id=7777777777
Here's what I was messing around with https://regex101.com/r/WYz5dn/2
(?<=https://m\.)([^\s]+)(?=\?sfnsn)

You could use a capturing group instead of a positive lookbehind and match either ?sfnsn or the end of the string.
https://m\.(\S*?)(?:\?sfnsn|$)
Regex demo
Using the lookarounds, the pattern could be:
(?<=https://m\.)\S*?(?=\?sfnsn|$)
Regex demo

Putting a ? at the end works, since the last grouped lookahead may or may not exist, we put a question mark after it:
(?<=https://m\.)([^\s]+)(?=\?sfnsn)?

Replace second and last second characters, using re.sub

I have a string "F(foo)", and I'd like to replace that string with "F('foo')". I know we can also use regular expression in the second parameter and do this replacement using re.sub(r"F\(foo\)", r"F\('foo'\)",str). But the problem here is, foo is a dynamic string variable. It is different every time we want to do this replacement. Is it possible by some sort of regex, to do such replacement in a cleaner way?
I remember one way to extract foo using () and then .group(1). But this would require me to define one more temporary variable just to store foo. I'm curious if there is a way by which we can replace "F(foo)" with "F('foo')" in a single line or in other words in a more cleaner way.
Examples :
F(name) should be replaced with F('name').
F(id) should be replaced with F('id').
G(name) should not be replaced.
So, the regex would be r"F\((\w)+\)" to find such strings.

Using re.sub
Ex:
import re
s = "F(foo)"
print(re.sub(r"\((.*)\)", r"('\1')", s))
Output:
F('foo')

The following regex encloses valid [Python|C|Java] identifiers after F and in parentheses in single quotation marks:
re.sub(r"F\(([_a-z][_a-z0-9]+)\)", r"F('\1')", s, flags=re.I)
#"F('foo')"

There are several ways, depending on what foo actually is.
If it can't contain ( or ), you can just replace ( with (' and ) with '). Otherwise, try using
re.sub(r"F\((.*)\)", r"F('\1')", yourstring)
where the \1 in the replacement part will reference the (.*) capture group in the search regex

In your pattern F\((\w)+\) you are almost there, you just need to put the quantifier + after the \w to repeat matching 1+ word characters.
If you put it after the capturing group, you repeat the group which will give you the value of the last iteration in the capturing group which would be the second o in foo.
You could update your expression to:
F\((\w+)\)
And in the replacement refer to the capturing group using \1
F('\1')
For example:
import re
str = "F(foo)"
print(re.sub(r"F\((\w+)\)", r"F('\1')", str)) # F('foo')
Python demo

[FORKING]Python Regex - Re.Sub and Re.Findall Interesting Challenges

Not sure if this is something that should be a bounty. II just want to understand regex better.
I checked the responses in the Regex to match pattern.one skip newlines and characters until pattern.two and Regex to match if given text is not found and match as little as possible threads and read about Tempered Greedy Token Solutions and Explicit Greedy Alternation Solutions on RexEgg, but admittedly the explanations baffled me.
I spent the last day fiddling mainly with re.sub (and with findall) because re.sub's behaviour is odd to me.
.
Problem 1:
Given Strings below with characters followed by / how would I produce a SINGLE regex (using only either re.sub or re.findall) that uses alternating capture groups which must use [\S]+/ to get the desired output
>>> string_1 = 'variety.com/2017/biz/news/tax-march-donald-trump-protest-1202031487/'
>>> string_2 = 'variety.com/2017/biz/the/life/of/madam/green/news/tax-march-donald-trump-protest-1202031487/'
>>> string_3 = 'variety.com/2017/biz/the/life/of/news/tax-march-donald-trump-protest-1202031487/the/days/of/our/lives'
Desired Output Given the Conditions(!!)
tax-march-donald-trump-protest-
CONDITIONS: Must use alternating capture groups which must capture ([\S]+) or ([\S]+?)/ to capture the other groups but ignore them if they don't contain -
I'M WELL AWARE that it would be better to use re.findall('([\-]*(?:[^/]+?\-)+)[\d]+', string) or something similar but I want to know if I can use [\S]+ or ([\S]+) or ([\S]+?)/ and tell regex that if those are captured, ignore the result if it contains / or doesn't contain - While also having used an alternating capture group
I KNOW I don't need to use [\S]+ or ([\S]+) but I want to see if there is an extra directive I can use to make the regex reject some characters those two would normally capture.

Posted per request:
(?:(?!/)[\S])*-(?:(?!/)[\S])*
https://regex101.com/r/azrwjO/1
Explained
(?: # Optional group
(?! / ) # Not a forward slash ahead
[\S] # Not whitespace class
)* # End group, do 0 to many times
- # A dash must exist
(?: # Optional group, same as above
(?! / )
[\S]
)*

You could use
/([-a-z]+)-\d+
and take the first capturing group, see a demo on regex101.com.

python regular expression of a string

I have python string
wrong_data_type is not one of the allowed values `([one_two, two_three, three_four])`
and I have a regexp:
\w+ is not one of the allowed values`\(\[\w,+\)\]`
However, it is not correct? Any help?

The regexp should be
\w+ is not one of the allowed values `\(\[(?:\w+, )*\w+\]\)`
Fixes:
Added space after values.
\]\) at the end instead of \)\].
Inside the brackets, need to allow multiple \w, so it should be \w+.
Need to have a space after ,.
Need a group around \w+, to match multiple comma-separated words using the * quantifier.
Then have to match a single last word with no comma after it.

data = re.search(r'\(\[[\w,\s]+\]\)', string).group()

You can use the following:
\w+ is not one of the allowed values `\(\[[\w,\s]+\]\)`