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Regular expression match / split

I am having some trouble trying to figure out how to use regular expressions in python. Ultimately I am trying to do what sscanf does for me in C.
I am trying to match given strings that look like so:
12345_arbitrarystring_2020_05_20_10_10_10.dat
I (seem) to be able to validate this format by calling match on the following regular expression
regex = re.compile('[0-9]{5}_.+_[0-9]{4}([-_])[0-9]{2}([-_])[0-9]{2}([-_])[0-9]{2}([:_])[0-9]{2}([:_])[0-9]{2}\\.dat')
(Note that I do allow for a few other separators then just '_')
I would like to split the given string on these separators so I do:
regex = re.compile('[_\\-:.]+')
parts = regex.split(given_string)
This is all fine .. the problem is that I would like my 'arbitrarystring' part to include '-' and '_' and the last split currently, well, splits them.
Other than manually cutting the timestamp and the first 5 digits off that given string, what can I do to get that arbitrarystring part?

You could use a capturing group to get the arbitrarystring part and omit the other capturing groups.
You could for example use a character class to match 1+ word characters or a hyphen using [\w-]+
If you still want to use split, you could add capturing groups for the first and the second part, and split only those groups.
^[0-9]{5}_([\w-]+)_[0-9]{4}[-_][0-9]{2}[-_][0-9]{2}[-_][0-9]{2}[:_][0-9]{2}[:_][0-9]{2}\.dat$
^^^^^^^^
Regex demo

It seems to be possible to cut down your regex to validate the whole pattern to:
^\d{5}_(.+?)_\d{4}[-_](?:\d{2}[-_]){2}(?:\d{2}[:_]){2}\d{2}\.dat$
Refer to group 1 for your arbitrary string.
Online demo
Quick reminder: You didn't seem to have used raw strings, but instead escaping with a double backslash. Python has raw strings which makes you don't have to escape backslashes nomore.

Regular expressions: distinguish strings including/excluding a given word

I'm working in Python and try to handle StatsModel's GLM output. I'm relatively new to regular expressions.
I have strings such as
string_1 = "C(State)[T.Kansas]"
string_2 = "C(State, Treatment('Alaska'))[T.Kansas]"
I wrote the following regex:
pattern = re.compile('C\((.+?)\)\[T\.(.+?)\]')
print(pattern.search(string_1).group(1))
#State
print(pattern.search(string_2).group(1))
#State, Treatment('Alaska')
So both of these strings match the pattern. But we want to get State in both cases. Basically we want to get read of everything after comma (including it) inside first brackets.
How can we distinguish the string_2 pattern from string_1's and extract only State without , Treatment?

You can add an optional non-capturing group instead of just allowing all characters:
pattern = re.compile('C\((.+?)(?:, .+?)?\)\[T\.(.+?)\]')
(?:...) groups the contents together without capturing it. The trailing ? makes the group optional.

You may use this regex using negative character classes:
C\((\w+)[^[]*\[T\.([^]]+)\]
RegEx Demo

Python regex to find all case of curly brackets, inclusive of brackets

I would to like find and replace strings within double curly brackets, inclusive of the brackets themselves.
For example:
{{hello}}
should ideally return:
{{hello}}
I found this expression: {{(.*?)}} here
However this only returns text between the brackets. Ie. With this expression, the example above would return: "hello", rather than "{{hello}}"

I'm assuming you're using re.findall() which only returns the contents of capturing groups, if they are present in the regex.
Since you don't want them, just remove the capturing parentheses:
matches = re.findall("{{.*?}}", mystring)
When using re.sub() for replacing, the original regex will work just fine.

You can just move the ( and ) to include the brackets within the group.
Try: ({{.*?}}). I highly recommend a regex tester for this kind of work.

Match everything expect a specific string

I am using Python 2.7 and have a question with regards to regular expressions. My string would be something like this...
"SecurityGroup:Pub HDP SG"
"SecurityGroup:Group-Name"
"SecurityGroup:TestName"
My regular expression looks something like below
[^S^e^c^r^i^t^y^G^r^o^u^p^:].*
The above seems to work but I have the feeling it is not very efficient and also if the string has the word "group" in it, that will fail as well...
What I am looking for is the output should find anything after the colon (:). I also thought I can do something like using group 2 as my match... but the problem with that is, if there are spaces in the name then I won't be able to get the correct name.
(SecurityGroup):(\w{1,})

Why not just do
security_string.split(':')[1]
To grab the second part of the String after the colon?

You could use lookbehind:
pattern = re.compile(r"(?<=SecurityGroup:)(.*)")
matches = re.findall(pattern, your_string)
Breaking it down:
(?<= # positive lookbehind. Matches things preceded by the following group
SecurityGroup: # pattern you want your matches preceded by
) # end positive lookbehind
( # start matching group
.* # any number of characters
) # end matching group
When tested on the string "something something SecurityGroup:stuff and stuff" it returns matches = ['stuff and stuff'].
Edit:
As mentioned in a comment, pattern = re.compile(r"SecurityGroup:(.*)") accomplishes the same thing. In this case you are matching the string "SecurityGroup:" followed by anything, but only returning the stuff that follows. This is probably more clear than my original example using lookbehind.

Maybe this:
([^:"]+[^\s](?="))
Regex live here.

Backreferencing in Python: findall() method output for HTML string

I am trying to learn some regular expressions in Python. The following does not produce the output I expected:
with open('ex06-11.html') as f:
a = re.findall("<div[^>]*id\\s*=\\s*([\"\'])header\\1[^>]*>(.*?)</div>", f.read())
# output: [('"', 'Some random text')]
The output I was expecting (same code, but without the backreference):
with open('ex06-11.html') as f:
print re.findall("<div[^>]*id\\s*=\\s*[\"\']header[\"\'][^>]*>(.*?)</div>", f.read())
# output: ['Some random text']
The question really boils down to: why is there a quotation mark in my first output, but not in my second? I thought that ([abc]) ... //1 == [abc] ... [abc]. Am I incorrect?

From the docs on re.findall:
If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group.
If you want the entire match to be returned, remove the capturing groups or change them to non-capturing groups by adding ?: after the opening paren. For example you would change (foo) in your regex to (?:foo).
Of course in this case you need the capturing group for the backreference, so your best bet is to keep your current regex and then use a list comprehension with re.finditer() to get a list of only the second group:
regex = re.compile(r"""<div[^>]*id\s*=\s*(["'])header\1[^>]*>(.*?)</div>""")
with open('ex06-11.html') as f:
a = [m.group(2) for m in regex.finditer(f.read())
A couple of side notes, you should really consider using an HTML parser like BeautifulSoup instead of regex. You should also use triple-quoted strings if you need to include single or double quotes within you string, and use raw string literals when writing regular expressions so that you don't need to escape the backslashes.

The behaviour is clearly documented. See re.findall:
Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found.
If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group. Empty matches are included in the result unless they touch the beginning of another match.
So, if you have a capture group in your regex pattern, then findall method returns a list of tuple, containing all the captured groups for a particular match, plus the group(0).
So, either you use a non-capturing group - (?:[\"\']), or don't use any group at all, as in your 2nd case.
P.S: Use raw string literals for your regex pattern, to avoid escaping your backslashes. Also, compile your regex outside the loop, so that is is not re-compiled on every iteration. Use re.compile for that.

When I asked this question I was just starting with regular expressions. I have since read the docs completely, and I just wanted to share what I found out.
Firstly, what Rohit and F.J suggested, use raw strings (to make the regex more readable and less error-prone) and compile your regex beforehand using re.compile. To match an HTML string whose id is 'header':
s = "<div id='header'>Some random text</div>"
We would need a regex like:
p = re.compile(r'<div[^>]*id\s*=\s*([\"\'])header\1[^>]*>(.*?)</div>')
In the Python implementation of regex, a capturing group is made by enclosing part of your regex in parentheses (...). Capturing groups capture the span of text that they match. They are also needed for backreferencing. So in my regex above, I have two capturing groups: ([\"\']) and (.*?). The first one is needed to make the backreference \1 possible. The use of a backreferences (and the fact that they reference back to a capturing group) has consequences, however. As pointed out in the other answers to this question, when using findall on my pattern p, findall will return matches from all groups and put them in a list of tuples:
print p.findall(s)
# [("'", 'Some random text')]
Since we only want the plain text from between the HTML tags, this is not the output we're looking for.
(Arguably, we could use:
print p.findall(s)[0][1]
# Some random text
But this may be a bit contrived.)
So in order to return only the text from between the HTML tags (captured by the second group), we use the group() method on p.search():
print p.search(s).group(2)
# Some random text
I'm fully aware that all but the most simple HTML should not be handled by regex, and that instead you should use a parser. But this was just a tutorial example for me to grasp the basics of regex in Python.