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Python Regex to extract meeting invite from Gmail Subject

I'm trying to extract the meeting date / time from meeting invites within Gmail's subject. Below is an example of a subject for a meeting invite:
Invitation: Bob / Carol Meeting # Tue Oct 25, 2022 11:30am - 12pm (CST) (bob#example.org)
What I would like to extract:
Tue Oct 25, 2022 11:30am - 12pm (CST)
I think the pattern could simply start with the space after the "#" and end with the ")". My Regex is very rusty so would appreciate any help :)
Many thanks!

Try this - it should match everything after the "# " and up to the end of the timezone ")"
import re
string = (
'Invitation: Bob / Carol Meeting # Tue Oct 25, 2022 11:30am - 12pm (CST) (bob#example.org)'
)
pattern = re.compile(r'(?<=# )[^)]+\)')
matches = re.findall(pattern, string)
print(matches)
# => 'Tue Oct 25, 2022 11:30am - 12pm (CST)'
See here for a breakdown of the RegEx I used. Bear in mind that re.findall returns a list of matches, which is helpful if you want to scan a long multiline string of text and get all the matches at once. If you only care about the 1st match, you can get it by index e.g. print(matches[0]).

It looks like you don't technically need regex for this.
Try the following:
>>> s = 'Invitation: Bob / Carol Meeting # Tue Oct 25, 2022 11:30am - 12pm (CST) (bob#example.org)'
>>> s[s.index('#') + 1 : s.rindex('(')].strip()
'Tue Oct 25, 2022 11:30am - 12pm (CST)'

How to find the words correspond to month and replace it with numerical?

How to find the words that correspond to the month "January, February, March,.. etc." and replace them with numerical "01, 02, 03,.."
I tried the code below
def transformMonths(string):
rep = [("May", "05"), ("June", "06")]
for pat, repl in rep:
s = re.sub(pat, repl, string)
return s
print( transformMonths('I was born on June 24 and my sister was born on May 17') )
My code provides this result ('I was born on 06 24 and my sister was born on May 17')
However, I want the output to be like this ('I was born on 06 24 and my sister was born on 05 17')

You are performing the replacement on the initial (unmodified) string at each iteration so you end up with only one month name being replaced. You can fix that by assigning string instead of s in the loop (and return string at the end).
Note that your approach does not require a regular expression and could use a simple string replace: string = string.replace(pat,repl).
In both cases, because the replacement does not take into account word boundaries, the function would replace partial words such as:
"Mayor Smith was elected on May 25" --> "05or Smith was elected on 05 25".
You can fix that in your regular expression by adding \b before and after each month name. This will ensure that the month names are only found if they are between word boundaries.
The re.sub can perform multiple replacements with varying values if you give it a function instead of a fixed string. So you can build a combined regular expression that will find all the months and replace the words that are found using a dictionary:
import re
def numericMonths(string):
months = {"January":"01", "Ffebruary":"02","March":"03", "April":"04",
"May":"05", "June":"06", "July":"07", "August":"08",
"September":"09","October":"10", "November":"11","December":"12"}
pattern = r"\b("+"|".join(months)+r")\b" # all months as distinct words
return re.sub(pattern,lambda m:months[m.group()],string)
output:
numericMonths('I was born on June 24 and my sister was born on May 17')
'I was born on 06 24 and my sister was born on 05 17'

Regex to extract date and time from email text

I've got a file that has a ton of text in it. Some of it looks like this:
X-DSPAM-Processed: Fri Jan 4 18:10:48 2008
X-DSPAM-Confidence: 0.6178
X-DSPAM-Probability: 0.0000
Details: http://source.sakaiproject.org/viewsvn/?view=rev&rev=39771
Author: louis#media.berkeley.edu
Date: 2008-01-04 18:08:50 -0500 (Fri, 04 Jan 2008)
New Revision: 39771
Modified:
bspace/site-manage/sakai_2-4-x/site-manage-tool/tool/src/bundle/sitesetupgeneric.properties
bspace/site-manage/sakai_2-4-x/site-manage-tool/tool/src/java/org/sakaiproject/site/tool/SiteAction.java
Log:
BSP-1415 New (Guest) user Notification
I need to pull out only dates that follow this pattern:
2008-01-04 18:08:50 -0500
Here's what I tried:
import re
text = open('mbox-short.txt')
for line in text:
dates = re.compile('\d{4}(?P<sep>[-/])\d{2}(?P=sep)\d{2}\s\d{2}:\d{2}:]\d{2}\s[-/]\d{4}')
print(dates)
text.close()
The return I got was hundreds of:
\d{4}(?P<sep>[-/])\d{2}(?P=sep)\d{2}\s\d{2}:\d{2}:]\d{2}\s[-/]\d{4}

Two things:
First, the regex itself:
regex = re.compile(r'\b\d{4}[-/]\d{2}[-/]\d{2}\s\d{2}:\d{2}:\d{2}\s[-+]\d{4}\b')
Secondly, you need to call regex.findall(file) where file is a string:
>>> regex.findall(file)
['2008-01-04 18:08:50 -0500']
re.compile() produces a compiled regular expression object. findall is one of several methods of this object that let you do the actual searching/matching/finding.
Lastly: you're currently using named capturing groups. ((?P<sep>[-/])) From your question, "I need to pull out only dates that follow this pattern," it doesn't seem like you need these. You want to extract the entire expression, not capture the "separators," which is what capturing groups are designed for.
Full code block:
>>> import re
>>> regex = re.compile(r'\b\d{4}[-/]\d{2}[-/]\d{2}\s\d{2}:\d{2}:\d{2}\s[-+]\d{4}\b')
>>> with open('mbox-short.txt') as f:
... print(regex.findall(f.read()))
...
['2008-01-04 18:08:50 -0500']

Here's another solution.
import re
numberExtractRegex = re.compile(r'(\d\d\d\d[-]\d\d[-]\d\d\s\d\d[:]\d\d[:]\d\d\s[-]\d\d\d\d)')
print(numberExtractRegex.findall('Date: 2008-01-04 18:08:50 -0500 (Fri, 04 Jan 2008), Date: 2010-01-04 18:08:50 -0500 (Fri, 04 Jan 2010)'))

Extracting words next to a location or Duration in python

How can i extract words next to a location or Duration? What is the best possible regex in python to do this action?
Example:-
Kathick Kumar, Bangalore who was a great person and lived from 29th March 1980 - 21 Dec 2014.
In the above example i want to extract the words before location and the words before duration. Here the location and duration is not fixed, what will be the best possible regex for this in python? Or can we do this using nltk?
Desired output:-
Output-1: Karthick Kumar (Keyword here is Location)
Output-2: who was a great person and lived from (Keyword here is duration)

I suggest using Lookaheads.
In your example, assuming you want the words before Bangalore and 29th March 1980 - 21 Dec 2014, you could use lookaheads( and lookbehinds) to get the relevant match.
I've used this regex: (.*)(?>Bangalore)(.+)(?=29th March 1980 - 21 Dec 2014) and captured the text in parentheses, which can be accessed by using \1 and \2.
DEMO

Python string split without common delimiter

I am fairly new to Python. An external simulation software I use gives me reports which include data in the following format:
1 29 Jan 2013 07:33:19.273 29 Jan 2013 09:58:10.460 8691.186
I am looking to split the above data into four strings namely;
'1', '29 Jan 2013 07:33:19.273', '29 Jan 2013 09:58:10.460', '8691.186'
I cannot use str.split since it splits out the date into multiple strings. There appears to be four white spaces between 1 and the first date and between the first and second dates. I don't know if this is four white spaces or tabs.
Using '\t' as a delimiter on split doesn't do much. If I specify ' ' (4 spaces) as a delimiter, I get the first three strings. I also then get an empty string and leading spaces in the final string. There are 10 spaces between the second date and the number.
Any suggestions on how to deal with this would be much helpful!
Thanks!

You can split on more than one space with a simple regular expression:
import re
multispace = re.compile(r'\s{2,}') # 2 or more whitespace characters
fields = multispace.split(inputline)
Demonstration:
>>> import re
>>> multispace = re.compile(r'\s{2,}') # 2 or more whitespace characters
>>> multispace.split('1 29 Jan 2013 07:33:19.273 29 Jan 2013 09:58:10.460 8691.186')
['1', '29 Jan 2013 07:33:19.273', '29 Jan 2013 09:58:10.460', '8691.186']

If the data is fixed width you can use character addressing in the string
n=str[0]
d1=str[2:26]
d2=str[27:51]
l=str[52:]
However, if Jan 02 is shown as Jan 2 this may not work as the width of the string may be variable