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how to deliminate string with different and multiple deliminater

I have a string (python), that has the date, model, make, and the year it is below:
string = "Mar 17 1997 H569, CAT: 2022"
I want to write a program that will ask the user to enter the string, and the program will automatically do something like:
date: data
model:data
make: data
year: data
The question, how can I deliminate,since I have space, comma, colon, etc. If I use characters then the problem will be not all makes and model have the same number of characters. What I am trying to do is to deliminate a string with more than once deliminater randomly mixed, in python?

One option is to use a regex:
import re
regex = re.compile('(\w+ \d+ \d+) (\w+), (\w+): (\d+)')
string = "Mar 17 1997 H569, CAT: 2022"
regex.findall(string)
output: [('Mar 17 1997', 'H569', 'CAT', '2022')]

string = input('date :') #"Mar 17 1997 H569, CAT: 2022"
mounth,day,dyear,model,make,year=string.split()
print(f'date:{day}/{mounth}/{year}')
print(f'model:{model.strip(",")}')
print(f'make:{make.strip(":")}')
print(f'year:{year}')

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)'))

Unexpected result in regex - what am I missing?

I am trying to extract immunization records of this form:
Immunization: Tetanus
Other: Booster
Method: Injection
Date Received: 07 Jan 2013
and also of this form:
Immunization: TETANUS DIPTHERIA (TD-ADULT)
Date Received: 07 Dec 2012 # 1155
Location: PORTLAND (OR) VAMC
Reaction:* None Reported
Comments: 1234567
Here is my pattern string:
"Immunization:(.*?)\n[.\n*?]*?Date Received:(.*?)\n"
This is identifying the second pattern and extracting vaccination name and date but not the first pattern. I thought that [.\n*?]*? would take care of the two possibilities (that there are other fields between vaccination name and vaccination date...or not...but this doesn't seem to be doing the trick. What is wrong with my regex and how cna I fix it?

You can use:
import re
matches = re.findall(r"Immunization:\s+(.*?)\s+.*?Date Received:\s+(.*?)$", subject, re.IGNORECASE | re.DOTALL | re.MULTILINE)
Regex Demo | Python Demo
Regex Explanation:

Tested this on pythex with MULTILINE and DOTALL:
Input
Immunization: Tetanus
Other: Booster
Method: Injection
Date Received: 07 Jan 2013
Immunization: TETANUS DIPTHERIA (TD-ADULT)
Date Received: 07 Dec 2012 # 1155
Location: PORTLAND (OR) VAMC
Reaction:* None Reported
Comments: 1234567
Pattern: Immunization:\s+(\w+).*?Date Received:\s+([^\n]+)
Match 1
Tetanus
07 Jan 2013
Match 2
TETANUS
07 Dec 2012 # 1155
Pythex
Pythex with different grouping

The . in [.\n] is taken as a literal '.', not as a symbol for any-character. This is why the date line immediately following the immunisation is accepted but you fail to jump across a character that is not a newline or a dot.
(.*\n)* comes to mind to help you out in the closest way to what you already have. However, it is a bit unfortunate to have so many nested * since this means a long breath for parsing the record and as a human I also find it more difficult to understand. It may be preferable to start every loop with a literal to help the decision making if a loop shall be entered/continued at all.
If I did not mess it up then
Immunization:(.*?)(\n.*)*\nDate Received:(.*)\n
would do without left recursion and "Date Received" would only be detected at the beginning of the line.

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