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need to find python regex to get only the last digit

I have a huge pdf that is all very basic text on pages for invoices, I need to create a regex or 2 so when I split it I get the customer number and the invoice number to use in the file name. I am using python 3 and pypdf2 currently
text example of 2 of the pages:
Detailed Invoice Report
Starting 8/12/2015 and ending 8/11/2022
Company: (Multiple Companies) Printed by Robert S on 8/11/2022 1:26:46PM
Donna Contact Cust# Name: Customer A 1234
Customer A Invoice Date Invoice Name 8/12/2015 241849
Item Description Qty Price Extended Price
Credit ($810.00) 1 ($810.00) 1
Due Paid Total Total Taxes Subtotal
($810.00) ($810.00) $0.00 ($810.00)
Balance: ($810.00) $0.00 $0.00
8/11/2022 1:26:46PM Page 1 of 340977
Detailed Invoice Report
Starting 8/12/2015 and ending 8/11/2022
Company: (Multiple Companies) Printed by Robert S on 8/11/2022 1:26:46PM
Customer B Cust# Name: Customer B 45678
Customer B Invoice Date Invoice Name 8/12/2015 241850
Item Description Qty Price Extended Price
credit ($49.99) 1 ($49.99) 1
Due Paid Total Total Taxes Subtotal
($49.99) ($49.99) $0.00 ($49.99)
Balance: ($49.99) $0.00 $0.00
8/11/2022 1:26:46PM Page 2 of 340977
currently I have these 2 regex filters to get each one kind of but I do not know how to only keep the last groups match from them.
Note: the firstmatch regex is broken if the customer name has a number in it which is an edge case but not uncommon in the data
firstmatch=r"(Name:)(\D*)(\d+)"
secondmatch=r"(Name )(\d*.\d*.\d*..)(\d*)"
Each one is its own page and I would like the regex to be able to pull from the first one 1234 241849 and the second one 45678 241850

You could get both values using a capture matching the last digits on the line.
For the first pattern:
\bName:.*?\b(\d+)[^\d\n]*$
Explanation
\bName: Match Name: preceded by a word boundary
.*? Match any character without a newline, as least as possible
\b(\d+) A word boundary, then capture 1+ digits in group 1
[^\d\n]* Optionally match any character except digits or a newline
$ End of string
Regex demo
For the second pattern you can make it a bit more specific, where [^\S\n]+ matches 1+ whitespace chars without newlines:
\bName[^\S\n]+\d+/\d+/\d+[^\S\n]+(\d+)[^\d\n]*$
Regex demo
Or if the lines are right behind each other, you can also use 1 pattern with 2 capture groups and match the newline at the end of the first line:
\bName:.*?\b(\d+)[^\d\n]*\n\b.*?Name[^\S\n]+\d+/\d+/\d+[^\S\n]+(\d+)[^\d\n]*$
Regex demo

Returning empty string for missing capture group Python regex

I'm working on parsing string text containing information on university, year, degree field, and whether or not a person graduated. Here are two examples:
ex1 = 'BYU: 1990 Bachelor of Arts Theater (Graduated):BYU: 1990 Bachelor of Science Mathematics (Graduated):UNIVERSITY OF VIRGINIA: 1995 Master of Science Mechanical Engineering (Graduated):MICHIGAN STATE UNIVERSITY: 2008 Master of Fine Arts INDUSTRIAL DESIGN (Graduated)'
ex2 = 'UCSD: 2001 Bachelor of Arts English:UCLA: 2005 Bachelor of Science Economics (Graduated):UCSD 2010 Master of Science Economics'
What I am struggling to accomplish is to have an entry for each school experience regardless of whether specific information is missing. In particular, imagine I wanted to pull whether each degree was finished from ex1 and ex2 above. When I try to use re.findall I end up with something like the following for ex1:
# Code:
re.findall('[A-Z ]+: \d+ Bachelor [^:]+\((Graduated)', ex1)
# Output:
['Graduated', 'Graduated']
which is what I want, two entries for two Bachelor's degrees. For ex2, however, one of the Bachelor's degrees was unfinished so the text does not contain "(Graduated)", so the output is the following:
# Code:
re.findall('[A-Z ]+: \d+ Bachelor [^:]+\((Graduated)', ex2)
# Output:
['Graduated']
# Desired Output:
['', 'Graduated']
I have tried making the capture group optional or including the colon after graduated and am not making much headway. The example I am using is the "Graduated" information, but in principle the more general question remains if there is an identifiable degree but it is missing one or two pieces of information (like graduation year or university). Ultimately I am just looking to have complete information on each degree, including whether certain pieces of information are missing. Thank you for any help you can provide!

You can use the ?-Quantifier to match "Graduated" (and the paranthesis () between 0 and n times.
re.findall('[A-Z ]+: \d+ Bachelor [^:()]*\(?(Graduated)?', ex2)
Output:
>>> re.findall('[A-Z ]+: \d+ Bachelor [^:()]*\(?(Graduated)?', ex2)
['', 'Graduated']

How about this?
[re.sub('[(:)]', '', t) for t in [re.sub('^[^\(]+','', s) for s in re.findall('[A-Z ]+: \d+ Bachelor [^:]+:', ex1)]]
# output ['Graduated', 'Graduated']
[re.sub('[(:)]', '', t) for t in [re.sub('^[^\(]+','', s) for s in re.findall('[A-Z ]+: \d+ Bachelor [^:]+:', ex2)]]
# output ['', 'Graduated']

Add a single space and comma between words that are connected using regex

I have a nested list_3 which looks like:
[['Company OverviewCompany: HowSector: SoftwareYear Founded: 2010One Sentence Pitch: Easily give and request low-quality feedback with your team to achieve more togetherUniversity Affiliation(s): Duke$ Raised: $240,000Investors: Friends & familyTraction to Date: 10% of monthly active users (MAU) are also active weekly'], [['Company OverviewCompany: GrubSector: SoftwareYear Founded: 2018One Sentence Pitch: Find food you likeUniversity Affiliation(s): Stanford$ Raised: $340,000Investors: Friends & familyTraction to Date: 40% of monthly active users (MAU) are also active weekly']]]
I would like to use regex to add a comma followed by a single space between each joined word ie(HowSector:, SoftwareYear, 2010One), So far I have tried to write a re.sub code to do, by selecting all the characters without whitespace and replacing this, but have run into some issues:
for i, list in enumerate(list_3):
list_3[i] = [re.sub('r\s\s+', ', ', word) for word in list]
list_33.append(list_3[i])
print(list_33)
error:
return _compile(pattern, flags).sub(repl, string, count)
TypeError: expected string or bytes-like object
I would like the output to be:
[['Company Overview, Company: How, Sector: Software, Year Founded: 2010, One Sentence Pitch: Easily give and request low-quality feedback with your team to achieve more together University, Affiliation(s): Duke, $ Raised: $240,000, Investors: Friends & family, Traction to Date: 10% of monthly active users (MAU) are also active weekly'],[...]]
Any ideas how I can use regex to do this?

The main problem is that your nested list has no constant levels. Sometimes it has 2 levels and sometimes it has 3 levels. This is why you are getting the above error. In the case the list has 3 levels, re.sub receives a list as the third argument instead of a string.
The second problem is that the regex you are using is not the correct regex. The most naive regex we can use here should (at the very least) be able to find a non-whitespace charcter followed by a capital letter.
In the below example code, I'm using re.compile (since the same regex will be used over and over again, we might as well pre-compile it and gain some performance boost) and I'm just printing the output. You'll need to figure out a way to get the output in the format you want.
regex = re.compile(r'(\S)([A-Z])')
replacement = r'\1, \2'
for inner_list in nested_list:
for string_or_list in inner_list:
if isinstance(string_or_list, str):
print(regex.sub(replacement, string_or_list))
else:
for string in string_or_list:
print(regex.sub(replacement, string))
Outputs
Company Overview, Company: How, Sector: Software, Year Founded: 2010, One Sentence Pitch: Easily give and request low-quality feedback with your team to achieve more together, University Affiliation(s): Duke$ Raised: $240,000, Investors: Friends & family, Traction to Date: 10% of monthly active users (, MA, U) are also active weekly
Company Overview, Company: Grub, Sector: Software, Year Founded: 2018, One Sentence Pitch: Find food you like, University Affiliation(s): Stanford$ Raised: $340,000, Investors: Friends & family, Traction to Date: 40% of monthly active users (, MA, U) are also active weekly

I believe you can use the following Python code.
rgx = r'(?<=[a-z\d])([A-Z$][A-Za-z]*(?: +\S+?)*)*:'
rep = r', \1:'
re.sub(rgx, rep, s)
where s is the string.
Start your engine! | Python code
Python's regex engine performs the following operations when matching.
(?<= : begin positive lookbehind
[a-z\d] : match a letter or digit
) : end positive lookbehind
( : begin capture group 1
[A-Z$] : match a capital letter or '$'
[A-Za-z]* : match 0+ letters
(?: +\S+?) : match 1+ spaces greedily, 1+ non-spaces
non-greedily in a non-capture group
* : execute non-capture group 0+ times
) : end capture group
: : match ':'
Note that the positive lookbehind and permissible characters for each token in the capture group may need to be adjusted to suit requirements.
The regular expression employed to construct replacement strings (, \1:) creates the string ', ' followed by the contents of capture group 1 followed by a colon.

If your list of lists is arbitrary deep, you can recursively traverse it and process (with THIS regex) the strings and yield the same structure:
import re
from collections.abc import Iterable
def process(l):
for el in l:
if isinstance(el, Iterable) and not isinstance(el, (str, bytes)):
yield type(el)(process(el))
else:
yield ', '.join(re.split(r'(?<=[a-z])(?=[A-Z])', el))
Given your example as LoL here is the result:
>>> list(process(LoL))
[['Company Overview, Company: How, Sector: Software, Year Founded: 2010One Sentence Pitch: Easily give and request low-quality feedback with your team to achieve more together, University Affiliation(s): Duke$ Raised: $240,000Investors: Friends & family, Traction to Date: 10% of monthly active users (MAU) are also active weekly'], [['Company Overview, Company: Grub, Sector: Software, Year Founded: 2018One Sentence Pitch: Find food you like, University Affiliation(s): Stanford$ Raised: $340,000Investors: Friends & family, Traction to Date: 40% of monthly active users (MAU) are also active weekly']]]

Regex replace in Python picking a specific substring

Here's what I want to happen:
input = "asdsad,200200-12964,0009,""TREASURY SETTLEMENT NON-COMPLIANCE ASSESSMENT FOR CPD2020-01-21 USD 589,037.17"" 0.00000000,1.000000"
output = "asdsad,200200-12964,0009,""TREASURY SETTLEMENT NON-COMPLIANCE ASSESSMENT FOR CPD2020-01-21 USD 589.037.17"" 0.00000000,1.000000"
How can I change the comma (,) to a dot (.) between ""...589,037.17..."" in Python using regex.
Extra: 589,037.17 => 589.037.17
I tried:
print(re.sub(r'(?<=\d),', '.', input))
But my output was:
output = "asdsad,200200-12964.0009,""TREASURY SETTLEMENT NON-COMPLIANCE ASSESSMENT FOR CPD2020-01-21 USD 589.037.17"" 0.00000000,1.000000"

First, don't call a variable input, because it overwrites the the built-in function input(). Also you repeated strings are just one string in Python.
i = 'asdsad,200200-12964,0009,TREASURY SETTLEMENT NON-COMPLIANCE ASSESSMENT FOR CPD2020-01-21 USD 589,037.17 0.00000000,1.000000'
To solve your specific case, you could match a the country code followed by 3 numbers in the first bit of the price before the comma. That works for this, but probably isn't generic enough for any country code and any price, as look-behinds must be of fixed width.
print(re.sub(r'(?<=USD \d{3}),', '.', i))
I would use a look-behind for the country code and space, then group the first bit of the number and replace it with a backreference.
print(re.sub(r'(?<=[A-Z]{3} )(\d+),', r'\1.', i))

import re
input = "asdsad,200200-12964,0009,""TREASURY SETTLEMENT NON-COMPLIANCE ASSESSMENT FOR CPD2020-01-21 USD 589,037.17"" 0.00000000,1.000000"
print(input)
print(re.sub(r'USD (\d+),(\d+)', r'USD \1.\2', input))
Output:
asdsad,200200-12964,0009,TREASURY SETTLEMENT NON-COMPLIANCE ASSESSMENT FOR CPD2020-01-21 USD 589,037.17 0.00000000,1.000000
asdsad,200200-12964,0009,TREASURY SETTLEMENT NON-COMPLIANCE ASSESSMENT FOR CPD2020-01-21 USD 589.037.17 0.00000000,1.000000
You can go through this Search and Replace and this link for documenation on this.

How to remove html commands

I have a .htm document. By using text_content(), I extracted the text from the document.
Here is the text:
'PART II \xa0\r\nThe Company\x92s common stock is traded on the over-the-counter market and is quoted on the NASDAQ Global Select Market under the symbol\r\nAAPL and on the Frankfurt Stock Exchange under the symbol APCD. Price Range of Common Stock The price range per share of common stock presented below represents the highest and lowest sales prices for the Company\x92s common stock\r\non the NASDAQ Global Select Market during each quarter of the two most recent years. \xa0\r\nHolders As of October\xa016, 2009, there were 30,573 shareholders of record. Dividends\r\n The Company did not declare or pay cash dividends in either 2009 or 2008. The Company anticipates that for the foreseeable\r\nfuture it will retain any earnings for use in the operation of its business. Purchases of Equity Securities by the Issuer and Affiliated\r\nPurchasers None. \xa0\r\n 33 '
With this text, I need to remove a heading which is preceded and followed by a blank line. Thus, lines of the following form should be removed:
\n
some text here\n
\n
I have a code that does that for .txt version of document. However, from .htm document, I realized that some weird things like \xa0\r\n are used to make words capital (for example). Is there any way to remove all of these things and correctly remove only the headings?
Here is the function that does remove the heading:
def clean_text_passage(a_text_string):
"""REMOVE /n: take a list of strings (some passage of text)
and remove noise which is defined as lines that are preceded
by a blank line and followed by a blank line that is lines of
this form will not be in the output
\n
some text here\n
\n
"""
new_passage=[]
p=[line+'\n' for line in a_text_string.split('\n')]
passage = [w.lower().replace('</b>\n', '\n') for w in p]
if len(passage[0].strip())>0:
if len(passage[1].strip())>0:
new_passage.append(passage[0])
for counter, text_line in enumerate(passage[:-1]):
len_line_before=len(passage[counter-1].strip())
len_line_after=len(passage[counter+1].strip())
if len_line_before==len_line_after==0:
continue
if len(text_line.strip())!=0:
new_passage.append(text_line)
if len(passage[-2].strip())!=0:
if len(passage[-1].strip())!=0:
new_passage.append(passage[-1])
return new_passage
I guess the key is to identify the heading in htm document.
Thank you so much for your time and help.