Develop Reference

Match words using this regex pattern, only if these words do not appear within a list of substrings

import re
input_text = "a áshgdhSdah saasas a corrEr, assasass a saltó sasass a sdssaa" #example
list_verbs_in_this_input = ["serías", "serían", "sería", "ser", "es", "corré", "corrió", "corría", "correr", "saltó", "salta", "salto", "circularías", "circularía", "circulando", "circula", "consiste", "consistían", "consistía", "consistió", "ladró", "ladrando", "ladra", "visualizar", "ver", "vieron", "vió"]
noun_pattern = r"((?:\w+))" # pattern that doesnt tolerate whitespace in middle
imput_text = re.sub(r"(?:^|\s+)a\s+" + noun_pattern,
"\(\g<0>\)",
input_text, re.IGNORECASE)
print(repr(input_text)) # --> output
I need the regex to identify and replace a substring containing no whitespaces in between "((?:\w+))" when it is at the beginning of the line or preceded by "a", "(?:^|\s+)a\s+", only if "((?:\w+))" does not match any of the strings that are inside the list list_verbs_in_this_input or a dot . , using a regex pattern similar to this re.compile(r"(?:" + rf"({'|'.join(list_verbs_in_this_input)})" + r"|[.;\n]|$)", flags = re.IGNORECASE)
And the correct output should look like this:
'(áshgdhSdah) saasas a corrEr, assasass a saltó sasass (sdssaa)'
Note that the substrings "a corrEr" and "a saltó" were not modified, since they contained substring(words) that are in the list_verbs_in_this_input list

To exclude some words, you can use a negative look ahead assertion when at the start of the word you're about to match.
A few things to correct:
re.sub takes the flags as 5th argument, not 4th
"\(" is not an escape sequence, so you should just do "(\g<0>)" without "escaping" the parentheses -- they have no special meaning in that string.
r"(?:^|\s+)a\s+" will always require the a to be there. From your description I understood that the a could be optional when the word is at the start of a line, so r"(?:\ba\s|^)\s*"
In the regex that should match the forbidden words, make sure to require that the word ends right after the match, so add \b in the pattern.
Here is what you could do:
import re
input_text = "a áshgdhSdah saasas a corrEr, assasass a saltó sasass a sdssaa" #example
list_verbs_in_this_input = ["serías", "serían", "sería", "ser", "es", "corré", "corrió", "corría", "correr", "saltó", "salta", "salto", "circularías", "circularía", "circulando", "circula", "consiste", "consistían", "consistía", "consistió", "ladró", "ladrando", "ladra", "visualizar", "ver", "vieron", "vió"]
noun_pattern = r"\w+"
exclude = rf"(?!\b(?:{'|'.join(list_verbs_in_this_input)})\b)"
article = r"(?:\ba\s|^)\s*"
regex = article + exclude + noun_pattern
input_text = re.sub(regex, "(\g<0>)", input_text, flags=re.I|re.U)
print(repr(input_text))

Remove Characters From A String Until A Specific Format is Reached

So I have the following strings and I have been trying to figure out how to manipulate them in such a way that I get a specific format.
string1-itd_jan2021-internal
string2itd_mar2021-space
string3itd_feb2021-internal
string4-itd_mar2021-moon
string5itd_jun2021-internal
string6-itd_feb2021-apollo
I want to be able to get rid of any of the last string so I am just left with the month and year, like below:
string1-itd_jan2021
string2itd_mar2021
string3itd_feb2021
string4-itd_mar2021
string5itd_jun2021
string6-itd_feb2021
I thought about using string.split on the - but then realized that for some strings this wouldn't work. I also thought about getting rid of a set amount of characters by putting it into a list and slicing but the end is varying characters length?
Is there anything I can do it with regex or any other python module?

Use str.rsplit with the appropriate maxsplit parameter:
s = s.rsplit("-", 1)[0]
You could also use str.split (even though this is clearly the worse choice):
s = "-".join(s.split("-")[:-1])
Or using regular expressions:
s = re.sub(r'-[^-]*$', '', s)
# "-[^-]*" a "-" followed by any number of non-"-"

With a regex:
import re
re.sub(r'([0-9]{4}).*$', r'\1', s)

Use re.sub like so:
import re
lines = '''string1-itd_jan2021-internal
string2itd_mar2021-space
string3itd_feb2021-internal
string4-itd_mar2021-moon
string5itd_jun2021-internal
string6-itd_feb2021-apollo'''
for old in lines.split('\n'):
new = re.sub(r'[-][^-]+$', '', old)
print('\t'.join([old, new]))
Prints:
string1-itd_jan2021-internal string1-itd_jan2021
string2itd_mar2021-space string2itd_mar2021
string3itd_feb2021-internal string3itd_feb2021
string4-itd_mar2021-moon string4-itd_mar2021
string5itd_jun2021-internal string5itd_jun2021
string6-itd_feb2021-apollo string6-itd_feb2021
Explanation:
r'[-][^-]+$' : Literal dash (-), followed by any character other than a dash ([^-]) repeated 1 or more times, followed by the end of the string ($).

Replacing when a word is in another word but with special circumstances

My program replaces tokens with values when they are in a file. When reading in a certain line it gets stuck here is an example:
1.1.1.1.1.1.1.1.1.1 Token100.1 1.1.1.1.1.1.1Token100a
The two tokens in the example are Token100 and Token100a. I need a way to only replace Token100 with its data and not replace Token100a with Token100's data with an a afterwards. I can't look for spaces before and after because sometimes they are in the middle of lines. Any thoughts are appreciated. Thanks.

You can use regex:
import re
line = "1.1.1.1.1.1.1.1.1.1 Token100.1 1.1.1.1.1.1.1Token100a"
match = re.sub("Token100a", "data", line)
print(match)
Outputs:
1.1.1.1.1.1.1.1.1.1 Token100.1 1.1.1.1.1.1.1data
More about regex here:
https://www.w3schools.com/python/python_regex.asp

You can use a regular expression with a negative lookahead to ensure that the following character is not an "a":
>>> import re
>>> test = '1.1.1.1.1.1.1.1.1.1 Token100.1 1.1.1.1.1.1.1Token100a'
>>> re.sub(r'Token100(?!a)', 'data', test)
'1.1.1.1.1.1.1.1.1.1 data.1 1.1.1.1.1.1.1Token100a'

Python - Regex to avoid matching duplicates

My string looks like this:
bo_1
bo_1
bo_2
bo_2
bo_3
bo_3
bo_4
bo_4
bo_5
bo_5
bo_6
bo_6
bo_7
bo_7
bo_8
bo_8
bo_9
bo_9
bo_10
bo_10
I want to match the first instance of each digit and ignore the next duplicate line. My regex is as follows:
(bo_\d)(?![\s\S]*\1)
which returns the following:
'bo_2'
'bo_3'
'bo_4'
'bo_5'
'bo_6'
'bo_7'
'bo_8'
'bo_9'
'bo_1'
How would I modify the regex to return a result like this instead (to include 'bo_1' at the start and 'bo_10' at the end):
'bo_1'
'bo_2'
'bo_3'
'bo_4'
'bo_5'
'bo_6'
'bo_7'
'bo_8'
'bo_9'
'bo_10'

Technically you don't need regex for that (you can use set() for instance):
>>> # Assume your string is in the variable called "text"
>>> result = set(text.split('\n'))
>>> result
{'bo_7', 'bo_3', 'bo_1', 'bo_6', 'bo_5', 'bo_8', 'bo_9', 'bo_2', 'bo_4', 'bo_10'}
Anyway, the issue with your regex is that bo_1 is also matching bo_10, so it will be seen as a duplicate by the regex. You can solve it using word boundaries to ensure that the full 'word' is tested for a match:
\b(bo_\d+)\b(?![\s\S]*\b\1\b)
regex101 demo

Use
(bo_\d+$)(?![\s\S]*^\1$)
Since you want to include bo_10, you should use \d+ and not just \d in the initial group. Then, in your negative lookahead, put the backrefrence between start-of-line and end-of-line anchors, so that, for example, bo_1 does not get excluded because it's followed by a bo_10.
https://regex101.com/r/8khbcc/1

Python: strip function definition using regex

I am a very beginner of programming and reading the book "Automate the boring stuff with Python'. In Chapter 7, there is a project practice: the regex version of strip(). My code below does not work (I use Python 3.6.1). Could anyone help?
import re
string = input("Enter a string to strip: ")
strip_chars = input("Enter the characters you want to be stripped: ")
def strip_fn(string, strip_chars):
if strip_chars == '':
blank_start_end_regex = re.compile(r'^(\s)+|(\s)+$')
stripped_string = blank_start_end_regex.sub('', string)
print(stripped_string)
else:
strip_chars_start_end_regex = re.compile(r'^(strip_chars)*|(strip_chars)*$')
stripped_string = strip_chars_start_end_regex.sub('', string)
print(stripped_string)

You can also use re.sub to substitute the characters in the start or end.
Let us say if the char is 'x'
re.sub(r'^x+', "", string)
re.sub(r'x+$', "", string)
The first line as lstrip and the second as rstrip
This just looks simpler.

When using r'^(strip_chars)*|(strip_chars)*$' string literal, the strip_chars is not interpolated, i.e. it is treated as a part of the string. You need to pass it as a variable to the regex. However, just passing it in the current form would result in a "corrupt" regex because (...) in a regex is a grouping construct, while you want to match a single char from the define set of chars stored in the strip_chars variable.
You could just wrap the string with a pair of [ and ] to create a character class, but if the variable contains, say z-a, it would make the resulting pattern invalid. You also need to escape each char to play it safe.
Replace
r'^(strip_chars)*|(strip_chars)*$'
with
r'^[{0}]+|[{0}]+$'.format("".join([re.escape(x) for x in strip_chars]))
I advise to replace * (zero or more occurrences) with + (one or more occurrences) quantifier because in most cases, when we want to remove something, we need to match at least 1 occurrence of the unnecessary string(s).
Also, you may replace r'^(\s)+|(\s)+$' with r'^\s+|\s+$' since the repeated capturing groups will keep on re-writing group values upon each iteration slightly hampering the regex execution.

#! python
# Regex Version of Strip()
import re
def RegexStrip(mainString,charsToBeRemoved=None):
if(charsToBeRemoved!=None):
regex=re.compile(r'[%s]'%charsToBeRemoved)#Interesting TO NOTE
return regex.sub('',mainString)
else:
regex=re.compile(r'^\s+')
regex1=re.compile(r'$\s+')
newString=regex1.sub('',mainString)
newString=regex.sub('',newString)
return newString
Str=' hello3123my43name is antony '
print(RegexStrip(Str))
Maybe this could help, it can be further simplified of course.