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I have a string "hello\n1hello123\n2yahoo". Want to split it with \n[integer value]

I have a string in python.
str1 = "hello\n1hello123\n2yahoo"
I would like to split this with \n[integer value] to get a list that looks like:
[hello, hello123, yahoo]
Can anyone please help?

As someone who goes too far in avoiding regular expressions (avoiding them whenever possible rather than simply avoiding them when they are inappropriate), I would lean towards splitting on \n and processing the resulting list element by element:
from string import digits
result = [x.lstrip(digits) for x in str1.split("\n")]
If you are less regex-averse than I, and as recommended in the comments,
from re import split
from string import digits
results = split(f'\n[{digits}]*', str1)

regular expression package
enter code here
str1 = "hello\n1hello123\n2yahoo"
import re
print(re.split(r"\n[1-9]", str1))

How to get all the string after and before two specific words?

I want to replace all the string after "my;encoded;image:" (which is the base64 data of the image) and i want to stop before the word "END" , but the following code is replacing also the two strings "my;encoded;image:" and "END". Any suggestions?
import re
re.sub("my;encoded;image:.*END","random_words",image,flags=re.DOTALL)
NB : a simple way could be to use replacement but i want to use regex in my case Thanks

You can use a non-greedy regex to split the string into three groups. Then replace the second group with your string:
import re
x = re.sub(r'(.*my;encoded;image:)(.*?)(END.*)', r"\1my string\3", image)
print(x)
You can use f-strings with Python 3.6 and higher:
replacement = "hello"
x = re.sub(r'(.*my;encoded;image:)(.*?)(END.*)', fr'\1{replacement}\3', image)

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'

Separating Strings and other values with comma as a delimiter

I'm working with a project, where there will be variable holding any data types just separated with a comma.
I need to separate all these things and I also need to define which type it is.
For e.g:
data='"Hello, Hey",123,10.04'
I used split() function to separate, but it separates the comma even within "Hello,Hey", outputing:
['"Hello','Hey"','123','10'.'04']
I don't need it like this, all i need is to separate the values by commas but not the ones inside other quotes. The output should be like this:
['"Hello, Hey"','123','10.04']
I killed my brain, but it is still a problem for me. Because I'm a beginner.
Thanks in Advance

I'm struggling to understand your question - it seems you have a string with data inside the string, separated by commas:
data='"Hello, Hey",123,10.04'
You can use the shlex module to split it respecting the quotes
>>> import shlex
>>> s = shlex.shlex(data)
>>> s.whitespace = ','
>>> s.wordchars += '.'
>>> print(list(s))
['"Hello, Hey"', '123', '10.04']

You may use the re module like so:
[m.group(1) or m.group(2) for m in re.finditer(r'"([^"]*)",?|([^,]*),?', '"Hello, Hey",123,10.04')]

You can use re.findall with regex pattern "[^"]+"|[^,]+:
import re
print(re.findall(r'"[^"]+"|[^,]+', '"Hello, Hey",123,10.04'))
This outputs:
['"Hello, Hey"', '123', '10.04']

Just use the shlex module
import shlex
data = '"Hello, Hey",123,10.04'
data = shlex.split(data)
print(data)
Output:
["Hello, Hey", "123" , "10.04"]

You can use re.split to split on a combination of either a double quote before a comma or a comma followed by a digit
import re
data='"Hello, Hey",123,10.04'
re.split(r'(?<="),|,(?=\d)', data)
['"Hello, Hey"', '123', '10.04']

Print only alphabetics in a string using Regular Expression

Goal : i want only alphabets to be printed in a string
#Input
#======
string = ' 529Wind3#. '
#Neededoutput
#============
'Wind'
I tried coding for this using the below code
import re
string=re.sub('[^a-z]+[^A-Z]',' ',string)
print(string)
The output i'm getting is
ind
But this code only applies for lowercase
Can you please tell me how to write code for both upper and lowercase

Try using a list comprehension to check if each character is in string.ascii_letters or not, if it is, it will be stored:
import string
String = ' 529Wind3#. '
print(''.join([i for i in String if i in string.ascii_letters]))
Output:
Wind

I agree with #U8-Forward's point but I think you may also want to know why your regular expression isn't working. This
[^a-z]+[^A-Z]
doesn't do what you want because W matches [^a-z]+ and so gets removed.
Put all of the characters you don't want in a single character class:
[^a-zA-Z]+

You need to write [^a-zA-Z] instead of [^a-z]+[^A-Z]. The + operator is for detecting repetitive characters and not to combine multiple conditions.
Try the below code for your requirement:
import re
string=re.sub('[^a-zA-Z]',' ',string)
print(string)

you can use re.findall
import re
String = ' 529Wind3#. '
string = re.findall('[a-zA-Z]+', String)
print(''.join(string))

print re.sub('[^a-zA-Z]','',string)