replace Regex expression with the match [duplicate] - python

This question already has answers here:
How to use python regex to replace using captured group? [duplicate]
(4 answers)
Closed 3 years ago.
I have the following string: message = 'hi <#ABC> and <#DEF>',
And the following Regex: exp = '<#(.*?)>', so that re.findall(exp, message) outputs ['ABC', 'DEF']. How can I replace the original message matches with those outputs so I get 'hi ABC and DEF'?


import re
line = re.sub(
r"<#(.*?)>",
r"\1",
line
)

Related

about Regex in Python [duplicate]

This question already has answers here:
Python extract pattern matches
(10 answers)
Closed 1 year ago.
I made this code:
import re
match = re.search(r'[DER]\d+[Y]', 'DER1234Y' )
print(match.group())
and it prints this :
R1234Y
I want the code to only print the numbers and nothing else. How to do that ?

It's basically regex. So would this work?: re.sub('[^0-9]+', '', 'DER1234Y')
[^0-9]+ = everything that is not a numeric value (0-9).

Using regex to match two specific characters [duplicate]

This question already has answers here:
Regular Expressions: Is there an AND operator?
(14 answers)
Closed 2 years ago.
I have a list of strings like:
1,-102a
1,123-f
1943dsa
-da238,
-,dwjqi92
How can I make a Regex expression in Python that matches as long as the string contains the characters , AND - regardless of the order or the pattern in which they appear?

I would use the following regex alternation:
,.*-|-.*,
Sample script:
inp = ['1,-102a', '1,123-f', '1943dsa', '-da238,', '-,dwjqi92']
output = [x for x in inp if re.search(r',.*-|-.*,', x)]
print(output)
This prints:
['1,-102a', '1,123-f', '-da238,', '-,dwjqi92']

How to delete certain words from string without spaces? [duplicate]

This question already has answers here:
How to remove substring from string in Python 3
(2 answers)
Closed 2 years ago.
Is there I way to delete words from a string in Python if it doesn't have spaces. For example, if you have the string "WUBHELLOWUB" I want to remove "WUB". I tried
s = 'WUBHELLOWUB'
while 'WUB' in s:
ind = s.find('WUB')
s = s[:ind] + s[ind+1:]
print(s)
but it did not work.

You can use regex
import re
data=r"\S*WUB\S*"
re.sub(data, '','WUBWUBHELLO')

How can I split a string maintaining the punctuation? (Python) [duplicate]

This question already has answers here:
Splitting a string into words and punctuation
(11 answers)
Closed 3 years ago.
How can I split a string in python taking into account the punctuation in the result?
The following code:
s = "Hello, my name is Robert."
s_splitted = s.split()
will give as output:
["Hello,","my","name","is","Robert."]
How can I obtain the following result?
["Hello",",","my","name","is","Robert","."]

Regex can handle this.
import re
s = "Hello, my name is Robert."
s_splitted = [part for part in re.split(r'\b|\s', s) if part != '']
# ['Hello', ',', 'my', 'name', 'is', 'Robert']

Does this answer your question?
So in your case:
import re
s = "Hello, my name is Robert."
items = re.findall(r"[\w']+|[.,!?;]", s)

Splitting on a lookahead [duplicate]

This question already has answers here:
Decode HTML entities in Python string?
(6 answers)
Closed 6 years ago.
I'm trying to split on a lookahead, but it doesn't work for the last occurrence. How do I do this?
my_str = 'HRC’s'
import re
print(re.split(r'.(?=&)', my_str))
My output:
['HR', '&#226', '&#128', '™s']
My desired output:
['HRC', '&#226', '&#128', '&#153', 's']

The solution using re.findall() function:
my_str = 'HRC’s'
result = re.findall(r'\w+|&#\d+(?=;)', my_str)
print(result)
The output:
['HRC', '&#226', '&#128', '&#153', 's']

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