语句未在python3中实现

Question

I have tried to remove the punctuation and the numeric values from the string. I have tried the following statement:

name_check = ''.join([letter for letter in name_check if letter not in string.punctuation or not letter.isdigit()])

when I give input as tom.alter99999 , it returns the exactly same value, whereas I expect to give the value as tomalter .

Just me what I need to change in the statement so that I get the desired output.

Answer 1

You used an or conditional, which means that it'll always be true that any given letter is either not punctuation or not a digit.

Changing your conditional to an and will work.

name_check = ''.join([letter for letter in name_check
                      if letter not in string.punctuation
                      and not letter.isdigit()])

You can also use str.maketrans and transliterate your string, replacing all punctuation and digits in a string with nothing.

import string

translator = str.maketrans('', '', string.punctuation + string.digits)

print('tom.alter99999'.translate(translator))

Answer 2

您需要and or

name_check = ''.join([letter for letter in name_check if letter not in string.punctuation or not letter.isdigit()])

Answer 3

You could use the re library

import re
name='tom.alter99999'
res = re.findall(r'[a-zA-Z]',name)
print("".join(res))

Answer 4

The statement I solved with or condition too. Here is what I did:

name_check = ''.join([letter for letter in name_check if not (letter in string.punctuation or letter.isdigit())])

I got the output as: tomalter