How to remove tuple from the list if the tuple contains punctuation mark

Question

I have list of tuple:

tupl = [('0', 'Hey'),('1', ','),('2', 'I'), ('3', 'feel'),('4', 'you'), ('5', '!')]

I want to remove any tuple containing a punctuation mark.

I already tried with the following code, but it is working for '!' , only because I don't know how to put multiple condition in this code.

out_tup = [i for i in tupl if '!' not in i]
print out_tup

How can I remove all tuples containing punctuation marks (eg ',' )?

Answer 1

Using any

Ex:

import string

tupl = [('0', 'Hey'),('1', ','),('2', 'I'), ('3', 'feel'),('4', 'you'), ('5', '!')]
print([i for i in tupl if not any(p in i for p in string.punctuation)])

#or
print([i for i in tupl if not any(p in i for p in [",", "!"])])

Answer 2

We can change the condition if '!' not in i if '!' not in i to if '!' not in i and ',' not in i if '!' not in i and ',' not in i .

tupl = [('0', 'Hey'),('1', ','),('2', 'I'), ('3', 'feel'),('4', 'you'), ('5', '!')]
out_tup = [i for i in tupl if '!' not in i and ',' not in i]
print(out_tup)

Answer 3

Add and ',' not in i

Full code is below:

tupl = [('0', 'Hey'),('1', ','),('2', 'I'), ('3', 'feel'),('4', 'you'), ('5', '!')]
out_tup = [i for i in tupl if '!' not in i and ',' not in i]
print(out_tup)

Answer 4

import re

tupl = [('0', 'Hey'),('1', ','),('2', 'I'), ('3', 'feel'),('4', 'you'), ('5', '!')]
out_tup = [i for i in tupl if not re.search(r"[!,]", i[1])]
print(out_tup) # [('0', 'Hey'), ('2', 'I'), ('3', 'feel'), ('4', 'you')]

where

r"[!,]"

can be expanded with the punctuations you don't want

Answer 5

您可以使用此：

out_tup = [i for i in tupl if not i[1] in [',','!']]

Answer 6

Using the regular expression:

import re, string

tupl = [('0', 'Hey'), ('1', ','), ('2', 'I'), ('3', 'feel'), ('4', 'you'), ('5', '!')]
rx = re.compile('[{}]'.format(re.escape(string.punctuation)))
print(list(filter(lambda t: not rx.search(t[1]), tupl)))

Output:

[('0', 'Hey'), ('2', 'I'), ('3', 'feel'), ('4', 'you')]

Demo on Rextester .