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[英]Remove a tuple from a list if the tuple contains a certain element by given index
[英]How to remove tuple from the list if the tuple contains punctuation mark
我有元组列表:
tupl = [('0', 'Hey'),('1', ','),('2', 'I'), ('3', 'feel'),('4', 'you'), ('5', '!')]
我想删除任何包含标点符号的元组。
我已经尝试使用以下代码,但是它适用于'!'
,只是因为我不知道如何在此代码中添加多个条件。
out_tup = [i for i in tupl if '!' not in i]
print out_tup
如何删除所有包含标点符号(例如','
)的元组?
使用any
例如:
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 [",", "!"])])
if '!' not in i
我们可以更改条件if '!' not in i
if '!' not in i
, 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)
加and ',' not in i
完整代码如下:
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)
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')]
哪里
r"[!,]"
可以使用不需要的标点符号进行扩展
您可以使用此:
out_tup = [i for i in tupl if not i[1] in [',','!']]
使用正则表达式:
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)))
输出:
[('0', 'Hey'), ('2', 'I'), ('3', 'feel'), ('4', 'you')]
在Rextester上的演示。
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