我有一个代码的问题,不知道如何继续前进。
tweet = "I am tired! I like fruit...and milk"
clean_words = tweet.translate(None, ",.;@#?!&$")
words = clean_words.split()
print tweet
print words输出:
['I', 'am', 'tired', 'I', 'like', 'fruitand', 'milk']我想要的是用空格替换标点符号,但不知道什么函数或循环使用。有谁能帮帮我吗?
发布于 2017-05-12 18:05:16
通过像这样改变你的"maketrans“很容易实现:
import string
tweet = "I am tired! I like fruit...and milk"
translator = string.maketrans(string.punctuation, ' '*len(string.punctuation)) #map punctuation to space
print(tweet.translate(translator))它可以在我的运行python 3.5.2和2.x的机器上运行。希望它也能在你的应用中起作用。
发布于 2016-01-21 19:35:52
这是一个基于正则表达式的解决方案,已经在Python 3.5.1下进行了测试。我认为它既简单又简洁。
import re
tweet = "I am tired! I like fruit...and milk"
clean = re.sub(r"""
[,.;@#?!&$]+ # Accept one or more copies of punctuation
\ * # plus zero or more copies of a space,
""",
" ", # and replace it with a single space
tweet, flags=re.VERBOSE)
print(tweet + "\n" + clean)结果:
I am tired! I like fruit...and milk
I am tired I like fruit and milk紧凑版本:
tweet = "I am tired! I like fruit...and milk"
clean = re.sub(r"[,.;@#?!&$]+\ *", " ", tweet)
print(tweet + "\n" + clean)发布于 2016-01-21 01:34:09
有几种方法可以解决这个问题。我有一个可以工作的,但我认为它是次优的。希望更了解正则表达式的人会出现并改进答案或提供更好的答案。
您的问题被标记为python-3.x,但是您的代码是python 2.x,所以我的代码也是2.x。我包含了一个在3.x中工作的版本。
#!/usr/bin/env python
import re
tweet = "I am tired! I like fruit...and milk"
# print tweet
clean_words = tweet.translate(None, ",.;@#?!&$") # Python 2
# clean_words = tweet.translate(",.;@#?!&$") # Python 3
print(clean_words) # Does not handle fruit...and
regex_sub = re.sub(r"[,.;@#?!&$]+", ' ', tweet) # + means match one or more
print(regex_sub) # extra space between tired and I
regex_sub = re.sub(r"\s+", ' ', regex_sub) # Replaces any number of spaces with one space
print(regex_sub) # looks goodhttps://stackoverflow.com/questions/34860982
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