Python 用空格替换标点符号
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/34860982/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
replace the punctuation with whitespace
提问by oceano22
I have a problem with the code and can not figure out how to move forward.
我的代码有问题，无法弄清楚如何继续前进。
tweet = "I am tired! I like fruit...and milk"
clean_words = tweet.translate(None, ",.;@#?!&$")
words = clean_words.split()
print tweet
print words
Output:
输出：
['I', 'am', 'tired', 'I', 'like', 'fruitand', 'milk']
What I would like is to replace the punctuation with white space but do not know what function or cycle use. Can anyone help me please?
我想要的是用空格替换标点符号但不知道使用什么功能或循环。有人可以帮我吗？
回答by Bryan
There are a few ways to approach this problem. I have one that works, but believe it is suboptimal. Hopefully someone who knows regex better will come along and improve the answer or offer a better one.
有几种方法可以解决这个问题。我有一个有效的，但相信它是次优的。希望更了解正则表达式的人会出现并改进答案或提供更好的答案。
Your question is labeled python-3.x, but your code is python 2.x, so my code is 2.x as well. I include a version that works in 3.x.
你的问题被标记为python-3.x，但你的代码是python 2.x，所以我的代码也是2.x。我包括一个适用于 3.x 的版本。
#!/usr/bin/env python
import re
tweet = "I am tired! I like fruit...and milk"
# print tweet
clean_words = tweet.translate(None, ",.;@#?!&$") # 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 good
回答by Thomas Baruchel
I am not sure I fully understand your requirements, but did you consider adding only one more line to your current code like:
我不确定我是否完全理解您的要求，但是您是否考虑过在当前代码中再添加一行，例如：
>>> a=['I', 'am', 'tired', 'I', 'like', 'fruitand', 'milk']
>>> " ".join(a)
'I am tired I like fruitand milk'
Is it what you are asking or do you need something more specific? Regards.
这是您要问的还是您需要更具体的东西？问候。
回答by pivanchy
If you're using Python 2.x you could try:
如果您使用的是 Python 2.x，您可以尝试：
import string
tweet = "I am tired! I like fruit...and milk"
clean_words = tweet.translate(string.maketrans("",""), string.punctuation)
print clean_words
For Python 3.x it works:
对于 Python 3.x，它有效：
import string
tweet = "I am tired! I like fruit...and milk"
transtable = str.maketrans('', '', string.punctuation)
clean_words = tweet.translate(transtable)
print(clean_words)
These parts of code removes all the punctuation symbols from string.
这些代码部分从字符串中删除了所有标点符号。
回答by Jonathan
Here is a regex based solution that has been tested under Python 3.5.1. I think it is both simple and succinct.
这是一个基于正则表达式的解决方案，已在 Python 3.5.1 下测试过。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"""
[,.;@#?!&$]+ # Accept one or more copies of punctuation
\ * # plus zero or more copies of a space,
""",
" ", # and replace it with a single space
tweet, flags=re.VERBOSE)
print(tweet + "\n" + clean)
Results:
结果：
I am tired! I like fruit...and milk
I am tired I like fruit and milk
Compact version:
精简版：
tweet = "I am tired! I like fruit...and milk"
clean = re.sub(r"[,.;@#?!&$]+\ *", " ", tweet)
print(tweet + "\n" + clean)
回答by YuanzhiKe
It is easy to achieve by changing your "maketrans" like this:
通过像这样更改“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))
It works on my machine running python 3.5.2 and 2.x. Hope that it works on yours too.
它适用于我运行 python 3.5.2 和 2.x 的机器。希望它也适用于你的。
