繁体   English   中英

Python-如何通过空格将标点符号与单词分开,在标点符号和单词之间仅留一个空格?

[英]Python - How do I separate punctuation from words by white space leaving only one space between the punctuation and the word?

我有以下字符串:

input = "I love programming with Python-3.3! Do you? It's great... I give it a 10/10. It's free-to-use, no $$$ involved!"

所有标点符号都应与单词“ /”,“'',“-”,“ +”和“ $”除外。

因此输出应为:

"I love programming with Python-3 . 3 ! Do you ? It's great . . . I give it a 10/10. It's free-to-use , no $$$ involved !"

我使用以下代码:

for x in string.punctuation:
    if x == "/":
        continue
    if x == "'":
        continue
    if x == "-":
        continue
    if x == "+":
        continue
    if x == "$":
        continue
    input = input.replace(x," %s " % x)

我得到以下输出:

I love programming with Python-3 . 3 !  Do you ?  It's great .  .  .  I give it a 10/10 .  It's free-to-use ,  no $$$ involved ! 

它可以工作,但问题是它有时在标点符号和单词之间留下两个空格,例如句子中的第一个感叹号和单词“ Do”之间。 这是因为它们之间已经存在空间。

使用以下命令也会发生此问题:input =“ Hello。(hi)”。 输出为:

" Hello .  ( hi ) "

注意左方括号前的两个空格。

除了上面提到的5个标点符号与单词没有分隔之外,我需要在任何标点符号和单词之间仅留一个空格的输出。 我怎样才能解决这个问题? 还是有更好的方法使用正则表达式来做到这一点?

提前致谢。

看起来re可以为您做...

>>> import re
>>> re.sub(r"([\w/'+$\s-]+|[^\w/'+$\s-]+)\s*", r"\1 ", input)
"I love programming with Python-3 . 3 ! Do you ? It's great ... I give it a 10/10 . It's free-    to-use , no $$$ involved ! "

>>> re.sub(r"([\w/'+$\s-]+|[^\w/'+$\s-]+)\s*", r"\1 ", "Hello. (hi)")
'Hello . ( hi ) '

如果尾随空格有问题, .rtrim(theresult, ' ')应该为您解决:-)

我可以这样尝试吗:

>>> import string
>>> input = "I love programming with Python-3.3! Do you? It's great... I give it a 10/10. It's free-to-use, no $$$ involved!"
>>> ls = []
>>> for x in input:
...     if x in string.punctuation:
...         ls.append(' %s' % x)
...     else:
...         ls.append(x)
...
>>> ''.join(ls)
"I love programming with Python -3 .3 ! Do you ? It 's great . . . I give it a 10 /10 . It 's free -to -use , no  $ $ $ involved !"
>>>

由于缺乏声誉而无法发表评论,但是在这种情况下,这里

在句子中的第一个感叹号和单词“ Do”之间

好像有两个空格,因为它们之间已经有一个空格了! 和做

因此,如果标点符号后已经有一个空格,请不要再放置其他空格。

另外,这里也有类似的问题: python regex在标点和字母之间插入空格

因此,也许考虑使用re

在我看来,否定的字符类更简单:

import re

input_string = "I love programming with Python-3.3! Do you? It's great... I give it a 10/10. It's free-to-use, no $$$ involved!"

print re.sub(r"\s?([^\w\s'/\-\+$]+)\s?", r" \1 ", input_string)

输出:

I love programming with Python-3 . 3 ! Do you ? It's great ... I give it a 10/10 . It's free-to-use , no $$$ involved ! 
# Approach 1

import re

sample_input = "I love programming with Python-3.3! Do you? It's great... I give it a 10/10. It's free-to-use, no $$$ involved!"

sample_input = re.sub(r"([^\s])([^\w\/'+$\s-])", r'\1 \2', sample_input)
print(re.sub(r"([^\w\/'+$\s-])([^\s])", r'\1 \2', sample_input))

# Approach 2

import string

sample_input = "I love programming with Python-3.3! Do you? It's great... I give it a 10/10. It's free-to-use, no $$$ involved!"

punctuation = string.punctuation.replace('/', '').replace("'", '') \
        .replace('-', '').replace('+', '').replace('$', '')

i = 0

while i < len(sample_input):
    if sample_input[i] not in punctuation:
        i += 1
        continue

    if i > 0 and sample_input[i-1] != ' ':
        sample_input = sample_input[:i] + ' ' + sample_input[i:]
        i += 1

    if i + 1 < len(sample_input) and sample_input[i+1] != ' ':
        sample_input = sample_input[:i+1] + ' ' + sample_input[i+1:]
        i += 1

    i += 1

print(sample_input)

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM