>>> import re
>>> re.sub(' +', ' ', 'The     quick brown    fox')
'The quick brown fox'

def unPretty(S):
   # Given a dictionary, JSON, list, float, int, or even a string...
   # return a string stripped of CR, LF replaced by space, with multiple spaces reduced to one.
   return ' '.join(str(S).replace('\n', ' ').replace('\r', '').split())

>>> import re
>>> re.sub(' +', ' ', 'The     quick brown    fox')
'The quick brown fox'

一行代码，删除句子之前、之后和内部所有多余的空格:

sentence = "  The   fox jumped   over    the log.  "
sentence = ' '.join(filter(None,sentence.split(' ')))

解释:

将整个字符串拆分为一个列表。 从列表中过滤空元素。 用一个空格重新连接剩下的元素*

*其余的元素应该是单词或带有标点符号的单词等。我没有对此进行广泛测试，但这应该是一个很好的起点。祝你一切顺利!

在某些情况下，需要将每个空格字符的连续出现替换为该字符的单个实例。你可以使用带有反向引用的正则表达式来实现这一点。

(\s)\1{1，}匹配任何空格字符，后面跟着一个或多个该字符。现在，您所需要做的就是指定第一个组(\1)作为匹配的替换。

将其包装在函数中:

import re

def normalize_whitespace(string):
    return re.sub(r'(\s)\1{1,}', r'\1', string)

>>> normalize_whitespace('The   fox jumped   over    the log.')
'The fox jumped over the log.'
>>> normalize_whitespace('First    line\t\t\t \n\n\nSecond    line')
'First line\t \nSecond line'

另一个选择:

>>> import re
>>> str = 'this is a            string with    multiple spaces and    tabs'
>>> str = re.sub('[ \t]+' , ' ', str)
>>> print str
this is a string with multiple spaces and tabs