一个简单的灵魂

>>> import re
>>> s="The   fox jumped   over    the log."
>>> print re.sub('\s+',' ', s)
The fox jumped over the log.

一个简单的灵魂

>>> import re
>>> s="The   fox jumped   over    the log."
>>> print re.sub('\s+',' ', s)
The fox jumped over the log.

import re

Text = " You can select below trims for removing white space!!   BR Aliakbar     "
  # trims all white spaces
print('Remove all space:',re.sub(r"\s+", "", Text), sep='') 
# trims left space
print('Remove leading space:', re.sub(r"^\s+", "", Text), sep='') 
# trims right space
print('Remove trailing spaces:', re.sub(r"\s+$", "", Text), sep='')  
# trims both
print('Remove leading and trailing spaces:', re.sub(r"^\s+|\s+$", "", Text), sep='')
# replace more than one white space in the string with one white space
print('Remove more than one space:',re.sub(' +', ' ',Text), sep='') 

结果:作为代码

"Remove all space:Youcanselectbelowtrimsforremovingwhitespace!!BRAliakbar"
"Remove leading space:You can select below trims for removing white space!!   BR Aliakbar"     
"Remove trailing spaces: You can select below trims for removing white space!!   BR Aliakbar"
"Remove leading and trailing spaces:You can select below trims for removing white space!!   BR Aliakbar"
"Remove more than one space: You can select below trims for removing white space!! BR Aliakbar" 

类似于前面的解决方案，但更具体:用一个空格替换两个或多个空格:

>>> import re
>>> s = "The   fox jumped   over    the log."
>>> re.sub('\s{2,}', ' ', s)
'The fox jumped over the log.'

" ".join(foo.split())对于所问的问题不太正确，因为它也完全删除了单个前导和/或尾随空格。所以，如果它们也将被1个空白替换，你应该像下面这样做:

" ".join(('*' + foo + '*').split()) [1:-1]

当然，它没有那么优雅。

你也可以在Pandas DataFrame中使用字符串分割技术，而不需要使用.apply(..)，如果你需要对大量字符串快速执行操作，这是非常有用的。这是一行话:

df['message'] = (df['message'].str.split()).str.join(' ')