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" 

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

Python开发人员的解决方案:

import re

text1 = 'Python      Exercises    Are   Challenging Exercises'
print("Original string: ", text1)
print("Without extra spaces: ", re.sub(' +', ' ', text1))

输出: 原始字符串:Python练习是具有挑战性的练习 没有额外的空格:Python练习是具有挑战性的练习

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'

你能得到的用户生成字符串的最快速度是:

if '  ' in text:
    while '  ' in text:
        text = text.replace('  ', ' ')

短路使它比pythonlarry的综合回答略快。如果你追求效率，并严格要求去除单个空格的额外空白，那么可以使用这种方法。

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

or

re.sub("\s\s+", " ", s)

正如用户Martin Thoma在评论中提到的，在PEP 8中，逗号前的空格被列为令人讨厌的地方。