假设这个字符串:
The fox jumped over the log.
变成:
The fox jumped over the log.
在不分割和进入列表的情况下,最简单的实现方法(1-2行)是什么?
假设这个字符串:
The fox jumped over the log.
变成:
The fox jumped over the log.
在不分割和进入列表的情况下,最简单的实现方法(1-2行)是什么?
当前回答
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
其他回答
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。
令人惊讶的是,没有人发布一个简单的函数,它会比所有其他发布的解决方案快得多。是这样的:
def compactSpaces(s):
os = ""
for c in s:
if c != " " or (os and os[-1] != " "):
os += c
return os
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())
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
这样做,并将这样做::)
# python... 3.x
import operator
...
# line: line of text
return " ".join(filter(lambda a: operator.is_not(a, ""), line.strip().split(" ")))