Remove substring only at the end of string

I have a bunch of strings, some of them have ' rec'. I want to remove that only if those are the last 4 characters.

So in other words I have

somestring = 'this is some string rec'

and I want it to become

somestring = 'this is some string'

What is the Python way to approach this?

def rchop(s, suffix):
    if suffix and s.endswith(suffix):
        return s[:-len(suffix)]
    return s

somestring = 'this is some string rec'
rchop(somestring, ' rec')  # returns 'this is some string'

由于len(trailing)无论如何都必须得到（如果trailing要删除的字符串，要删除的字符串在哪里），我建议避免.endswith这种情况下可能引起的工作重复。当然，代码的证明是在时间上，因此，让我们做一些测量（在受访者提出功能后命名它们）：

import re

astring = 'this is some string rec'
trailing = ' rec'

def andrew(astring=astring, trailing=trailing):
    regex = r'(.*)%s$' % re.escape(trailing)
    return re.sub(regex, r'\1', astring)

def jack0(astring=astring, trailing=trailing):
    if astring.endswith(trailing):
        return astring[:-len(trailing)]
    return astring

def jack1(astring=astring, trailing=trailing):
    regex = r'%s$' % re.escape(trailing)
    return re.sub(regex, '', astring)

def alex(astring=astring, trailing=trailing):
    thelen = len(trailing)
    if astring[-thelen:] == trailing:
        return astring[:-thelen]
    return astring

假设我们已经命名了这个python文件a.py，它在当前目录中；现在，...：

$ python2.6 -mtimeit -s'import a' 'a.andrew()'
100000 loops, best of 3: 19 usec per loop
$ python2.6 -mtimeit -s'import a' 'a.jack0()'
1000000 loops, best of 3: 0.564 usec per loop
$ python2.6 -mtimeit -s'import a' 'a.jack1()'
100000 loops, best of 3: 9.83 usec per loop
$ python2.6 -mtimeit -s'import a' 'a.alex()'
1000000 loops, best of 3: 0.479 usec per loop

如您所见，基于RE的解决方案被“绝望地淘汰了”（当一个“过度杀伤”一个问题时经常发生-这可能是RE在Python社区中表现不佳的原因之一！），尽管@Jack的评论比@Andrew的原始评论好得多。正如预期的那样，基于字符串的解决方案endswith令人生畏，我的-避免方案比@Jack的解决方案具有微不足道的优势（仅提高了15％）。因此，这两种纯字符串的想法都是好的（以及简洁明了的想法）-我更喜欢我的变体，仅是因为我是一个节俭的人（有人可能会说小气；-）。 。 “浪费不可”！-）

如果速度不重要，请使用正则表达式：

import re

somestring='this is some string rec'

somestring = re.sub(' rec$', '', somestring)

这是杰克·凯利（Jack Kelly）的答案及其同级的单线版本：

def rchop(s, sub):
    return s[:-len(sub)] if s.endswith(sub) else s

def lchop(s, sub):
    return s[len(sub):] if s.startswith(sub) else s

从开始Python 3.9，您可以使用removesuffix：

'this is some string rec'.removesuffix(' rec')
# 'this is some string'

您也可以使用正则表达式：

from re import sub

str = r"this is some string rec"
regex = r"(.*)\srec$"
print sub(regex, r"\1", str)

作为一种班轮发电机的加入：

test = """somestring='this is some string rec'
this is some string in the end word rec
This has not the word."""
match = 'rec'
print('\n'.join((line[:-len(match)] if line.endswith(match) else line)
      for line in test.splitlines()))
""" Output:
somestring='this is some string rec'
this is some string in the end word 
This has not the word.
"""

使用more_itertools，我们可以rstrip传递通过谓词的字符串。

安装

> pip install more_itertools

码

import more_itertools as mit


iterable = "this is some string rec".split()
" ".join(mit.rstrip(iterable, pred=lambda x: x in {"rec", " "}))
# 'this is some string'

" ".join(mit.rstrip(iterable, pred=lambda x: x in {"rec", " "}))
# 'this is some string'

在这里，我们传递希望从末尾剥离的所有尾随项​​目。

另请参阅more_itertools文档以了解详细信息。

借鉴@ David Foster的灵感，我会做

def _remove_suffix(text, suffix):
    if text is not None and suffix is not None:
        return text[:-len(suffix)] if text.endswith(suffix) else text
    else:
        return text

参考：Python字符串切片

def remove_trailing_string(content, trailing):
    """
    Strip trailing component `trailing` from `content` if it exists.
    """
    if content.endswith(trailing) and content != trailing:
        return content[:-len(trailing)]
    return content

采用：

somestring.rsplit(' rec')[0]