Python：还有什么 compact/efficient 方法可以从列表中删除后缀字符串？

Question

是否有一种更紧凑、更有效的方法来从给定列表中去除任何后缀的字符串，即：

sfxs = ['suffix1', 'sfx2', 'suffix333']
s = 'string-to-process-sfx2'
for sfx in sfxs:
    i = s.find(sfx)
    if not i == -1:
        s = s[:i]
        break

后缀长度不同

Answer 1

您可以使用re.sub。

>>> import re
>>> sfxs = ['suffix1', 'sfx2', 'suffix333']
>>> s = 'string-to-process-sfx2'
>>> re.sub(r'(' + '|'.join(sfxs) + r')$', '',s)
'string-to-process-'
>>> re.sub(r'\b(' + '|'.join(sfxs) + r')$', '',s)
'string-to-process-'

>>> re.sub(r'-(' + '|'.join(sfxs) + r')$', '',s)
'string-to-process'

>>> re.sub(r'(' + '|'.join(sorted(sfxs, key=lambda x:len(x), reverse=True)) + r')$', '',s)
'string-to-process-'

Answer 2

sfxs = ['suffix1', 'sfx2', 'suffix333']
s = 'string-to-process-sfx2'
for sfx in sfxs:
    if sfx in s:
        s.replace(sfx, "")

应该做的。检查以确保后缀在字符串中，如果是则将其删除。

Answer 3

sfxs = ['suffix1', 'sfx2', 'suffix333']
s = 'string-to-process-sfx2'
for sfx in sfxs:
    if s.endswith(sfx):
        s = s[:-len(sfx)]
        break
print(s)

效率稍微高一些，因为字符串比较仅查看字符串的末尾。

Python：还有什么 compact/efficient 方法可以从列表中删除后缀字符串？

Python: what's a more compact/efficient way to strip string of a suffix from list?

python

string

strip