Python：从列表中剥离后缀字符串的更紧凑/更有效的方法是什么？

Question

Is there a more compact and efficient way of stripping a string of any suffix from a given list, ie:

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

Suffixes are of different lengths

Answer 1

You may use 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'

'|'.join(sfxs) helps to join the suffix list with | as delimiter. So r'(' + '|'.join(sfxs) + r')$' would form a regex like (suff1|suff2|suff3|..)$ . Note that $ anchor, which matches the end of the line. So this would do matching only at the end.

>>> 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 s.endswith(sfx):
        s = s[:-len(sfx)]
        break
print(s)

is slightly more efficient because the string comparison looks at the end of the string only.

Answer 3

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

Should do it. Check to ensure the suffix is in the string, then remove it if it is.