删除字符串中的字符列表 [英] Removing a list of characters in string

问题描述

我想在python中删除字符串中的字符:

I want to remove characters in a string in python:

string.replace(',', '').replace("!", '').replace(":", '').replace(";", '')...

但是我必须删除许多字符.我想到了一个清单

But I have many characters I have to remove. I thought about a list

list = [',', '!', '.', ';'...]

但是如何使用list替换string中的字符?

But how can I use the list to replace the characters in the string?

推荐答案

如果您使用的是python2，而您的输入是字符串(不是unicodes)，则绝对最佳的方法是

If you're using python2 and your inputs are strings (not unicodes), the absolutely best method is str.translate:

>>> chars_to_remove = ['.', '!', '?']
>>> subj = 'A.B!C?'
>>> subj.translate(None, ''.join(chars_to_remove))
'ABC'

否则，要考虑以下选项:

Otherwise, there are following options to consider:

A.按字符迭代主题char，省略不需要的字符和 join 结果列表:

A. Iterate the subject char by char, omit unwanted characters and join the resulting list:

>>> sc = set(chars_to_remove)
>>> ''.join([c for c in subj if c not in sc])
'ABC'

(请注意，生成器版本''.join(c for c ...)的效率较低).

(Note that the generator version ''.join(c for c ...) will be less efficient).

B.动态创建正则表达式，并使用 re.sub 空字符串:

B. Create a regular expression on the fly and re.sub with an empty string:

>>> import re
>>> rx = '[' + re.escape(''.join(chars_to_remove)) + ']'
>>> re.sub(rx, '', subj)
'ABC'

( re.escape 确保类似或]不会破坏正则表达式).

(re.escape ensures that characters like ^ or ] won't break the regular expression).

C.使用translate 的映射变体:

C. Use the mapping variant of translate:

>>> chars_to_remove = [u'δ', u'Γ', u'ж']
>>> subj = u'AжBδCΓ'
>>> dd = {ord(c):None for c in chars_to_remove}
>>> subj.translate(dd)
u'ABC'

完整的测试代码和时间安排:

Full testing code and timings:

#coding=utf8

import re

def remove_chars_iter(subj, chars):
    sc = set(chars)
    return ''.join([c for c in subj if c not in sc])

def remove_chars_re(subj, chars):
    return re.sub('[' + re.escape(''.join(chars)) + ']', '', subj)

def remove_chars_re_unicode(subj, chars):
    return re.sub(u'(?u)[' + re.escape(''.join(chars)) + ']', '', subj)

def remove_chars_translate_bytes(subj, chars):
    return subj.translate(None, ''.join(chars))

def remove_chars_translate_unicode(subj, chars):
    d = {ord(c):None for c in chars}
    return subj.translate(d)

import timeit, sys

def profile(f):
    assert f(subj, chars_to_remove) == test
    t = timeit.timeit(lambda: f(subj, chars_to_remove), number=1000)
    print ('{0:.3f} {1}'.format(t, f.__name__))

print (sys.version)
PYTHON2 = sys.version_info[0] == 2

print ('\n"plain" string:\n')

chars_to_remove = ['.', '!', '?']
subj = 'A.B!C?' * 1000
test = 'ABC' * 1000

profile(remove_chars_iter)
profile(remove_chars_re)

if PYTHON2:
    profile(remove_chars_translate_bytes)
else:
    profile(remove_chars_translate_unicode)

print ('\nunicode string:\n')

if PYTHON2:
    chars_to_remove = [u'δ', u'Γ', u'ж']
    subj = u'AжBδCΓ'
else:
    chars_to_remove = ['δ', 'Γ', 'ж']
    subj = 'AжBδCΓ'

subj = subj * 1000
test = 'ABC' * 1000

profile(remove_chars_iter)

if PYTHON2:
    profile(remove_chars_re_unicode)
else:
    profile(remove_chars_re)

profile(remove_chars_translate_unicode)

结果:

2.7.5 (default, Mar  9 2014, 22:15:05) 
[GCC 4.2.1 Compatible Apple LLVM 5.0 (clang-500.0.68)]

"plain" string:

0.637 remove_chars_iter
0.649 remove_chars_re
0.010 remove_chars_translate_bytes

unicode string:

0.866 remove_chars_iter
0.680 remove_chars_re_unicode
1.373 remove_chars_translate_unicode

---

3.4.2 (v3.4.2:ab2c023a9432, Oct  5 2014, 20:42:22) 
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)]

"plain" string:

0.512 remove_chars_iter
0.574 remove_chars_re
0.765 remove_chars_translate_unicode

unicode string:

0.817 remove_chars_iter
0.686 remove_chars_re
0.876 remove_chars_translate_unicode

(作为附带说明，remove_chars_translate_bytes的数字可能为我们提供了一个线索，说明为什么该行业这么长时间不愿采用Unicode).

(As a side note, the figure for remove_chars_translate_bytes might give us a clue why the industry was reluctant to adopt Unicode for such a long time).

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