使用 Python 从字符串中去除所有非字母数字字符的最佳方法是什么?
这个问题的 PHP 变体中提供的解决方案可能会通过一些小的调整来工作,但对我来说似乎不是很“pythonic”。
作为记录,我不仅想去掉句号和逗号(和其他标点符号),还想去掉引号、括号等。
使用 Python 从字符串中去除所有非字母数字字符的最佳方法是什么?
这个问题的 PHP 变体中提供的解决方案可能会通过一些小的调整来工作,但对我来说似乎不是很“pythonic”。
作为记录,我不仅想去掉句号和逗号(和其他标点符号),还想去掉引号、括号等。
出于好奇,我只是对一些功能进行了计时。在这些测试中,我从字符串中删除了非字母数字字符string.printable
(内置string
模块的一部分)。使用编译'[\W_]+'
后pattern.sub('', str)
发现速度最快。
$ python -m timeit -s \
"import string" \
"''.join(ch for ch in string.printable if ch.isalnum())"
10000 loops, best of 3: 57.6 usec per loop
$ python -m timeit -s \
"import string" \
"filter(str.isalnum, string.printable)"
10000 loops, best of 3: 37.9 usec per loop
$ python -m timeit -s \
"import re, string" \
"re.sub('[\W_]', '', string.printable)"
10000 loops, best of 3: 27.5 usec per loop
$ python -m timeit -s \
"import re, string" \
"re.sub('[\W_]+', '', string.printable)"
100000 loops, best of 3: 15 usec per loop
$ python -m timeit -s \
"import re, string; pattern = re.compile('[\W_]+')" \
"pattern.sub('', string.printable)"
100000 loops, best of 3: 11.2 usec per loop
正则表达式的救援:
import re
re.sub(r'\W+', '', your_string)
通过 Python 定义
'\W
==[^a-zA-Z0-9_]
,它排除了 allnumbers
,letters
并且_
使用str.translate()
方法。
假设你会经常这样做:
一次,创建一个包含您要删除的所有字符的字符串:
delchars = ''.join(c for c in map(chr, range(256)) if not c.isalnum())
每当您想压缩字符串时:
scrunched = s.translate(None, delchars)
设置成本可能与re.compile
; 边际成本要低得多:
C:\junk>\python26\python -mtimeit -s"import string;d=''.join(c for c in map(chr,range(256)) if not c.isalnum());s=string.printable" "s.translate(None,d)"
100000 loops, best of 3: 2.04 usec per loop
C:\junk>\python26\python -mtimeit -s"import re,string;s=string.printable;r=re.compile(r'[\W_]+')" "r.sub('',s)"
100000 loops, best of 3: 7.34 usec per loop
注意:使用string.printable
作为基准数据会给模式带来'[\W_]+'
不公平的优势;所有非字母数字字符都在一堆......在典型数据中,将有不止一个替换:
C:\junk>\python26\python -c "import string; s = string.printable; print len(s),repr(s)"
100 '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!"#$%&\'()*+,-./:;=>?@[\\]^_`{|}~ \t\n\r\x0b\x0c'
如果您re.sub
多做一些工作,会发生以下情况:
C:\junk>\python26\python -mtimeit -s"d=''.join(c for c in map(chr,range(256)) if not c.isalnum());s='foo-'*25" "s.translate(None,d)"
1000000 loops, best of 3: 1.97 usec per loop
C:\junk>\python26\python -mtimeit -s"import re;s='foo-'*25;r=re.compile(r'[\W_]+')" "r.sub('',s)"
10000 loops, best of 3: 26.4 usec per loop
你可以试试:
print ''.join(ch for ch in some_string if ch.isalnum())
>>> import re
>>> string = "Kl13@£$%[};'\""
>>> pattern = re.compile('\W')
>>> string = re.sub(pattern, '', string)
>>> print string
Kl13
怎么样:
def ExtractAlphanumeric(InputString):
from string import ascii_letters, digits
return "".join([ch for ch in InputString if ch in (ascii_letters + digits)])
这通过使用列表推导来生成字符列表(InputString
如果它们存在于组合字符串中)ascii_letters
。digits
然后它将列表连接成一个字符串。
sent = "".join(e for e in sent if e.isalpha())
作为此处其他一些答案的衍生,我提供了一种非常简单且灵活的方法来定义您希望将字符串内容限制为的一组字符。在这种情况下,我允许使用字母数字加上破折号和下划线。PERMITTED_CHARS
只需根据您的用例 添加或删除我的字符即可。
PERMITTED_CHARS = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ_-"
someString = "".join(c for c in someString if c in PERMITTED_CHARS)
使用随机的 ASCII 可打印字符串计时:
from inspect import getsource
from random import sample
import re
from string import printable
from timeit import timeit
pattern_single = re.compile(r'[\W]')
pattern_repeat = re.compile(r'[\W]+')
translation_tb = str.maketrans('', '', ''.join(c for c in map(chr, range(256)) if not c.isalnum()))
def generate_test_string(length):
return ''.join(sample(printable, length))
def main():
for i in range(0, 60, 10):
for test in [
lambda: ''.join(c for c in generate_test_string(i) if c.isalnum()),
lambda: ''.join(filter(str.isalnum, generate_test_string(i))),
lambda: re.sub(r'[\W]', '', generate_test_string(i)),
lambda: re.sub(r'[\W]+', '', generate_test_string(i)),
lambda: pattern_single.sub('', generate_test_string(i)),
lambda: pattern_repeat.sub('', generate_test_string(i)),
lambda: generate_test_string(i).translate(translation_tb),
]:
print(timeit(test), i, getsource(test).lstrip(' lambda: ').rstrip(',\n'), sep='\t')
if __name__ == '__main__':
main()
结果(Python 3.7):
Time Length Code
6.3716264850008880 00 ''.join(c for c in generate_test_string(i) if c.isalnum())
5.7285426190064750 00 ''.join(filter(str.isalnum, generate_test_string(i)))
8.1875841680011940 00 re.sub(r'[\W]', '', generate_test_string(i))
8.0002205439959650 00 re.sub(r'[\W]+', '', generate_test_string(i))
5.5290945199958510 00 pattern_single.sub('', generate_test_string(i))
5.4417179649972240 00 pattern_repeat.sub('', generate_test_string(i))
4.6772285089973590 00 generate_test_string(i).translate(translation_tb)
23.574712151996210 10 ''.join(c for c in generate_test_string(i) if c.isalnum())
22.829975890002970 10 ''.join(filter(str.isalnum, generate_test_string(i)))
27.210196289997840 10 re.sub(r'[\W]', '', generate_test_string(i))
27.203713296003116 10 re.sub(r'[\W]+', '', generate_test_string(i))
24.008979928999906 10 pattern_single.sub('', generate_test_string(i))
23.945240008994006 10 pattern_repeat.sub('', generate_test_string(i))
21.830899796994345 10 generate_test_string(i).translate(translation_tb)
38.731336012999236 20 ''.join(c for c in generate_test_string(i) if c.isalnum())
37.942474347000825 20 ''.join(filter(str.isalnum, generate_test_string(i)))
42.169366310001350 20 re.sub(r'[\W]', '', generate_test_string(i))
41.933375883003464 20 re.sub(r'[\W]+', '', generate_test_string(i))
38.899814646996674 20 pattern_single.sub('', generate_test_string(i))
38.636144253003295 20 pattern_repeat.sub('', generate_test_string(i))
36.201238164998360 20 generate_test_string(i).translate(translation_tb)
49.377356811004574 30 ''.join(c for c in generate_test_string(i) if c.isalnum())
48.408927293996385 30 ''.join(filter(str.isalnum, generate_test_string(i)))
53.901889764994850 30 re.sub(r'[\W]', '', generate_test_string(i))
52.130339455994545 30 re.sub(r'[\W]+', '', generate_test_string(i))
50.061149017004940 30 pattern_single.sub('', generate_test_string(i))
49.366573111998150 30 pattern_repeat.sub('', generate_test_string(i))
46.649754120997386 30 generate_test_string(i).translate(translation_tb)
63.107938601999194 40 ''.join(c for c in generate_test_string(i) if c.isalnum())
65.116287978999030 40 ''.join(filter(str.isalnum, generate_test_string(i)))
71.477421126997800 40 re.sub(r'[\W]', '', generate_test_string(i))
66.027950693998720 40 re.sub(r'[\W]+', '', generate_test_string(i))
63.315361931003280 40 pattern_single.sub('', generate_test_string(i))
62.342320287003530 40 pattern_repeat.sub('', generate_test_string(i))
58.249303059004890 40 generate_test_string(i).translate(translation_tb)
73.810345625002810 50 ''.join(c for c in generate_test_string(i) if c.isalnum())
72.593953348005020 50 ''.join(filter(str.isalnum, generate_test_string(i)))
76.048324580995540 50 re.sub(r'[\W]', '', generate_test_string(i))
75.106637657001560 50 re.sub(r'[\W]+', '', generate_test_string(i))
74.681338128997600 50 pattern_single.sub('', generate_test_string(i))
72.430461594005460 50 pattern_repeat.sub('', generate_test_string(i))
69.394243567003290 50 generate_test_string(i).translate(translation_tb)
str.maketrans
&str.translate
最快,但包括所有非 ASCII 字符。
re.compile
&pattern.sub
较慢,但在某种程度上比''.join
&快filter
。
对于一个简单的单行(Python 3.0):
''.join(filter( lambda x: x in '0123456789abcdefghijklmnopqrstuvwxyz', the_string_you_want_stripped ))
对于 Python < 3.0:
filter( lambda x: x in '0123456789abcdefghijklmnopqrstuvwxyz', the_string_you_want_stripped )
注意:如果需要,您可以将其他字符添加到允许的字符列表中(例如,'0123456789abcdefghijklmnopqrstuvwxyz.,_')。
使用与@John Machin 的答案相同的方法,但针对 Python 3 进行了更新:
translate
。现在假定 Python 代码以 UTF-8 编码
(来源:PEP 3120)
这意味着包含您要删除的所有字符的字符串会变得更大:
del_chars = ''.join(c for c in map(chr, range(1114111)) if not c.isalnum())
并且该translate
方法现在需要使用我们可以创建的转换表maketrans()
:
del_map = str.maketrans('', '', del_chars)
现在,和以前一样,s
您想要“揉搓”的任何字符串:
scrunched = s.translate(del_map)
使用@Joe Machin 的最后一个计时示例,我们可以看到它仍然re
以一个数量级的速度节拍:
> python -mtimeit -s"d=''.join(c for c in map(chr,range(1114111)) if not c.isalnum());m=str.maketrans('','',d);s='foo-'*25" "s.translate(m)"
1000000 loops, best of 5: 255 nsec per loop
> python -mtimeit -s"import re;s='foo-'*25;r=re.compile(r'[\W_]+')" "r.sub('',s)"
50000 loops, best of 5: 4.8 usec per loop
for char in my_string:
if not char.isalnum():
my_string = my_string.replace(char,"")
一个简单的解决方案,因为这里的所有答案都很复杂
filtered = ''
for c in unfiltered:
if str.isalnum(c):
filtered += c
print(filtered)
我用perfplot (我的一个项目)检查了结果,发现对于短字符串,
"".join(filter(str.isalnum, s))
最快。对于长字符串(200+ 字符)
re.sub("[\W_]", "", s)
最快。
重现情节的代码:
import perfplot
import random
import re
import string
pattern = re.compile("[\W_]+")
def setup(n):
return "".join(random.choices(string.ascii_letters + string.digits, k=n))
def string_alphanum(s):
return "".join(ch for ch in s if ch.isalnum())
def filter_str(s):
return "".join(filter(str.isalnum, s))
def re_sub1(s):
return re.sub("[\W_]", "", s)
def re_sub2(s):
return re.sub("[\W_]+", "", s)
def re_sub3(s):
return pattern.sub("", s)
b = perfplot.bench(
setup=setup,
kernels=[string_alphanum, filter_str, re_sub1, re_sub2, re_sub3],
n_range=[2**k for k in range(10)],
)
b.save("out.png")
b.show()
如果我理解正确,最简单的方法是使用正则表达式,因为它为您提供了很大的灵活性,但另一种简单的方法是使用 for 循环以下是示例代码,我还计算了单词的出现并存储在字典中。
s = """An... essay is, generally, a piece of writing that gives the author's own
argument — but the definition is vague,
overlapping with those of a paper, an article, a pamphlet, and a short story. Essays
have traditionally been
sub-classified as formal and informal. Formal essays are characterized by "serious
purpose, dignity, logical
organization, length," whereas the informal essay is characterized by "the personal
element (self-revelation,
individual tastes and experiences, confidential manner), humor, graceful style,
rambling structure, unconventionality
or novelty of theme," etc.[1]"""
d = {} # creating empty dic
words = s.split() # spliting string and stroing in list
for word in words:
new_word = ''
for c in word:
if c.isalnum(): # checking if indiviual chr is alphanumeric or not
new_word = new_word + c
print(new_word, end=' ')
# if new_word not in d:
# d[new_word] = 1
# else:
# d[new_word] = d[new_word] +1
print(d)
如果这个答案有用,请给这个评分!