12

检查字符串是否包含列表中任何项目中的某些字符的最快方法是什么？

目前，我正在使用这种方法：

lestring = "Text123"

lelist = ["Text", "foo", "bar"]

for x in lelist:
    if lestring.count(x):
        print 'Yep. "%s" contains characters from "%s" item.' % (lestring, x)

有没有什么方法可以在没有迭代的情况下做到这一点（我想这会让它更快。）？

4

4 回答 4

22

您可以尝试使用成员资格检查进行列表理解

>>> lestring = "Text123"
>>> lelist = ["Text", "foo", "bar"]
>>> [e for e in lelist if e in lestring]
['Text']

与您的实现相比，虽然 LC 有一个隐式循环，但它更快，因为没有像您的情况那样显式调用函数count

与 Joe 的实现相比，您的实现要快得多，因为 filter 函数需要在一个循环中调用两个函数，lambda并且count

>>> def joe(lelist, lestring):
    return ''.join(random.sample(x + 'b'*len(x), len(x)))

>>> def uz(lelist, lestring):
    for x in lelist:
        if lestring.count(x):
            return 'Yep. "%s" contains characters from "%s" item.' % (lestring, x)


>>> def ab(lelist, lestring):
    return [e for e in lelist if e in lestring]

>>> t_ab = timeit.Timer("ab(lelist, lestring)", setup="from __main__ import lelist, lestring, ab")
>>> t_uz = timeit.Timer("uz(lelist, lestring)", setup="from __main__ import lelist, lestring, uz")
>>> t_joe = timeit.Timer("joe(lelist, lestring)", setup="from __main__ import lelist, lestring, joe")
>>> t_ab.timeit(100000)
0.09391469893125759
>>> t_uz.timeit(100000)
0.1528471407273173
>>> t_joe.timeit(100000)
1.4272649857800843

Jamie 的评论解决方案对于较短的字符串来说较慢。这是测试结果

>>> def jamie(lelist, lestring):
    return next(itertools.chain((e for e in lelist if e in lestring), (None,))) is not None

>>> t_jamie = timeit.Timer("jamie(lelist, lestring)", setup="from __main__ import lelist, lestring, jamie")
>>> t_jamie.timeit(100000)
0.22237164127909637

如果您需要布尔值，对于较短的字符串，只需修改上面的 LC 表达式

[e in lestring for e in lelist if e in lestring]

或者对于更长的字符串，您可以执行以下操作

>>> next(e in lestring for e in lelist if e in lestring)
True

或者

>>> any(e in lestring for e in lelist)
于 2013-01-19T06:08:58.030 回答
1 
filter(lambda x: lestring.count(x), lelist)

这将返回您尝试查找的所有字符串作为列表。

于 2013-01-19T06:08:24.120 回答
0

如果测试是查看是否有任何共同的字符（不是单词或段），请从列表中的字母创建一个集合，然后再检查字符串中的字母：

char_list = set(''.join(list_of_words))
test_set = set(string_to_teat)
common_chars = char_list.intersection(test_set)

但是我假设你正在寻找一个共同的角色......

于 2013-01-19T06:20:15.930 回答
0

esmre库可以解决问题。在您的情况下，更简单的esm（esmre 的一部分）就是您想要的。

https://pypi.python.org/pypi/esmre/

https://code.google.com/p/esmre/

他们有很好的文档和示例：取自他们的示例：

>>> import esm
>>> index = esm.Index()
>>> index.enter("he")
>>> index.enter("she")
>>> index.enter("his")
>>> index.enter("hers")
>>> index.fix()
>>> index.query("this here is history")
[((1, 4), 'his'), ((5, 7), 'he'), ((13, 16), 'his')]
>>> index.query("Those are his sheep!")
[((10, 13), 'his'), ((14, 17), 'she'), ((15, 17), 'he')]
>>>

我进行了一些性能测试：

import random, timeit, string, esm

def uz(lelist, lestring):
    for x in lelist:
        if lestring.count(x):
            return 'Yep. "%s" contains characters from "%s" item.' % (lestring, x)



def ab(lelist, lestring):
    return [e for e in lelist if e in lestring]


def use_esm(index, lestring):
    return index.query(lestring)

for TEXT_LEN in [5, 50, 1000]:
    for SEARCH_LEN in [5, 20]:
        for N in [5, 50, 1000, 10000]:
            if TEXT_LEN < SEARCH_LEN:
                continue

            print 'TEXT_LEN:', TEXT_LEN, 'SEARCH_LEN:', SEARCH_LEN, 'N:', N

            lestring = ''.join((random.choice(string.ascii_uppercase + string.digits) for _ in range(TEXT_LEN)))
            lelist = [''.join((random.choice(string.ascii_uppercase + string.digits) for _ in range(SEARCH_LEN))) for _
                      in range(N)]

            index = esm.Index()
            for i in lelist:
                index.enter(i)
            index.fix()

            t_ab = timeit.Timer("ab(lelist, lestring)", setup="from __main__ import lelist, lestring, ab")
            t_uz = timeit.Timer("uz(lelist, lestring)", setup="from __main__ import lelist, lestring, uz")
            t_esm = timeit.Timer("use_esm(index, lestring)", setup="from __main__ import index, lestring, use_esm")

            ab_time = t_ab.timeit(1000)
            uz_time = t_uz.timeit(1000)
            esm_time = t_esm.timeit(1000)

            min_time = min(ab_time, uz_time, esm_time)
            print '  ab%s: %f' % ('*' if ab_time == min_time else '', ab_time)
            print '  uz%s: %f' % ('*' if uz_time == min_time else '', uz_time)
            print '  esm%s %f:' % ('*' if esm_time == min_time else '', esm_time)

得到的结果主要取决于一个人正在寻找的项目数量（在我的例子中，'N'）：

TEXT_LEN: 1000 SEARCH_LEN: 20 N: 5
  ab*: 0.001733
  uz: 0.002512
  esm 0.126853:

TEXT_LEN: 1000 SEARCH_LEN: 20 N: 50
  ab*: 0.017564
  uz: 0.023701
  esm 0.079925:

TEXT_LEN: 1000 SEARCH_LEN: 20 N: 1000
  ab: 0.370371
  uz: 0.489523
  esm* 0.133783:

TEXT_LEN: 1000 SEARCH_LEN: 20 N: 10000
  ab: 3.678790
  uz: 4.883575
  esm* 0.259605:
于 2014-05-12T10:32:43.033 回答