Python - 检查字符串是否包含列表中任何项目中的特定字符的最快方法 [英] Python - Fastest way to check if a string contains specific characters in any of the items in a list

问题描述

检查字符串是否包含列表中任何项目的某些字符的最快方法是什么？

What is the fastest way to check if a string contains some characters from any items of a list?

目前，我正在使用此方法：

Currently, I'm using this method:

lestring = "Text123"

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

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

是有没有办法没有迭代（这会让我觉得它更快。）？

Is there any way to do it without iteration (which will make it faster I suppose.)?

推荐答案

你可以试试列表理解会员支票

You can try list comprehension with membership check

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

比较对你的实现，虽然LC有一个隐式循环，但它更快，因为没有显式函数调用，如你的情况下 count

Compared to your implementation, though LC has an implicit loop but its faster as there is no explicit function call as in your case with count

与Joe的实现相比，你的实现更快，因为过滤器函数需要在循环中调用两个函数， lambda 和 count

Compared to Joe's implementation, yours is way faster, as the filter function would require to call two functions in a loop, lambda and 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速度较慢。以下是测试结果

---

Jamie's commented solution is slower for shorter string's. Here is the test result

>>> 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

---

If you need Boolean values, for shorter strings, just modify the above LC expression

[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

Or for longer strings, you can do the following

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

或

>>> any(e in lestring for e in lelist)

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