有效地检查一个元素是否在列表中至少出现 n 次 [英] Efficiently check if an element occurs at least n times in a list

问题描述

如何最好地编写 Python 函数 (check_list) 以有效测试元素 (x) 是否在一个元素中至少出现 n 次列表 (l)?

How to best write a Python function (check_list) to efficiently test if an element (x) occurs at least n times in a list (l)?

我的第一个想法是:

def check_list(l, x, n):
    return l.count(x) >= n

但是一旦 x 被找到 n 次并且总是 O(n)，这不会短路.

But this doesn't short-circuit once x has been found n times and is always O(n).

一种简单的短路方法是:

A simple approach that does short-circuit would be:

def check_list(l, x, n):
    count = 0
    for item in l:
        if item == x:
            count += 1
            if count == n:
                return True
    return False

我还有一个更紧凑的带发电机的短路解决方案:

I also have a more compact short-circuiting solution with a generator:

def check_list(l, x, n):
    gen = (1 for item in l if item == x)
    return all(next(gen,0) for i in range(n))

还有其他好的解决方案吗?什么是最有效的方法?

Are there other good solutions? What is the best efficient approach?

谢谢

推荐答案

而不是通过设置 range 对象并使用必须测试的 all 来产生额外的开销每个项目的真实性，你可以使用itertools.islice 使生成器前进 n 步，如果切片存在或默认 False，则返回切片中的 next 项 如果不是:

Instead of incurring extra overhead with the setup of a range object and using all which has to test the truthiness of each item, you could use itertools.islice to advance the generator n steps ahead, and then return the next item in the slice if the slice exists or a default False if not:

from itertools import islice

def check_list(lst, x, n):
    gen = (True for i in lst if i==x)
    return next(islice(gen, n-1, None), False)

请注意，与 list.count 一样，itertools.islice 也以 C 速度运行.这具有处理非列表的可迭代对象的额外优势.

Note that like list.count, itertools.islice also runs at C speed. And this has the extra advantage of handling iterables that are not lists.

一些时间:

In [1]: from itertools import islice

In [2]: from random import randrange

In [3]: lst = [randrange(1,10) for i in range(100000)]

In [5]: %%timeit # using list.index
   ....: check_list(lst, 5, 1000)
   ....:
1000 loops, best of 3: 736 µs per loop

In [7]: %%timeit # islice
   ....: check_list(lst, 5, 1000)
   ....:
1000 loops, best of 3: 662 µs per loop

In [9]: %%timeit # using list.index
   ....: check_list(lst, 5, 10000)
   ....:
100 loops, best of 3: 7.6 ms per loop

In [11]: %%timeit # islice
   ....: check_list(lst, 5, 10000)
   ....:
100 loops, best of 3: 6.7 ms per loop

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