查找元素是否在python列表中连续出现n次

问题描述

myList = [True, True, False, False, True, False, True, False, False]

我想确定True是否连续出现3次.

I want to find if True appears 3 times in a row.

我可以通过以下方法找到它:

I can find it by doing:

for x0, x1, x2 in zip(myList, myList[1:], myList[2:]):
    if x0 == True and x1 == True and x2 == True:
        print True

有更好的方法吗?

推荐答案

使用 itertools.groupby() 对元素进行分组，然后对每个组进行计数.使用 any()函数，如果匹配，您可以提早退出循环被发现:

Use itertools.groupby() to group elements, then count each group. Using the any() function lets you exit the loop early if a match was found:

from itertools import groupby, islice

print any(sum(1 for _ in islice(g, 3)) == 3 for k, g in groupby(myList) if k)

if k过滤组以仅计算True值的组.

The if k filters the groups to only count the groups of True values.

itertools.islice()函数确保我们仅查看组的前三个元素，而忽略该组的其余部分.这样，您就不必为了确定至少发现3个而计算下一个True个下一个值.

演示:

>>> from itertools import groupby, islice
>>> myList = [True, True, False, False, True, False, True, False, False]
>>> [sum(1 for _ in islice(g, 3)) for k, g in groupby(myList) if k]
[2, 1, 1]
>>> any(sum(1 for _ in islice(g, 3)) == 3 for k, g in groupby(myList) if k)
False
>>> myList = [True, True, False, False, True, True, True, True, False, True, False, False]
>>> [sum(1 for _ in islice(g, 3)) for k, g in groupby(myList) if k]
[2, 3, 1]
>>> any(sum(1 for _ in islice(g, 3)) == 3 for k, g in groupby(myList) if k)
True

我使用列表推导来显示组的大小(仅计算True个组)，以显示为什么any()调用首先返回False，然后返回True的原因；第二个示例包含一组4个连续的True值.

I used a list comprehension to show the group sizes (counting only True groups) to show why the any() call returns False first, then True; the second example has a group of 4 consecutive True values.

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