在满足条件之前迭代列表的最pythonic方法是什么? [英] What is the most pythonic way to iterate over a list until a condition is met?

问题描述

我有一个实际值列表:

values = [0.1, 2.9, 1.4, 5.7, 9.2, 3.8]

我希望迭代项目的数量，直到满足某些条件为止.例如，如果条件为item > 5，则结果为3.

I want to have the number of items iterated over, until some condition was met. For example, if the condition was item > 5, the result would be 3.

最简单，最直接的方法是:

The easiest way, and perhaps most straight-forward way to do this would be:

for index, item in enumerate(values):
    if (item > 5):
        print(index)
        break

还有其他方法更多pythonic，最好是单行吗?

推荐答案

您可以将itertools.takewhile与sum结合使用，该元素将使用元素，直到满足条件为止，并使用我们懒惰地求值的生成器表达式:

You can use itertools.takewhile with sum which will take elements until the condition is met, using a generator expression we lazily evaluate:

values = [0.1, 2.9, 1.4, 5.7, 9.2, 3.8]

from itertools import takewhile

print(sum(1 for _ in takewhile(lambda x: x< 5,values)))

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