如何返回与条件匹配的列表的子集 [英] How to return a subset of a list that matches a condition

问题描述

假设我有一个 int s的列表:

Let's say I have a list of ints:

listOfNumbers = range(100)

我想返回一个满足一定条件的元素的列表，说:

And I want to return a list of the elements that meet a certain condition, say:

def meetsCondition(element):
    return bool(element != 0 and element % 7 == 0)

在 list 中为其返回 meetsCondition(element)为 True的元素的子 list 的Python方法是什么??

What's a Pythonic way to return a sub-list of element in a list for which meetsCondition(element) is True?

幼稚的方法:

def subList(inputList):
    outputList = []

    for element in inputList:
        if meetsCondition(element):
            outputList.append(element)

    return outputList

divisibleBySeven = subList(listOfNumbers)

是否有一种简单的方法来执行此操作，也许具有列表理解功能或 set()函数，并且没有临时的 outputList ?

Is there a simple way to do this, perhaps with a list comprehension or set() functions, and without the temporary outputList?

推荐答案

使用列表理解，

divisibleBySeven = [num for num in inputList if num != 0 and num % 7 == 0]

或者您也可以使用 meetsCondition ，

divisibleBySeven = [num for num in inputList if meetsCondition(num)]

您实际上可以使用Python的真实的语义，例如

you can actually write the same condition with Python's truthy semantics, like this

divisibleBySeven = [num for num in inputList if num and num % 7]

---

或者，您可以在 meetsCondition 中使用 filter 函数，如下所示

divisibleBySeven = filter(meetsCondition, inputList)

％timeit

listOfNumbers = range(1000000)

%timeit [num for num in listOfNumbers if meetsCondition(num)]
[out]:
243 ms ± 4.51 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit list(filter(meetsCondition, listOfNumbers))
[out]:
211 ms ± 4.19 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

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