Python 返回符合条件的列表子集
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Return a subset of list that matches condition
提问by Will
Let's say I have a list of ints:
假设我有一个ints列表：
listOfNumbers = range(100)
And I want to return a list of the elements that meet a certain condition, say:
我想返回满足特定条件的元素列表，例如：
def meetsCondition(element):
return bool(element != 0 and element % 7 == 0)
What's a Pythonic way to return a sub-listof element in a listfor which meetsCondition(element)is True?
list在listfor which meetsCondition(element)is 中返回元素的子元素的 Pythonic 方法是True什么？
A naive approach:
一种幼稚的方法：
def subList(inputList):
outputList = []
for element in inputList:
if meetsCondition(element):
outputList.append(element)
return outputList
divisibleBySeven = subList(listOfNumbers)
Is there a simple way to do this, perhaps with a list comprehension or set()functions, and without the temporary outputList?
有没有一种简单的方法可以做到这一点，也许使用列表理解或set()函数，而没有临时的outputList？
采纳答案by thefourtheye
Use list comprehension,
使用列表理解，
divisibleBySeven = [num for num in inputList if num != 0 and num % 7 == 0]
or you can use the meetsConditionalso,
或者你也可以使用meetsCondition，
divisibleBySeven = [num for num in inputList if meetsCondition(num)]
you can actually write the same condition with Python's truthysemantics, like this
您实际上可以使用 Python 的真实语义编写相同的条件，如下所示
divisibleBySeven = [num for num in inputList if num and num % 7]
alternatively, you can use filterfunction with your meetsCondition, like this
或者，您可以将filterfunction 与您的 一起使用meetsCondition，就像这样
divisibleBySeven = filter(meetsCondition, inputList)
