列出Python中的理解和lambda

问题描述

我想创建一个lambda列表，但效果并不理想.

I wanted to create a list of lambdas, but it didn't quite work out as I hoped.

L = [(lambda x: x/y) for y in range(10)]

我希望列表中的每个函数都将其参数除以其索引，但是所有函数仅按最后一个索引除.

I expected every function in the list to divide its argument by its index, but all functions only divide by the last index.

>>> L[1](5)
0.5555555555555556
>>> L[5](5)
0.5555555555555556
>>> 5/9
0.5555555555555556

这是列表理解吗，每个lambda在Python中都有自己的y副本?

Is this kind of list comprehension, where every lambda has its own copy of ypossible in Python?

推荐答案

lambda中的y指的是y在其来源范围内的最后一个值，即9.

The y in your lambda refers to the last value that y had in the scope it came from, i.e., 9.

获得所需行为的最简单方法是在lambda中使用默认参数:

The easiest way to get the behavior you want is to use a default argument in your lambda:

lambda x, y=y: x/y

这将在定义lambda函数时捕获y的值.

This captures the value of y at the moment the lambda function is defined.

您还可以执行双lambda"，调用返回所需的lambda的函数，并传入所需的y值:

You can also do a "double-lambda", calling a function that returns the lambda you want, passing in the desired value of y:

(lambda y: lambda x: x/y)(y)

在这里，每次调用时，外部lambda都会提供一个新的作用域.

Here, the outer lambda provides a new scope each time you call it.

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