我想创建一个 lambda 列表，但结果并不如我所愿.
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，但所有函数仅除以最后一个索引.
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
这种列表推导式(每个 lambda 都有自己的 y副本)在 Python 中是否可行?
Is this kind of list comprehension, where every lambda has its own copy of ypossible in Python?
你的 lambda 中的 y 指的是 y 在它来自的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
这会在定义 lambda 函数时捕获 y 的值.
This captures the value of y at the moment the lambda function is defined.
你也可以做一个double-lambda"，调用一个返回你想要的lambda的函数，传入y的期望值:
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)
这里，每次调用外部 lambda 时都会提供一个新范围.
Here, the outer lambda provides a new scope each time you call it.
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