带有lambda函数的字典理解给出错误的结果 [英] Dictionary comprehension with lambda functions gives wrong results

问题描述

我在Python 3.5.1中尝试了以下代码:

I tried the following code in Python 3.5.1:

>>> f = {x: (lambda y: x) for x in range(10)}
>>> f[5](3)
9

很明显，这应该返回5.我不知道其他值来自何处，也找不到任何东西.

It's obvious that this should return 5. I don't understand where the other value comes from, and I wasn't able to find anything.

似乎与引用有关-它总是返回f[9]的答案，这是最后分配的功能.

It seems like it's something related to reference - it always returns the answer of f[9], which is the last function assigned.

这是什么错误，应该如何做才能使其正常工作?

What's the error here, and how should this be done so that it works properly?

推荐答案

Python作用域是词汇性的.闭包将引用变量的名称和范围，而不是变量的实际对象/值.

Python scoping is lexical. A closure will refer to the name and scope of the variable, not the actual object/value of the variable.

发生的情况是每个lambda都捕获变量x 不是值x.

What happens is that each lambda is capturing the variable x not the value of x.

在循环结束时，变量x绑定到9，因此每个lambda都将引用该值为9的x.

At the end of the loop the variable x is bound to 9, therefore every lambda will refer to this x whose value is 9.

为什么@ChrisP的答案有效:

Why @ChrisP's answer works:

make_func强制评估x的值(传递时) 转化为功能).因此，lambda的当前值为x 并且避免了上述范围界定问题.

make_func forces the value of x to be evaluated (as it is passed into a function). Thus, the lambda is made with value of x currently and we avoid the above scoping issue.

def make_func(x):
    return lambda y: x

f = {x: make_func(x) for x in range(10)}

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