lambda 函数的范围及其参数? [英] Scope of lambda functions and their parameters?

问题描述

我需要一个对于一系列 gui 事件几乎完全相同的回调函数.根据调用它的事件，该函数的行为会略有不同.对我来说似乎是一个简单的案例，但我无法弄清楚 lambda 函数的这种奇怪行为.

I need a callback function that is almost exactly the same for a series of gui events. The function will behave slightly differently depending on which event has called it. Seems like a simple case to me, but I cannot figure out this weird behavior of lambda functions.

所以我在下面有以下简化代码:

So I have the following simplified code below:

def callback(msg):
    print msg

#creating a list of function handles with an iterator
funcList=[]
for m in ('do', 're', 'mi'):
    funcList.append(lambda: callback(m))
for f in funcList:
    f()

#create one at a time
funcList=[]
funcList.append(lambda: callback('do'))
funcList.append(lambda: callback('re'))
funcList.append(lambda: callback('mi'))
for f in funcList:
    f()

这段代码的输出是:

mi
mi
mi
do
re
mi

我期望:

do
re
mi
do
re
mi

为什么使用迭代器会把事情搞砸?

Why has using an iterator messed things up?

我尝试过使用深层复制:

I've tried using a deepcopy:

import copy
funcList=[]
for m in ('do', 're', 'mi'):
    funcList.append(lambda: callback(copy.deepcopy(m)))
for f in funcList:
    f()

但这有同样的问题.

推荐答案

这里的问题是 m 变量(引用)从周围的作用域中获取.只有参数保存在 lambda 范围内.

The problem here is the m variable (a reference) being taken from the surrounding scope. Only parameters are held in the lambda scope.

要解决这个问题，您必须为 lambda 创建另一个作用域:

To solve this you have to create another scope for lambda:

def callback(msg):
    print msg

def callback_factory(m):
    return lambda: callback(m)

funcList=[]
for m in ('do', 're', 'mi'):
    funcList.append(callback_factory(m))
for f in funcList:
    f()

在上面的例子中，lambda 也使用了周围的作用域来查找 m，但是这个时间是 callback_factory 范围，每个 callback_factory 创建一次打电话.

In the example above, lambda also uses the surounding scope to find m, but this time it's callback_factory scope which is created once per every callback_factory call.

或者使用 functools.partial:

from functools import partial

def callback(msg):
    print msg

funcList=[partial(callback, m) for m in ('do', 're', 'mi')]
for f in funcList:
    f()

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