python 闭包/嵌套函数在分配给外部数组时失败

Question

如果符号出现，Python函数的闭包似乎有问题 引用已分配:

def outer():
    p = []
    def gen():
        def touch(e):
            if e[0] == 'add':
                p.append(e);
            elif e[0] == 'rem':
                p = [ x for x in p if not (x[1] == e[1]) ]
        return touch
    f = gen()
    for i in [["add","test1"],["add","test2"],["rem","test2"],["rem","test1"]]:
        f(i)      

outer();

结果是:

Traceback (most recent call last):
  File "b.py", line 22, in <module>
    outer();
  File "b.py", line 20, in outer
    f(i)      
  File "b.py", line 14, in touch
    p.append(e);
UnboundLocalError: local variable 'p' referenced before assignment

如果我只是为了测试替换:

 -       p = [ x for x in p if not (x[1] == e[1]logig is) ]                                                                                                                                
 +       a = [ x for x in p if not (x[1] == e[1]) ]                                                                                                                                

错误消失了，但是代码不是我想要的。 python 闭包/嵌套函数是否会出现上述行为？我是否需要将数组包装在对象内部进行修改并仅调用函数？

另一方面，这个有效:

class o():
    def __init__(self):
        self.p = []
    def add(self,e):
        self.p.append(e);
    def rem(self,e):
        self.p = [ x for x in self.p if not (x[1] == e[1]) ]

def outer():
    p = o()
    def gen():
        def touch(e):
            if e[0] == 'add':
                p.add(e);
            elif e[0] == 'rem':
                p.rem(e)
        return touch
    f = gen()
    for i in [["add","test1"],["add","test2"],["rem","test2"],["rem","test1"]]:
        f(i)      

outer();

Answer 1

因为您在 touch 内分配 p，所以它成为 touch 中的局部变量，并有效地“隐藏”所有其他名称 p 在封闭范围内。为了告诉 Python 您实际上想要引用 outer 中的 p，您应该使用 nonlocal p，例如:

def outer():
    p = []
    def touch(e):
        # The following line is required to refer to outer's p
        nonlocal p
        if e[0] == 'add':
            p.append(e)
        elif e[0] == 'rem':
            p = [ x for x in p if not (x[1] == e[1]) ]
    for i in [["add","test1"],["add","test2"],["rem","test2"],["rem","test1"]]:
        touch(i)
outer()

您的第二个示例之所以有效，是因为您在 touch 的两种情况下都引用了 p 的属性，而不是进行赋值 (p = ... )。

参见nonlocal in the Python reference documentation ，scopes 的引用文档，和PEP 3104其中提出了nonlocal 语法。 nonlocal 仅存在于 Python 3 中，但是 there is a workaround如果有必要使用Python 2。