分配给外部数组时，python闭包/嵌套函数失败

python closure/nested function fail when with assignment to outer array

本问题已经有最佳答案，请猛点这里访问。

如果符号引用分配：

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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：

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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而更换：

1 2	- 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]) ]

错误是不可模仿的，但是代码不是我想要的。上面的行为是用python闭包/嵌套函数实现的吗？是否需要包装数组以在对象内进行修改并只调用函数？

另一方面，这项工作：

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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();