直接‘show me the code’，懒得长篇大论一堆文字
1：赋值
>>> a = [1, [1]]
>>> b = a
>>> a[1][0] = 2
>>> print a, b
[1, [2]] [1, [2]]
>>> a[0] = 2
>>> print a, b
[2, [2]] [2, [2]]
>>> a.append(3)
>>> print a, b
[2, [2], 3] [2, [2], 3]
>>> a[1].append(4)
>>> print a, b
[2, [2, 4], 3] [2, [2, 4], 3]
赋值其实是对象的引用，而python通过传递引用处理对象，列表对象的引用分别赋值给a和b，也就是a和b都引用这相同的一个对象，所以无论怎么修改对象的内容，a和b结果都一样是修改后的内容，并没有一个原始，没有被修改的copy
2：切片，copy，list
>>> a = [1, [1]]
>>> b = a[:]
>>> a[1][0] = 2
>>> print a, b
[1, [2]] [1, [2]]
>>> a[0] = 2
>>> print a, b
[2, [2]] [1, [2]]
>>>
>>> import copy
>>> a = [1, [1]]
>>> b = copy.copy(a)
>>> a[1][0] = 2
>>> print a, b
[1, [2]] [1, [2]]
>>> a[0] = 2
>>> print a, b
[2, [2]] [1, [2]]
>>>
>>> a = [1, [1]]
>>> b = list(a)
>>> a[1][0] = 2
>>> print a, b
[1, [2]] [1, [2]]
>>> a[0] = 2
>>> print a, b
[2, [2]] [1, [2]]
这三种方法都是浅拷贝，虽然生成了一个新对象，但是对象内部的属性和内容依然引用了原对象，所以在给列表第二个元素列表元素重新赋值2的时候，a和b都改变了；但是通过a来进行第一个元素重新赋值2的时候，b却没有变，原因是整型是不可变类型，也就是说在进行a[0] = 2的时候，会重新创建一个整型对象2，而b缺没变
3：deepcopy
>>> import copy
>>> a = [1, [1]]
>>> b = copy.deepcopy(a)
>>> a[1][0] = 2
>>> print a, b
[1, [2]] [1, [1]]
>>> a[0] = 2
>>> print a, b
[2, [2]] [1, [1]]
这种方法就是从原来对象重新复制一个全新的对象，内部属性和内容也全部递归地修改了
