Something haunts me in Python

@1: 在查看"The Python Library Reference"(https://docs.python.org/2/library/stdtypes.html#sequence-types-str-unicode-list-tuple-bytearray-buffer-xrange)

的时候发现了这样的一段代码：

代码1：

>>> lists = [[]] * 3
>>> lists
[[], [], []]
>>> lists[0].append(3)
>>> lists

　　执行完lists[0].append(3)之后，程序将输出什么结果？ [[3], [0], [0]]？

　　正确答案是[[3], [3], [3]]，让我们来看看Reference上的解释：　　

　　This often haunts new Python programmers. What has happened is that [[]] is a one-element list containing

an empty list, so all three elements of [[]] * 3 are (pointers to) this single empty list. Modifying any of the elements

of lists modifies this single list. You can create a list of different lists this way:

代码2：

>>> lists = [[] for i in range(3)]
>>> lists[0].append(3)　　# 此时lists为[[3], [], []]
>>> lists[1].append(5)
>>> lists[2].append(7)
>>> lists
[[3], [5], [7]]

　　补充：代码1中lists的三个元素都指向同一个空list，是因为：s * n, n * s --- n shallow copies of s concatenated，

即Python中的*运算采用的是浅复制。

@2: Slicing & Slice Assignment(http://stackoverflow.com/questions/10623302/how-assignment-works-with-python-list-slice/10623352#10623352)

1. slicing:　　

b = a[0:2]

This makes a copy of the slice of a and assigns it to b.

2. slice assignment:

a[0:2] = b

This replaces the slice of a with the contents of b.

Although the syntax is similar (I imagine by design!), these are two different operations.

@3: 针对上面silce assignment的例子进行进一步分析：

>>> a = [1, 4, 3]
>>> b = [6, 7]
>>> a[1:3] = b
>>> a
[1, 6, 7]
>>> b
[6, 7]
>>> a[1] = 0
>>> a
[1, 0, 7]

此时b的值是多少？

>>> b
[6, 7]
>>> 

让我们继续：

代码1：

>>> a[0:3] = b     #长度不同，也允许
>>> a
[6, 7]
>>> b
[6, 7]
>>> a[1] = 1       #这种情况, 改变a不会影响b
>>> a
[6, 1]
>>> b
[6, 7]
>>> b[1] = 8       #这种情况, 改变b不会影响a
>>> b
[6, 8]
>>> a
[6, 1]

代码2：

>>> b = [6, 7]
>>> c = b
>>> c
[6, 7]
>>> b[0] = 0
>>> b
[0, 7]
>>> c
[0, 7]
>>> c[0] = 10
>>> b
[10, 7]
>>> c
[10, 7]

比较代码1和代码2结果的不同，进一步理解slice assignment。

代码3: slicing

>>> a = [1, 2, 3, 4]
>>> b = a[:2]
>>> a
[1, 2, 3, 4]
>>> b
[1, 2]
>>> b[0] = 9
>>> b
[9, 2]
>>> a
[1, 2, 3, 4]