@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。