#1.分割大列表为三个元素的小列表，不够三个元素的亦当成一个列表输出
In [17]: lst
Out[17]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [18]: for i in range(0,len(lst),3):
  ...:   print lst[i:i+3]
  ...:
[0, 1, 2]
[3, 4, 5]
[6, 7, 8]
[9]

#2.稍微改进下，用了个列表推导，结果都放到一个列表了
In [35]: lst
Out[35]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [36]: b = [lst[i:i+3] for i in range(0,len(lst),3)]

In [37]: b
Out[37]: [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

In [38]:

#3.不太喜欢的lambda，猛一看不懂，再细看逻辑还是上面的逻辑
In [10]: f = lambda a:map(lambda b:a[b:b+3],range(0,len(a),3))

In [11]: lst
Out[11]: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [12]: f(lst)
Out[12]: [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]