Python高级编程-如何根据字典中值得大小，对字典中的项排序

>>> from random import randint
>>> { x:randint (60,100) for x in 'xyondfuh'}
{'x': 91, 'y': 90, 'o': 67, 'n': 64, 'd': 70, 'f': 96, 'u': 72, 'h': 68}
>>> d={ x:randint (60,100) for x in 'xyondfuh'}
>>> sorted(d)
['d', 'f', 'h', 'n', 'o', 'u', 'x', 'y']
>>> d.keys()
dict_keys(['x', 'y', 'o', 'n', 'd', 'f', 'u', 'h'])
>>> d.values()
dict_values([94, 70, 61, 81, 80, 66, 62, 79])
>>> zip (d.values(),d.keys())
<zip object at 0x039D6918>
>>> sorted( zip (d.values(),d.keys()))
[(61, 'o'), (62, 'u'), (66, 'f'), (70, 'y'), (79, 'h'), (80, 'd'), (81, 'n'), (94, 'x')]
>>>
>>> d.items()
dict_items([('x', 94), ('y', 70), ('o', 61), ('n', 81), ('d', 80), ('f', 66), ('u', 62), ('h', 79)])
>>> sorted (d.items(),key=lambda x:x[1])
[('o', 61), ('u', 62), ('f', 66), ('y', 70), ('h', 79), ('d', 80), ('n', 81), ('x', 94)]