【python cookbook】【数据结构与算法】8.与字典有关的计算问题

问题：在字典上对数据执行各式各样的计算（比如求最小值、最大值、排序）。

解决方案：利用zip()将字典的键-值对“反转”为值-键对序列。

例如：如下字典存放的股票名称和对应的价格：

>>> prices = {
   ‘ACME‘: 45.23,
   ‘AAPL‘: 612.78,
   ‘IBM‘: 205.55,
   ‘HPQ‘: 37.20,
   ‘FB‘: 10.75
}
>>> prices
{‘HPQ‘: 37.2, ‘IBM‘: 205.55, ‘FB‘: 10.75, ‘ACME‘: 45.23, ‘AAPL‘: 612.78}
>>> min_price = min(zip(prices.values(), prices.keys()))  #注意zip(x,y)中参数的顺序
>>> max_price = max(zip(prices.values(), prices.keys()))
>>> min_price
(10.75, ‘FB‘)
>>> max_price
(612.78, ‘AAPL‘)
>>> prices_sorted = sorted(zip(prices.values(), prices.keys()))
>>> prices_sorted
[(10.75, ‘FB‘), (37.2, ‘HPQ‘), (45.23, ‘ACME‘), (205.55, ‘IBM‘), (612.78, ‘AAPL‘)]

>>> min_price2 = min(zip(prices)) #错误用法
>>> min_price2
(‘AAPL‘,)
>>> max_price2 = max(zip(prices))  #错误用法
>>> max_price2
(‘IBM‘,)
>>> min_price3 = min(zip(prices.keys(), prices.values())) #zip()参数顺序不对，获取错误的值
>>> min_price3
(‘AAPL‘, 612.78)
>>> max_price3 = max(zip(prices.keys(), prices.values())) #zip()参数顺序不对，获取错误的值
>>> max_price3
(‘IBM‘, 205.55)
>>> 

进行这些计算时，请注意zip()创建了一个迭代器，它的内容只能消费一次。例如：

>>> pirces_and_names=zip(prices.values(), prices.keys())
>>> pirces_and_names
<zip object at 0x023BDFA8>
>>> min(pirces_and_names)
(10.75, ‘FB‘)
>>> max(pirces_and_names)
Traceback (most recent call last):
  File "<pyshell#25>", line 1, in <module>
    max(pirces_and_names)
ValueError: max() arg is an empty sequence
>>> 

注意：当涉及（value,key）对的比较时，碰巧有多个条目拥有相同的value值，那么此时key将用来作为判定结果的依据。

>>> prices={‘AAA‘:45.23,‘ZZZ‘:45.23}
>>> min(zip(prices.values(), prices.keys()))
(45.23, ‘AAA‘)
>>> max(zip(prices.values(), prices.keys()))
(45.23, ‘ZZZ‘)
>>>