0

我有两个字典：

prices = {"banana": 4,"apple": 2}
stock = {"banana": 6,"apple": 0,}

如何从 1 和 2 字典（4*6+2*0 - 在本例中）中获取组合值的总和？

4

6 回答 6

5 
>>> prices = {"banana": 4,"apple": 2}
>>> stock = {"banana": 6,"apple": 0}
>>> {k: prices[k] * stock[k] for k in prices.viewkeys() & stock.viewkeys()}
{'banana': 24, 'apple': 0}
于 2013-06-11T09:41:16.033 回答
3

也许比其他解决方案更短更整洁。

>>> prices = {"banana": 4, "apple": 2}
>>> stock = {"banana": 6, "apple": 0}
>>> {key: value * stock.get(key, 0) for key, value in prices.items()}
{'banana': 24, 'apple': 0}

prices.iteritems()在 Python 2 中使用。

这避免了创建一个新集合，并且在技术上只使用一个字典查找。如果您只想要总和，请使用以下命令：

>>> sum(value * stock.get(key, 0) for key, value in prices.items())
24
于 2013-06-11T09:47:00.980 回答
2

不假设密钥是相同的

sum(prices[k] * stock.get(k, 0) for k in prices)
于 2013-06-11T10:22:55.297 回答
1

遍历键的交集：

mult = {}

for key in prices.viewkeys() & stock.viewkeys():
    mult[key] = prices[key] * stock[key]

这可以简化为 dict 理解：

mult = {key: prices[key] * stock[key] for key in prices.viewkeys() & stock.viewkeys()}

这使用dict.viewkeys()方法（dict.keys()在 Python 3 中），其作用类似于set; 该&运算符的作用类似于集合上的交集运算符，并为您提供两个字典中都存在的所有键。

要对总库存求和，请使用以下sum()函数：

total_stock = sum(prices[key] * stock[key] for key in prices.viewkeys() & stock.viewkeys())

演示：

>>> prices = {"banana": 4,"apple": 2}
>>> stock = {"banana": 6,"apple": 0}
>>> {key: prices[key] * stock[key] for key in prices.viewkeys() & stock.viewkeys()}
{'banana': 24, 'apple': 0}
>>> sum(prices[key] * stock[key] for key in prices.viewkeys() & stock.viewkeys())
24
于 2013-06-11T09:41:20.110 回答
0 
>>> prices = {"banana": 4,"apple": 2}
>>> stock = {"banana": 6,"apple": 0}
>>> sum( prices[k] * stock[k] for k in stock)
24
于 2013-06-11T09:41:18.080 回答
0 
prices = {"banana": 4,"apple": 2} 
stock = {"banana": 6,"apple": 0,}
t=0
for i in prices:
      t+=prices[i]*stock[i]
print t
于 2013-06-11T09:45:05.023 回答