python字典里的数求和_一个字典，其值为字典：尝试在python中对这些键求和

数据结构是一个字典&＃xff0c;每个值都是另一个字典&＃xff0c;如&＃xff1a;

>>> from lib import schedule

>>> schedule &＃61; schedule.Schedule()

>>> game &＃61; schedule.games[0]

>>> game.home

>>> game.home.lineup

{&＃39;guerv001&＃39;: {&＃39;HR&＃39;: 392, &＃39;1B&＃39;: 1297}, &＃39;kendh001&＃39;: {&＃39;HR&＃39;: 12, &＃39;1B&＃39;: 201}, &＃39;andeg001&＃39;: {&＃39;HR&＃39;: 272, &＃39;1B&＃39;: 1572}, &＃39;mattg002&＃39;: {&＃39;HR&＃39;: 104, &＃39;1B&＃39;: 632}, &＃39;figgc001&＃39;: {&＃39;HR&＃39;: 26, &＃39;1B&＃39;: 672}, &＃39;iztum001&＃39;: {&＃39;HR&＃39;: 16, &＃39;1B&＃39;: 253}, &＃39;huntt001&＃39;: {&＃39;HR&＃39;: 213, &＃39;1B&＃39;: 834}, &＃39;quinr002&＃39;: {&＃39;HR&＃39;: 23, &＃39;1B&＃39;: 200}, &＃39;napom001&＃39;: {&＃39;HR&＃39;: 46, &＃39;1B&＃39;: 96}}

希望Team有一个方法getTotals(self&＃xff0c;category)&＃xff0c;如果你调用了它&＃xff1a;

game.home.getTotals(&＃39;HR&＃39;)

在这种情况下&＃xff0c;该方法会产生&＃xff1a;

1104

基本上你可以看到我正在尝试做什么。有什么想法吗&＃xff1f;

更新&＃xff1a;我有两个列表推导工作&＃xff0c;但想把它归结为一个&＃xff1a;

def getTotals(self, category):

cats &＃61; [x for x in self.lineup.values()]

return sum([x[category] for x in cats])

另一次更新&＃xff1a;根据inspectorg4dget下面的有用反馈&＃xff0c;我已经知道了。谢谢&＃xff01;

def getTotals(self, category):

return sum(self.lineup[man][category] for man in self.lineup.keys())

最终更新&＃xff1a;根据Nadia的反馈&＃xff0c;这是另一种方法&＃xff1a;

def getTotals(self, category):

return sum(value.get(category, 0) for value in self.lineup.values())