Question

我有这样的事情：

A = {"Green":{"Small": 5, "Medium": 10, "Large": 15}, "Yellow": {"Small": 7, "Medium": 14, "Large": 21}}

B = {"Green":{"Small": 1, "Medium": 2, "Large": 3}, "Yellow": {"Small": 3, "Medium": 6, "Large": 9}}

我想写一个函数来获取字典＆＃34; C＆＃34;像这样：

C = {"Green":{"Small": 6, "Medium": 12, "Large": 18}, "Yellow": {"Small": 10, "Medium": 20, "Large": 30}}

这只是它键的每个值的总和（在所有词典中都是相同的。现在我的＆＃34;解决方案＆＃34;太长了，当然也不优雅。任何人都可以给关于我如何能够实现相对较短的功能的一些指示？

Answer 1

旧的循环：

C = {}
for k, v in A.items():
    inner = {}
    for inner_k, inner_v in v.items():
        inner[inner_k] = inner_v + B[k][inner_k]
    C[k] = inner

>>> C
{'Green': {'Large': 18, 'Medium': 12, 'Small': 6},
 'Yellow': {'Large': 30, 'Medium': 20, 'Small': 10}}

Answer 2

以OP形式接受任意数量字典的函数描述：

from pprint import pprint

def dsum(*args):
    return {
        k1: {
            k: sum(d.get(k1, dict()).get(k, 0) for d in args)
            for k in set.union(*(set(d.get(k1, dict())) for d in args))
        }
        for k1 in set.union(*(set(d) for d in args))
    }

A = {"Green":{"Small": 5, "Medium": 10, "Large": 15}, "Yellow": {"Small": 7, "Medium": 14, "Large": 21}}

B = {"Green":{"Small": 1, "Medium": 2, "Large": 3}, "Yellow": {"Small": 3, "Medium": 6, "Large": 9}}

C = {"Green":{"Small": 6, "Medium": 12, "Large": 18}, "Yellow": {"Small": 10, "Medium": 20, "Large": 30}}

testC = dsum(A, B)
assert C == testC
pprint(testC)

Answer 3

C={k: {kk: v + B[k][kk] for kk, v in a.items()} for k, a in A.items()}

更喜欢python2中的.iteritems()

Answer 4

缺少密钥的解决方案：

def merge_by_addition(d1, d2):
    def m2(d1_, d2_):
        keys_ = set(d1_.keys()) | set(d2_.keys())
        return {k_: (d1_.get(k_, 0) + d2_.get(k_, 0)) for k_ in keys_}
    keys = set(d1.keys()) | set(d2.keys())
    return {k: m2(d1.get(k, {}), d2.get(k, {})) for k in keys}

如何在python中的不同字典中添加相同键的值？

4 个答案: