两个字典的相交和区别

Question

给定两个字典，我想看看它们的交集和差异，并对与唯一元素相交并执行g的元素执行f函数，这就是我找出d1和d2是两个字典的唯一和相交元素的方法，如何将d_intersection和d_difference作为字典打印在元组中？输出应该看起来像这样（{相交的键，值}，{差异的键，值}），例如：

d1 = {1:30, 2:20, 3:30, 5:80}

d2 = {1:40, 2:50, 3:60, 4:70, 6:90}

输出应为 ({1: 70, 2: 70, 3: 90}, {4: 70, 5: 80, 6: 90})

dic = {}
d_intersect = set(d1) & set(d2)
d_difference =  set(d1) ^ set(d2)
for i in d_intersect:
    dic.update({i : f(d1[i],d2[i])})
for j in d_difference:
    dic.update({j : g(d1[j],d2[j])})

有人可以告诉我我哪里出了问题，为什么我的代码给出了关键错误4？

Answer 1

尽管可能存在更有效的方法，但这是一种方法。

d1 = {1:30, 2:20, 3:30, 5:80}
d2 = {1:40, 2:50, 3:60, 4:70, 6:90}

d_intersect = {} # Keys that appear in both dictionaries.
d_difference = {} # Unique keys that appear in only one dictionary.

# Get all keys from both dictionaries.
# Convert it into a set so that we don't loop through duplicate keys.
all_keys = set(d1.keys() + d2.keys()) # Python2.7
#all_keys = set(list(d1.keys()) + list(d2.keys())) # Python3.3

for key in all_keys:
    if key in d1 and key in d2:
        # If the key appears in both dictionaries, add both values
        # together and place it in intersect.
        d_intersect[key] = d1[key] + d2[key]
    else:
        # Otherwise find out the dictionary it comes from and place
        # it in difference.
        if key in d1:
            d_difference[key] = d1[key]
        else:
            d_difference[key] = d2[key]

输出：

{1：70，2：70，3：90}

{4：70，5：80，6：90}