使用字典视图来实现这一点；该dict.viewkeys()结果就像一组，让你做的十字路口和对称的区别：
def merge(A, B, f):
# Start with symmetric difference; keys either in A or B, but not both
merged = {k: A.get(k, B.get(k)) for k in A.viewkeys() ^ B.viewkeys()}
# Update with `f()` applied to the intersection
merged.update({k: f(A[k], B[k]) for k in A.viewkeys() & B.viewkeys()})
return merged
在Python 3中，该.viewkeys()方法已重命名为.keys()，取代了旧.keys()功能（在Python 2中会重现列表）。
上面的merge()方法是适用于任何给定的通用解决方案f()。
演示：
>>> def f(x, y):
... return x * y
...
>>> A = {1:1, 2:3}
>>> B = {7:3, 2:2}
>>> merge(A, B, f)
{1: 1, 2: 6, 7: 3}
>>> merge(A, B, lambda a, b: '{} merged with {}'.format(a, b))
{1: 1, 2: '3 merged with 2', 7: 3}
602392714
清零编程群