并行地迭代值或项将不起作用，因为键不一定匹配，也不一定以相同的顺序。
而是查找两个字典的公共键：

keys = set(A.keys()) & set(B.keys())

然后使用dict解析从每个dict中查找具有这些键的值，并将它们组装成结果的键-值对：

C = {A[k]:B[k] for k in keys}

您可以使用zip：

>>> A = {1: "IWillBeAKeySoon", 6: "MeToo"}
>>> B = {1: "ItSeemsIAmAValue",6: "Val"}
>>> C = dict(zip(A.values(), B.values()))
>>> print(c)
{'MeToo': 'Val', 'IWillBeAKeySoon': 'ItSeemsIAmAValue'}

在性能方面，dict(zip(...)))似乎是最快的：

import timeit

A = {x: '%s%12d' % ('a', x) for x in range(100)}
B = {x: '%s%12d' % ('b', x) for x in range(100)}

def v1():
    C={A[key]:value for key,value in B.items() if key in B}

def v2():
    C = dict(zip(A.values(), B.values()))

def v3():
    C = {x[0]: x[1] for x in zip(A.values(), B.values())}

print(timeit.timeit(v1))
print(timeit.timeit(v2))
print(timeit.timeit(v3))
18.561055098
9.385801745000002
17.763003313