如果键相同，则相交两个字典python以获取两个字典的共同元素的有效方法

Question

我正在尝试在两个字典中为每个键查找常用项。也许下面的代码可以更好地解释我的目的。我有太多记录，因此这段代码需要很长时间才能运行；有什么方法可以有效地编写它？

rcmd = {'1':{"A","B"},"2":{"A","C"},"3":{"B","C","D"}}   
rmv = {'1':{"C","B"},"2":{"A","C"},"3":{"B","C","A"},"4":{"A"}}

correct_rcmd = []
for i in range(len(rcmd)):
    for j in range(len(rmv)):
        if rcmd.keys()[i] == rmv.keys()[j]:
            correct_rcmd.append(rcmd.values() 
[i].intersection(rmv.values()[j]))
print correct_rcmd

Answer 1

您永远不应该在字典的键中使用索引 - 特别是如果您使用 python 2.x（正如您使用 print 而没有任何 () 所建议的那样）。

字典中键的顺序直到 3.6（仅针对作为实现细节的副作用的 CPython）或默认情况下从 python 3.7 开始才固定。

然后顺序是 插入顺序 - 如果您以 不同 顺序插入键，索引到 keys() 仍然会中断，因为您是混合不同键的值！。

rcmd = {'1':{"A","B"},"2":{"A","C"},"3":{"B","C","D"}}   
rmv = {'1':{"C","B"},"2":{"A","C"},"3":{"B","C","A"},"4":{"A"}}

# get all keys that are common in both dictionaries
# python 2 or 3:
keys_that_are_same = set(rcmd) & set(rmv)
# for python 3 better, as it uses the key-sets directly: 
# keys_that_are_same = rcmd.keys() & rmv.keys()

# loop over both keys and get the intersection into a new dict:
common = {}
for key in keys_that_are_same:
    common[key] = rcmd[key] & rmv[key]

# as dict comprehension - no loop needed: 
# common = {k : rcmd[k] & rmv[k] for k in keys_that_are_same} 

print(common)

输出：

{'1': set(['B']), '3': set(['C', 'B']), '2': set(['A', 'C'])} # py 2.7

{'1': {'B'}, '2': {'A', 'C'}, '3': {'C', 'B'}}                # py 3.6

这还具有使用更少需要构造和处理的对象以及使用集合来减少开始处理的键数量的好处。如果你在 python 2.