如何返回在不同字典中获得新值的字典键的值

Question

我的问题有点类似于：Replacing the value of a Python dictionary with the value of another dictionary that matches the other dictionaries key

但就我而言，我有两本字典

dict1 = {'foo' : ['val1' , 'val2' , 'val3'] , 'bar' : ['val4' , 'val5']}

dict2 = {'foo' : ['val2', 'val10', 'val11'] , 'bar' : ['val1' , 'val4']}

我要返回的是

dict3 = {'foo' : ['val10', 'val11'] , 'bar' : ['val1']}

反之

dict4 = {'foo' : ['val1', 'val3'] , 'bar' : ['val5']}

其中 dict3 返回键 'foo' 和 'bar' 在 dict2 中获得的值的字典，而 dict4 是键 'foo' 和 'bar' 在 dict2 中丢失的值的字典

我尝试解决的一种方法是：

 iterate over both dictionaries then
    if key of dict1 == key of dict2
       return the values of the key in dict1 and compare with the values in dict2
       return the values that aren't in both 
       as a dictionary of the key and those values

这个想法行不通，而且显然效率很低。我希望有一种更有效的工作方式来做到这一点

Answer 1

两个 dict 理解可以解决问题：

dict3 = {k: [x for x in v if x not in dict1[k]] for k, v in dict2.items()}
print(dict3)
# {'foo': ['val10', 'val11'], 'bar': ['val1']}

dict4 = {k: [x for x in v if x not in dict2[k]] for k, v in dict1.items()}
print(dict4)
# {'foo': ['val1', 'val3'], 'bar': ['val5']}

上述两种推导式基本上是从键中过滤掉另一个字典中不存在的值，这是双向完成的。

你也可以这样做没有 dict理解：

dict3 = {}
for k,v in dict2.items():
    dict3[k] = [x for x in v if x not in dict1[k]]

print(dict3)
# {'foo': ['val10', 'val11'], 'bar': ['val1']}

dict4 = {}
for k,v in dict1.items():
    dict4[k] = [x for x in v if x not in dict2[k]]

print(dict4)
# {'foo': ['val1', 'val3'], 'bar': ['val5']}