如果两个元组元素都不在其他字典键中，则删除元组键字典项

Question

我有一本以元组为键的字典。
和第二个具有单元素键的字典。
价值观并不重要

dictionary_1 = {("Apple","Banana") : 3,
                 ("Cat","Dog") : 5,
                 ("Spain", "Italy") : 10,
                 ("Chair","Sofa"): 23}

dictionary_2 = {"Denmark" : 4,
                "Apple" : 9,
                "Fish" : 7,
                "Sofa" : 8 }

如果键的任一元素是字典\u 2中的键之一，我想从字典\u 1中删除键

因此，解决办法是：

#Some code
print(dictionary_1)
#The remaining key value pairs would be:
{("Cat","Dog") : 5,
 ("Spain", "Italy") : 10}

谢谢

Answer 1

使用dictionary comprehension和sets可以很快解决这个问题：

>>> dictionary_1 = {fruits: n 
                    for fruits, n in dictionary_1.items()
                    if set(fruits).isdisjoint(dictionary_2)}
{('Cat', 'Dog'): 5, ('Spain', 'Italy'): 10}

上面说：

• 将dictionary_1重新分配到一个新的映射到数字的水果元组字典
• 从原始字典的项中获取键/值对
• 仅包括该对，因为水果元组和第二个字典的键之间没有重叠

希望这有帮助：-）

Answer 2

您可以使用for key in dict语法来迭代字典的键：

dictionary_1 = {("Apple","Banana") : 3,
                 ("Cat","Dog") : 5,
                 ("Spain", "Italy") : 10,
                 ("Chair","Sofa"): 23}

dictionary_2 = {"Denmark" : 4,
                "Apple" : 9,
                "Fish" : 7,
                "Sofa" : 8 }

new_dict = {}
for i in dictionary_1:
    if i[0] not in dictionary_2 and i[1] not in dictionary_2:
        new_dict[i] = dictionary_1[i]

print(new_dict) # {('Spain', 'Italy'): 10, ('Cat', 'Dog'): 5}

Answer 3

您可以制作一个新的字典，其中包含dict理解，并利用some()进行测试：

dictionary_1 = {("Apple","Banana") : 3,
                 ("Cat","Dog") : 5,
                 ("Spain", "Italy") : 10,
                 ("Chair","Sofa"): 23}

dictionary_2 = {"Denmark" : 4,
                "Apple" : 9,
                "Fish" : 7,
                "Sofa" : 8 }

{k:v for k,v in dictionary_1.items() 
 if not any(t in dictionary_2 for t in k)}
# {('Cat', 'Dog'): 5, ('Spain', 'Italy'): 10}