Question

我写的代码目前要求我检查一个词典中是否有任何项目（至少一个）存在于另一个词典中。

Answer 1

这个解决方案怎么样：

a = {"a":2, "b":4, "c":4, "d":4}
b = {"a":1, "e":1, "f":5}
print(any(a.items() & b.items()))

将产生输出： False

因为a和b中没有常见项目，而：

a = {"a":2, "b":4, "c":4, "d":4}
b = {"a":1, "b":4, "f":5}
print(any(a.items() & b.items()))

将产生输出： True

因为a和b

有一个共同的项目

它不会直接迭代字典，但正如juanpa-arrivillaga在评论中指出的那样，此解决方案在技术上使用迭代，因为在any()下进行迭代。

Answer 2

[item for item in dict1.items() if item in dict2.items()]

Answer 3

您可以使用set检查第二个中是否存在至少一个dict个键和/或值。你可以这样做：

a = {1:"a", 2:"b", 3:"c"}
b = {"foo":"hello", "bar":"hi", 2:"hoo"}
c = {"hello":"hoo", 1:"hii"}

def keys_exists(first:"dict", second:"dict") -> bool:
    # Or:
    # return not (set(first) - set(second)) == first.keys()
    return bool(set(first) & set(second))

def values_exists(first:"dict", second:"dict") -> bool:
    return bool(set(first.values()) & set(second.values()))


print("At least one of a keys exists in b keys: {0}".format(keys_exists(a,b)))
print("At least one of a keys exists in c keys: {0}".format(keys_exists(a,c)))
print("At least one of b keys exists in c keys: {0}".format(keys_exists(b,c)))

print("At least one of a values exists in b values: {0}".format(values_exists(a,b)))
print("At least one of a values exists in c values: {0}".format(values_exists(a,c)))
print("At least one of b values exists in c values: {0}".format(values_exists(b,c)))

输出：

At least one of a keys exists in b keys: True
At least one of a keys exists in c keys: True
At least one of b keys exists in c keys: False

At least one of a values exists in b values: False
At least one of a values exists in c values: False
At least one of b values exists in c values: True

如何在没有迭代的情况下检查一个字典中的任何项是否存在于另一个字典中？

3 个答案: