对于一个孤立的电话,您应该迭代dic和break按@ IvayloStrandjev的解决方案。对于重复的调用,最好反转字典以保持O(1)查找的复杂性。假设您的值不重叠:dic = {'key1': ["value1", "value2"], 'key2': ["value4", "value5"]}dic_reverse = {w: k for k, v in dic.items() for w in v}print(dic_reverse){'value1': 'key1', 'value2': 'key1', 'value4': 'key2', 'value5': 'key2'}然后,您可以通过检索密钥dic_reverse.get('value4', None)。