遍历字典并使用集合(set)来存储所有的键,然后找出重复的键。这种方法适用于较小的字典。 代码语言:python 代码运行次数:0 复制 deffind_duplicate_keys(dictionaries):all_keys=set()duplicate_keys=set()fordictionaryindictionaries:forkeyindictionary.keys():ifkeyinall_keys:duplicate_keys.add(key)else:all_key...
items(): if value == target_value: return key return None my_dict = {'a': 1, 'b': 2, 'c': 3} result = find_key_by_value(my_dict, 2) Output: b In this example, we define a function called find_key_by_value that takes a dictionary and a target value as arguments. ...
my_dict={'a':1,'b':2,'c':3,'d':1}value_to_find=2key=find_key_by_value(my_dict,value_to_find)ifkeyisnotNone:print(f"The key corresponding to value{value_to_find}is{key}.")else:print(f"The value{value_to_find}does not exist in the dictionary.") 1. 2. 3. 4. 5. 6...
def find_key_by_value(dictionary, search_value): for key, value in dictionary.items(): if value == search_value: return key rAIse ValueError("Value does not exist in the dictionary") 三、创建反向字典 当你需要频繁地通过值来查找键时,可以考虑创建一个反向字典,其中值作为键,原始键作为值。这样...
deffind_key(d, val):returnnext(keyforkey, valueind.items()ifvalue == val) d = {'Bob':1,'Mary':2,'Lisa':4,'Ken':5,'Vivi':2}print(find_key(d,2)) 输出: Mary 11、将两个列表组合为一个字典 这里有两个列表,第一个列表存放键,第二个列表存放值,要将这两个列表转换为一个字典。
File "<stdin>", line 1, in <module> RuntimeError: dictionary changed size during iteration #结果抛出异常了,两个0的元素,也只删掉一个。 >>> d {'a': 1, 'c': 1, 'd': 0} >>> d = {'a':1, 'b':0, 'c':1, 'd':0} ...
在Python中,字典(Dictionary)是一种通过键(key)来存储和访问值(value)的数据结构。然而,字典并不直接支持通过值来查找键。但你可以通过以下几种方法实现通过值查找键的功能: 1. 遍历字典 遍历字典的每一项,检查值是否匹配。如果找到匹配的值,则记录对应的键。 python def find_key_by_value(dictionary, value):...
# 新dictionary：value: key inverted_dict = {value: key for key, value in d.items()} find_...
As you can see, since the key did not exist, it added the key with some specified value. On comparing its length with the original dictionary we can find out whether the key existed or not. For example, 1 2 3 4 5 6 7 8 d = {"a": 10, "b": 2, 'c': 4} a = len(d)...
def find_key_in_list_of_dicts(target_value, list_of_dicts): for dictionary in list_of_dicts: if target_value in dictionary.values(): return dictionary.keys()[list(dictionary.values()).index(target_value)] return None # 示例用法 list_of_dicts = [ {"name": "Alice", "age": 25}, ...