我做了一个函数，它将在字典中查找年龄并显示匹配的名字:
dictionary = {'george' : 16, 'amber' : 19}
search_age = raw_input("Provide age")
for age in dictionary.values():
if age == search_age:
name = dictionary[age]
print name
我知道如何比较和查找年龄，只是不知道如何显示这个人的名字。此外，由于第5行，我得到了一个KeyError。我知道这是不正确的，但我不知道如何让它向后搜索。
当前回答
dict_a = {'length': 5, 'width': 9, 'height': 4}
# get the key of specific value 5
key_of_value = list(dict_a)[list(dict_a.values()).index(5)]
print(key_of_value) # length
# get the key of minimum value
key_min_value = list(dict_a)[list(dict_a.values()).index(sorted(dict_a.values())[0])]
print(key_min_value) # height
# get the key of maximum value
key_max_value = list(dict_a)[list(dict_a.values()).index(sorted(dict_a.values(), reverse=True)[0])]
print(key_max_value) # width
其他回答
my_dict = {'A': 19, 'B': 28, 'carson': 28}
search_age = 28
只拿一个
name = next((name for name, age in my_dict.items() if age == search_age), None)
print(name) # 'B'
获取多个数据
name_list = [name for name, age in filter(lambda item: item[1] == search_age, my_dict.items())]
print(name_list) # ['B', 'carson']
我发现这个答案很有效，但对我来说不太容易理解。
为了使它更清楚，您可以反转字典的键和值。这是使键值和值键，如这里所示。
mydict = {'george':16,'amber':19}
res = dict((v,k) for k,v in mydict.iteritems())
print(res[16]) # Prints george
或者Python 3，(谢谢@kkgarg)
mydict = {'george':16,'amber':19}
res = dict((v,k) for k,v in mydict.items())
print(res[16]) # Prints george
Also
print(res.get(16)) # Prints george
本质上和另一个答案是一样的。
有时可能需要int():
titleDic = {'Фильмы':1, 'Музыка':2}
def categoryTitleForNumber(self, num):
search_title = ''
for title, titleNum in self.titleDic.items():
if int(titleNum) == int(num):
search_title = title
return search_title
我瞥见所有的答案，没有提到简单地使用列表理解?
这个Python的单行解决方案可以返回任意数量的给定值的所有键(在Python 3.9.1中测试):
>>> dictionary = {'george' : 16, 'amber' : 19, 'frank': 19}
>>>
>>> age = 19
>>> name = [k for k in dictionary.keys() if dictionary[k] == age]; name
['george', 'frank']
>>>
>>> age = (16, 19)
>>> name = [k for k in dictionary.keys() if dictionary[k] in age]; name
['george', 'amber', 'frank']
>>>
>>> age = (22, 25)
>>> name = [k for k in dictionary.keys() if dictionary[k] in age]; name
[]
这是一个真正的“可逆字典”，基于Adam Acosta的解决方案，但强制val-to-key调用是唯一的，容易从值返回键:
from collections import UserDict
class ReversibleDict(UserDict):
def __init__(self, enforce_unique=True, *args, **kwargs):
super().__init__(*args, **kwargs)
self.val_to_keys = {}
self.check_val = self.check_unique if enforce_unique else lambda x: x
def __setitem__(self, key, value):
self.check_val(value)
super().__setitem__(key, value)
self.val_to_keys[value] = key
def __call__(self, value):
return self.val_to_keys[value]
def check_unique(self, value):
assert value not in self.val_to_keys, f"Non unique value '{value}'"
return value
如果你想强制字典值的唯一性，确保set enforce_unique=True。从值中获取键只需做rev_dict(value)，从键中调用值只需像往常一样做dict['key']，这里是一个用法示例:
rev_dict = ReversibleDict(enforce_unique=True)
rev_dict["a"] = 1
rev_dict["b"] = 2
rev_dict["c"] = 3
print("full dictinoary is: ", rev_dict)
print("value for key 'b' is: ", rev_dict["b"])
print("key for value '2' is: ", rev_dict(2))
print("tring to set another key with the same value results in error: ")
rev_dict["d"] = 1
