正如有人提到的，可能有多个键具有相同的值，如下面的my_dict。此外，可能没有匹配的键。

my_dict ={'k1':1,'k2':2, 'k3':1, 'k4':12, 'k5':1, 'k6':1, 'k7':12}

这里有三种找到钥匙的方法，一种用于最后一次敲击，两种用于第一次敲击。

def find_last(search_value:int, d:dict):
    
    return [x for x,y in d.items() if y==search_value].pop()

def find_first1(search_value:int, d:dict):
    return next(filter(lambda x: d[x]==search_value, d.keys()), None)

def find_first2(search_value:int, d:dict):
    return next(x for x,y in  d.items() if y==search_value)

在这些函数中，find_first1比其他函数快一点，如果没有匹配的键，它将返回None。

考虑使用Pandas。正如William McKinney的《Python for Data Analysis》中所述

另一种考虑级数的方法是固定长度的有序级数 Dict，因为它是索引值到数据值的映射。它可以是 在很多情况下，你可能会用到字典。

import pandas as pd
list = {'george':16,'amber':19}
lookup_list = pd.Series(list)

要查询您的系列，请执行以下操作:

lookup_list[lookup_list.values == 19]

收益率:

Out[1]: 
amber    19
dtype: int64

如果您需要对输出进行任何其他转换 回答成一个列表可能有用:

answer = lookup_list[lookup_list.values == 19].index
answer = pd.Index.tolist(answer)

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

我认为指出哪些方法是最快的，以及在什么情况下是最快的会很有趣:

以下是我在一台2012年的MacBook Pro上进行的一些测试

def method1(dict, search_age):
    for name, age in dict.iteritems():
        if age == search_age:
            return name

def method2(dict, search_age):
    return [name for name,age in dict.iteritems() if age == search_age]

def method3(dict, search_age):
    return dict.keys()[dict.values().index(search_age)]

profile.run()在每个方法上100,000次的结果:

方法1:

>>> profile.run("for i in range(0,100000): method1(dict, 16)")
     200004 function calls in 1.173 seconds

方法2:

>>> profile.run("for i in range(0,100000): method2(dict, 16)")
     200004 function calls in 1.222 seconds

方法3:

>>> profile.run("for i in range(0,100000): method3(dict, 16)")
     400004 function calls in 2.125 seconds

所以这表明，对于一个小字典，方法1是最快的。这很可能是因为它返回第一个匹配，而不是像方法2那样返回所有匹配(参见下面的注释)。

有趣的是，在我有2700个条目的字典上执行相同的测试，我得到了完全不同的结果(这次运行了10,000次):

方法1:

>>> profile.run("for i in range(0,10000): method1(UIC_CRS,'7088380')")
     20004 function calls in 2.928 seconds

方法2:

>>> profile.run("for i in range(0,10000): method2(UIC_CRS,'7088380')")
     20004 function calls in 3.872 seconds

方法3:

>>> profile.run("for i in range(0,10000): method3(UIC_CRS,'7088380')")
     40004 function calls in 1.176 seconds

这里，方法3要快得多。这表明字典的大小会影响你选择的方法。

注:

方法2返回所有名称的列表，而方法1和3只返回第一个匹配项。 我没有考虑内存使用情况。我不确定方法3是否创建了2个额外的列表(keys()和values())并将它们存储在内存中。

以下是我对这个问题的看法。：） 我刚刚开始学习Python，所以我称之为:

“初学者可以理解的”解决方案。

#Code without comments.

list1 = {'george':16,'amber':19, 'Garry':19}
search_age = raw_input("Provide age: ")
print
search_age = int(search_age)

listByAge = {}

for name, age in list1.items():
    if age == search_age:
        age = str(age)
        results = name + " " +age
        print results

        age2 = int(age)
        listByAge[name] = listByAge.get(name,0)+age2

print
print listByAge

.

#Code with comments.
#I've added another name with the same age to the list.
list1 = {'george':16,'amber':19, 'Garry':19}
#Original code.
search_age = raw_input("Provide age: ")
print
#Because raw_input gives a string, we need to convert it to int,
#so we can search the dictionary list with it.
search_age = int(search_age)

#Here we define another empty dictionary, to store the results in a more 
#permanent way.
listByAge = {}

#We use double variable iteration, so we get both the name and age 
#on each run of the loop.
for name, age in list1.items():
    #Here we check if the User Defined age = the age parameter 
    #for this run of the loop.
    if age == search_age:
        #Here we convert Age back to string, because we will concatenate it 
        #with the person's name. 
        age = str(age)
        #Here we concatenate.
        results = name + " " +age
        #If you want just the names and ages displayed you can delete
        #the code after "print results". If you want them stored, don't...
        print results

        #Here we create a second variable that uses the value of
        #the age for the current person in the list.
        #For example if "Anna" is "10", age2 = 10,
        #integer value which we can use in addition.
        age2 = int(age)
        #Here we use the method that checks or creates values in dictionaries.
        #We create a new entry for each name that matches the User Defined Age
        #with default value of 0, and then we add the value from age2.
        listByAge[name] = listByAge.get(name,0)+age2

#Here we print the new dictionary with the users with User Defined Age.
print
print listByAge

.

#Results
Running: *\test.py (Thu Jun 06 05:10:02 2013)

Provide age: 19

amber 19
Garry 19

{'amber': 19, 'Garry': 19}

Execution Successful!

就是我的答案和过滤器。

filter( lambda x, dictionary=dictionary, search_age=int(search_age): dictionary[x] == search_age  , dictionary )