你需要使用字典和字典的倒序。这意味着您需要另一种数据结构。如果你使用的是python 3，则使用enum模块;如果你使用的是python 2.7，则使用为python 2反向移植的enum34模块。

例子:

from enum import Enum

class Color(Enum): 
    red = 1 
    green = 2 
    blue = 3

>>> print(Color.red) 
Color.red

>>> print(repr(Color.red)) 
<color.red: 1=""> 

>>> type(Color.red) 
<enum 'color'=""> 
>>> isinstance(Color.green, Color) 
True 

>>> member = Color.red 
>>> member.name 
'red' 
>>> member.value 
1 

如果你想要名字和年龄，你应该使用.items()，它会给你key (key, value)元组:

for name, age in mydict.items():
    if age == search_age:
        print name

您可以在for循环中将元组解包为两个单独的变量，然后匹配年龄。

如果你通常要根据年龄查找，而且没有两个人的年龄相同，你还应该考虑颠倒字典:

{16: 'george', 19: 'amber'}

所以你可以通过这样做来查找这个名字

mydict[search_age]

我一直称它为mydict而不是list，因为list是内置类型的名称，你不应该将这个名称用于其他任何类型。

你甚至可以在一行中得到给定年龄的所有人的列表:

[name for name, age in mydict.items() if age == search_age]

或者如果每个年龄只有一个人:

next((name for name, age in mydict.items() if age == search_age), None)

如果没有这个年龄的人，就会给你None。

最后，如果字典很长并且你使用的是Python 2，你应该考虑使用.iteritems()而不是像Cat Plus Plus在他的回答中所做的那样使用.items()，因为它不需要复制列表。

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)

以下是我的看法。这对于显示多个结果很有好处，以防您需要一个结果。所以我也添加了这个列表

myList = {'george':16,'amber':19, 'rachel':19, 
           'david':15 }                         #Setting the dictionary
result=[]                                       #Making ready of the result list
search_age = int(input('Enter age '))

for keywords in myList.keys():
    if myList[keywords] ==search_age:
    result.append(keywords)                    #This part, we are making list of results

for res in result:                             #We are now printing the results
    print(res)

就是这样……

最后我用一个函数来做。这种方法可以避免进行完整的循环，直觉告诉我们，它应该比其他解决方案更快。

def get_key_from_value(my_dict, to_find):

    for k,v in my_dict.items():
        if v==to_find: return k

    return None

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!