for name in mydict:
    if mydict[name] == search_age:
        print(name) 
        #or do something else with it. 
        #if in a function append to a temporary list, 
        #then after the loop return the list

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 )

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!

这是你访问字典做你想做的事情的方式:

list = {'george': 16, 'amber': 19}
search_age = raw_input("Provide age")
for age in list:
    if list[age] == search_age:
        print age

当然，你们的名字太离谱了，看起来像是要打印一个年龄，但它确实打印了名字。因为你是通过名字来访问的，所以如果你这样写会更容易理解:

list = {'george': 16, 'amber': 19}
search_age = raw_input("Provide age")
for name in list:
    if list[name] == search_age:
        print name

更好的是:

people = {'george': {'age': 16}, 'amber': {'age': 19}}
search_age = raw_input("Provide age")
for name in people:
    if people[name]['age'] == search_age:
        print name

在我的情况下，最简单的方法是实例化字典在你的代码，然后你可以从它调用键如下

这是我们班有字典

class Config:

def local(self):
    return {
        "temp_dir": "/tmp/dirtest/",
        "devops": "Mansur",
    }

实例化你的字典

config =  vars.Config()
local_config = config.local()

最后调用你的字典键

patched = local_config.get("devops")

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)