试试下面的一行代码来反转字典:

reversed_dictionary = dict(map(reversed, dictionary.items()))

def get_Value(dic,value):
    for name in dic:
        if dic[name] == value:
            del dic[name]
            return name

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!

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

dict((v,k) for k,v in d.items()).get(16)

回显如下:

-> prints george

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。

我们可以通过以下方法得到dict的Key:

def getKey(dct,value):
     return [key for key in dct if (dct[key] == value)]