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

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

其他回答

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……

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
get_key = lambda v, d: next(k for k in d if d[k] is v)

一个简单的方法是:

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

这将返回值与search_age匹配的键的列表。如果需要,还可以将"list==search_age"替换为任何其他条件语句。

它被回答了,但它可以用一个奇特的“映射/减少”使用来完成,例如:

def find_key(value, dictionary):
    return reduce(lambda x, y: x if x is not None else y,
                  map(lambda x: x[0] if x[1] == value else None, 
                      dictionary.iteritems()))