Find keys through values in a dict for Python

NAMES  =  ['Alice', 'Bob','Cathy','Dan','Ed','Frank',
           'Gary','Helen','Irene','Jack', 'Kelly','Larry'] 
AGES  =  [20,21,18,18,19,20,20,19,19,19,22,19]

def nameage(a,b):
    nameagelist = [x for x in zip(a,b)]
    nameagedict = dict(na开发者_如何转开发meagelist)

    return nameagedict

def name(a):
    for x in nameage(NAMES,AGES):
        if a in nameage(NAMES,AGES).values():
            print nameage(NAMES,AGES).keys()

print name(19)

I am trying to return the names of the people aged 19. How do I search the dictionary by value and return the key?

print [key for (key,value) in nameagedict.items() if value == 19]

nameagedict.items() gives you a list of all items in the dictionary, as (key,value) tuples.

I had a similar problem. Here's my solution which returns the keys and values as an array of tuples.

filter(lambda (k,v): v==19, nameagedict.items())

Partial solution

names = {age:sorted([name for index,name in enumerate(NAMES) if age == AGES[index]]) for age in sorted(set(AGES))}
print(names)

Returns:

{18: ['Cathy', 'Dan'], 19: ['Ed', 'Helen', 'Irene', 'Jack', 'Larry'], 20: ['Alice', 'Frank', 'Gary'], 21: ['Bob'], 22: ['Kelly']}

Leave out sorted() if you wish. You could also add count() so the data returned something like

{18: [('Cathy',3), ('Dan',7), ...} 

if there were 3 Cathys and 7 Dans.

Although the above looks short I suspect an actual for loop would iterate over the original lists less.

Or even better:

def names(age):
    index = 0
    while index < len(AGES):
        if AGES[index] == age:
            yield NAMES[index]
        index += 1

print('19 year olds')
for name in names(19):
    print(name)