Find keys through values in a dict for Python

Question

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(nameagelist)

    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?

Your example data does not match your question. You start with two lists, not a dictionary. If your data starts in the type of a dict where key = name and value = age, do a reverse dictionary lookup: Some discussion here http://stackoverflow.com/questions/2568673/inverse-dictionary-lookup-python#2568673 — DevPlayer, Jul 24 '15 at 01:37
Also your function named "name" is misleading. Perhaps it should be named "names" as there are multiple names for most ages. Also, your data implies a potential problem. Even though the actual data may not be people's names nor people's ages, in data of any size both objects are very likely not unique. In your example there could often be more then one John Smith born on a certain day. — DevPlayer, Jul 24 '15 at 01:43

score 9 · Accepted Answer · answered Feb 03 '11 at 03:10

9

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.

answered Feb 03 '11 at 03:10

payne

13,833
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1

In Python 2, you often want `iteritems` rather than `items`. – Glenn Maynard Feb 03 '11 at 07:27

score 0 · Answer 2 · answered Nov 04 '13 at 17:39

0

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())

answered Nov 04 '13 at 17:39

UltramaticOrange

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1
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What I actually need is a reverse-lookup dictionary. Of course, this only works when Key:value is a 1 to 1 relationship (never 1 to many). `dict(map(lambda (k,v): (v,k), {1:2, 3:4, 5:6}.items()))` – UltramaticOrange Nov 06 '13 at 20:05

score 0 · Answer 3 · answered Jul 24 '15 at 01:57

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)

Find keys through values in a dict for Python

3 Answers3