Python dictionary — find the key by entering the value

Question

I have a little code but I don't know how to find the key in a dictionary by using the value. Here I have my code and what I want to do:

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 combine(list1,list2):
    dictionary = dict(zip(list1,list2))
    return dictionary

def people(age):
    if leeftijd in dictionary:
        return ['Bob']


print combine(NAMES, AGES)

print people(21) == ['Bob'] 
print people(22) == ['Kelly']
print people(23) == []

In this code I first put the list NAMES and AGES into a dictionary by using the combine function, but now I want to make a second function named people that will output all the names of the people with the age you enter.

For example: if I enter 21 into that function, it should output Bob since Bob is the only one that's 21 years old.

I just have no idea how to do this since the AGE isn't the key in my dictionary and I also can't use an integer as a key.

You absolutely *can* use an integer as a dictionary key - it's hashable and immutable. — jonrsharpe, Sep 18 '14 at 11:14
How would you deal with 2 people being 21? Also, see http://stackoverflow.com/questions/8023306/get-key-by-value-in-dictionary — fredtantini, Sep 18 '14 at 12:52
Is `leeftijd` in `people()` a typo (missed translation) of `age`? (Google translate does translate 'leeftijd' to 'age'.) — Jonathan Leffler, Sep 18 '14 at 12:52
AFAIK, if you don't just do a linear search through the dictionary, you'll have to maintain an inverse mapping (dictionary), which will need to maintain a list of keys for each value (as there are 2 eighteen year olds, 4 nineteen year olds, and 2 twenty year olds in the data). — Jonathan Leffler, Sep 18 '14 at 12:55
If you find yourself wanting to do this, you are probably using the wrong datastructure. You should either invert the dictionary, or use a bidirectional dictionary. For example: https://pypi.python.org/pypi/bidict/ — Jörg W Mittag, Sep 18 '14 at 19:48
Expanding on @JörgWMittag comment. Consider changing the data structure. You could save just one key per age and have a list of people with that age. e.g. `ppl = {18:['Cathy','Dan'], ...}`. querying would be trivial: `ppl[18]` — fluxens, Apr 15 '19 at 17:50

score 1 · Answer 1 · edited May 23 '17 at 10:33

As I noted in a comment, AFAIK, if you don't just do a linear search through the dictionary, you'll have to maintain an inverse mapping (dictionary), which will need to maintain a list of keys for each value (as there are 2 eighteen year olds, 5 nineteen year olds, and 3 twenty year olds in the data).

This code includes a (general purpose) function reverse_dictionary() which takes a given dictionary and creates the reverse mapping from it. The name and age variable names in the function are a giveaway that I wrote it for this problem, but change them to key and value and the code works on any dictionary where the values are hashable.

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 reverse_dictionary(fwd):
    rev = {}
    for name, age in fwd.items():
        if age not in rev:
            rev[age] = []
        rev[age].append(name)
    return rev

def combine(list1,list2):
    dictionary = dict(zip(list1,list2))
    return dictionary

def people(age, dictionary):
    if age not in dictionary:
        return []
    return dictionary[age]

name_age = combine(NAMES, AGES)
print name_age

age_name = reverse_dictionary(name_age)

for age in range(17, 24):
    print age, people(age, age_name)

Sample output:

{'Ed': 19, 'Dan': 18, 'Gary': 20, 'Alice': 20, 'Kelly': 22, 'Larry': 19, 'Jack': 19, 'Frank': 20, 'Cathy': 18, 'Bob': 21, 'Irene': 19, 'Helen': 19}
17 []
18 ['Dan', 'Cathy']
19 ['Ed', 'Larry', 'Jack', 'Irene', 'Helen']
20 ['Gary', 'Alice', 'Frank']
21 ['Bob']
22 ['Kelly']
23 []

Note that this does not keep the two dictionaries synchronized; if you add a new name and age to the name_age dictionary, you'd also need to add the age and name to the reverse dictionary. On the other hand, if you load the name/age map once and it then remains stable and you need to do multiple lookups based on age, then the reversed dictionary is potentially much more efficient than a solution that iterates through the name/age dictionary for each search.

score 0 · Answer 2 · answered Sep 18 '14 at 12:54

0

Actually there is no way as far as I know.

Just iterate through your dict:

people_array = combine(NAMES, AGES)
def people(age):
    for person in people_array:
        if people_array[person] == age:
            print(person)

answered Sep 18 '14 at 12:54

Igl3

4,900
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score 0 · Answer 3 · answered Sep 18 '14 at 12:54

0

It can be done this way

for name, age in mydict.items():
    if age == search_age:
        print name

answered Sep 18 '14 at 12:54

Omair Shamshir

2,126
13
23

score 0 · Answer 4 · answered Sep 18 '14 at 13:06

Try this function (key_for_value):

def key_for_value(d, value):
    for k, v in d.iteritems():
        if v == value:
            return k
    raise ValueError("{!r} not found".format(value))


mydict = {1: "a", 2: "b", 3: "c"}

print key_for_value(mydict, "a")
print key_for_value(mydict, "d")

score 0 · Answer 5 · answered Sep 18 '14 at 13:16

0

A nice way to do it would be:

def people(age):
    ages = [i for i,j in enumerate(combine.values()) if j == age]
    return [combine.keys()[i] for i in ages]

answered Sep 18 '14 at 13:16

Amit

1

by using combine, I meant the variable that is the result of combine(NAMES,AGES) – Amit Sep 18 '14 at 13:21

score 0 · Answer 6 · answered Sep 18 '14 at 13:18

0

try this:

def people(age):
    return map(lambda x: x[0], filter(lambda x: x[1]==age, dictionary.items()))

answered Sep 18 '14 at 13:18

Jason Hu

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Python dictionary — find the key by entering the value

6 Answers6