How to get key of a value in python dictionary?

Question

Dictionary values are lists:

dictionary_1 = {"ABC": [1,2,3], "DEF":[4,5,6]}

How do I get key of 5, which is "DEF" from dictionary_1 in python?

Do you mean you have an input of `5` and you want to return `"DEF"`? Or the other way around? — lc., Feb 18 '14 at 08:39
You probably want to use a two-way dictionary: https://pypi.python.org/pypi/bidict/0.1.1 — simonzack, Feb 18 '14 at 08:40
@lc. yes, I have `5`, and I want to get its key (which id "DEF"). — alwbtc, Feb 18 '14 at 08:43
@simonzack that question's answers don't look like answers posted below. because they are not duplicates. — alwbtc, Feb 18 '14 at 08:51
@alwbtc: yet they are closely related. Check the linked posts in the sidebar too. — Martijn Pieters, Feb 18 '14 at 08:55

ndpu · Answer 1 · 2014-02-18T08:48:21.250

2

You can create reverse dict:

>>> d = {i:k for k,v in dictionary_1.items() for i in v}
>>> d
{1: 'ABC', 2: 'ABC', 3: 'ABC', 4: 'DEF', 5: 'DEF', 6: 'DEF'}
>>> d[5]
'DEF'

edited Feb 18 '14 at 08:48

answered Feb 18 '14 at 08:40

ndpu

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Provided the values are unique.. – Martijn Pieters Feb 18 '14 at 08:41
This is a nice one because the op wanted to default to a key `GHJ` (in comments to my answer). So you can just call `d.get(5, "GHJ")` – msvalkon Feb 18 '14 at 09:01

msvalkon · Answer 2 · 2014-02-18T08:53:25.570

1

Maybe like this?

my_value = 5
for k, v in dictionary_1.iteritems():
    if my_value in v:
        print k
        break
else:
    print "No key, defaulting to GHJ"

Demo:

    In [12]: my_value = 8

    In [13]: for k, v in dictionary_1.iteritems():
        if my_value in v:
            print "Key for %s is %s" % (my_value, k)
            break
    else:
        print "No key found for %s, defaulting to GHJ" % my_value

    No key found for 8, defaulting to GHJ

edited Feb 18 '14 at 08:53

answered Feb 18 '14 at 08:39

msvalkon

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If item is not found in dictionary, then it should return a default value, like "GHJ". How would you do it above? – alwbtc Feb 18 '14 at 08:47
@alwbtc updated my answer. It's for/else pattern, if the key is not found, the code will never `break` and the `else` clause is executed. – msvalkon Feb 18 '14 at 08:54

score 1 · Answer 3 · answered Feb 18 '14 at 08:39

1

You'll have to search through the dictionary for that:

try:
    key = next(k for k, v in dictionary_1.iteritems() if 5 in v)
except StopIteration:
    raise KeyError('Key for 5 not found')

This assumes you are looking for a key. To find all keys you can use a list comprehension:

keys = [k for k, v in dictionary_1.iteritems() if 5 in v]

where the list could be empty.

answered Feb 18 '14 at 08:39

Martijn Pieters

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Haven't seen using `next` in this way..thanks. – rajpy Feb 19 '14 at 03:19
@alwbtc: If you want default value, use this `key = next((k for k, v in d.iteritems() if 9 in v), 'GHJ')` and remove try..except. – rajpy Feb 19 '14 at 03:22

score 0 · Answer 4 · answered Feb 18 '14 at 08:40

0

One way to do it is:

dictionary_1 = {"ABC": [1,2,3], "DEF":[4,5,6]}
for k, v in dictionary_1.items():
    if 5 in v:
        print(k)

answered Feb 18 '14 at 08:40

Douglas Denhartog

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score 0 · Answer 5 · answered Feb 18 '14 at 11:01

0

Assuming you can check if 5 is in [4,5,6] list, you can also try:

dictionary_1 = {"ABC": [1,2,3], "DEF":[4,5,6]}

print dictionary_1.keys()[dictionary_1.values().index([4,5,6])]

answered Feb 18 '14 at 11:01

tamiros

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How to get key of a value in python dictionary?

5 Answers5