How to iterate through a nested dict?

Question

I have a nested python dictionary data structure. I want to read its keys and values without using collection module. The data structure is like bellow.

d = {'dict1': {'foo': 1, 'bar': 2}, 'dict2': {'baz': 3, 'quux': 4}}

I was trying to read the keys in the dictionary using the bellow way but getting error.

Code

for key, value in d:
    print(Key)

Error

ValueError: too many values to unpack (expected 2)

So can anyone please explain the reason behind the error and how to iterate through the dictionary.

just write `d.items()`, it will work, by default on iterating the name of dict returns only the keys. Hope it help. :) — Sunil Lulla, May 03 '17 at 06:55
yes. I want to read all the keys and values that include the nested keys an values — Arijit Panda, May 03 '17 at 06:55
@Arijit Posted an answer for the requested output, have a look. — Sunil Lulla, May 03 '17 at 07:10
**See also:** [search for key in nested dict](https://stackoverflow.com/questions/7681301/search-for-a-key-in-a-nested-python-dictionary) -- [python dpath](https://stackoverflow.com/a/16508328/42223) — dreftymac, Oct 30 '17 at 20:17
General answer of this question isn't answered here but is answered elsewhere: https://stackoverflow.com/questions/10756427/loop-through-all-nested-dictionary-values — TimZaman, Aug 01 '18 at 23:46

score 32 · Answer 1 · answered Feb 13 '21 at 19:39

32

The following will work with multiple levels of nested-dictionary:

def get_all_keys(d):
    for key, value in d.items():
        yield key
        if isinstance(value, dict):
            yield from get_all_keys(value)


d = {'dict1': {'foo': 1, 'bar': 2}, 'dict2': {'dict3': {'baz': 3, 'quux': 4}}}
for x in get_all_keys(d):
    print(x)

This will give you:

dict1
foo
bar
dict2
dict3
baz
quux

answered Feb 13 '21 at 19:39

0x0

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Recursion can really save you a lot of time from manually specifying logic for each level of your nested dictionary – SuperCode Apr 07 '22 at 09:27

bharatk · Answer 2 · 2019-09-19T05:26:16.180

keys() method returns a view object that displays a list of all the keys in the dictionary

Iterate nested dictionary:

d = {'dict1': {'foo': 1, 'bar': 2}, 'dict2': {'baz': 3, 'quux': 4}}

for i in d.keys():
    print i
    for j in d[i].keys():
        print j

OR

for i in d:
    print i
    for j in d[i]:
        print j

output:

dict1 
foo
bar

dict2
baz 
quux

where i iterate main dictionary key and j iterate the nested dictionary key.

Sunil Lulla · Accepted Answer · 2018-09-26T10:49:04.217

9

As the requested output, the code goes like this

    d = {'dict1': {'foo': 1, 'bar': 2}, 'dict2': {'baz': 3, 'quux': 4}}

    for k1,v1 in d.iteritems(): # the basic way
        temp = ""   
        temp+=k1
        for k2,v2 in v1.iteritems():
           temp = temp+" "+str(k2)+" "+str(v2)
        print temp

In place of iteritems() you can use items() as well, but iteritems() is much more efficient and returns an iterator.

Hope this helps :)

edited Sep 26 '18 at 10:49

answered May 03 '17 at 07:10

Sunil Lulla

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The `iteritems` method does not exist for Python 3. – equaeghe Feb 13 '19 at 12:01

score 5 · Answer 4 · answered Aug 15 '20 at 10:46

you can iterate all keys and values of nested dictionary as following:

d = {'dict1': {'foo': 1, 'bar': 2}, 'dict2': {'baz': 3, 'quux': 4}}

for i in d:
    for j, k in d[i].items():
        print(j,"->", k)

Your output looks like this -

foo -> 1
bar -> 2
baz -> 3
quux -> 4

score 4 · Answer 5 · answered May 03 '17 at 06:47

4

To get keys and values you need dict.items():

for key, value in d.items():
    print(key)

If you want just the keys:

for key in d:
    print(key)

answered May 03 '17 at 06:47

Stephen Rauch

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For python2 you could prefer `dict.iteritems()` instead, it is faster for this use case. – jadsq May 03 '17 at 06:52
8

This is not an answer to "how to traverse a nested dict". – TimZaman Aug 01 '18 at 23:45

score 3 · Answer 6 · answered May 03 '17 at 06:48

Iterating through a dictionary only gives you the keys.

You told python to expect a bunch of tuples, and it tried to unpack something that wasn't a tuple (your code is set up to expect each iterated item to be of the form (key,value), which was not the case (you were simply getting key on each iteration).

You also tried to print Key, which is not the same as key, which would have led to a NameError.

for key in d:
    print(key)

should work.

score 2 · Answer 7 · edited Oct 29 '20 at 17:07

2

You could use benedict (a dict subclass) and the traverse utility method:

Installation: pip install python-benedict

from benedict import benedict

d = benedict({'dict1': {'foo': 1, 'bar': 2}, 'dict2': {'baz': 3, 'quux': 4}})

def traverse_item(dct, key, value):
   print('key: {} - value: {}'.format(key, value))

d.traverse(traverse_item)

Documentation: https://github.com/fabiocaccamo/python-benedict

Note: I am the author of this project.

edited Oct 29 '20 at 17:07

Martijn Pieters

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answered Oct 29 '19 at 16:20

Fabio Caccamo

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Checked it out and seems it's perfect for NoSql data parsing. Perfect Package!! – SuperCode Apr 07 '22 at 09:46

score 2 · Answer 8 · answered Apr 17 '20 at 01:45

if given dictionary pattern has monotone format and keys are known

dict_ = {'0': {'foo': 1, 'bar': 2}, '1': {'foo': 3, 'bar': 4}}
for key, val in dict_.items():
    if isinstance(val, dict):
        print(val.get('foo'))
        print(val.get('bar'))

in this case we can skip nested loop

How to iterate through a nested dict?

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