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.

Answer 1

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.

Answer 2

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 :)

Answer 3

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)

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

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.

Answer 8

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