python - Loop through all nested dictionary values?

Question

for k, v in d.iteritems():
    if type(v) is dict:
        for t, c in v.iteritems():
            print "{0} : {1}".format(t, c)

I'm trying to loop through a dictionary and print out all key value pairs where the value is not a nested dictionary. If the value is a dictionary I want to go into it and print out its key value pairs...etc. Any help?

EDIT

How about this? It still only prints one thing.

def printDict(d):
    for k, v in d.iteritems():
        if type(v) is dict:
            printDict(v)
        else:
            print "{0} : {1}".format(k, v)

Full Test Case

Dictionary:

{u'xml': {u'config': {u'portstatus': {u'status': u'good'}, u'target': u'1'},
      u'port': u'11'}}

Result:

xml : {u'config': {u'portstatus': {u'status': u'good'}, u'target': u'1'}, u'port': u'11'}

Answer 1

As said by Niklas, you need recursion, i.e. you want to define a function to print your dict, and if the value is a dict, you want to call your print function using this new dict.

Something like :

def myprint(d):
    for k, v in d.items():
        if isinstance(v, dict):
            myprint(v)
        else:
            print("{0} : {1}".format(k, v))

Answer 2

There are potential problems if you write your own recursive implementation or the iterative equivalent with stack. See this example:

    dic = {}
    dic["key1"] = {}
    dic["key1"]["key1.1"] = "value1"
    dic["key2"]  = {}
    dic["key2"]["key2.1"] = "value2"
    dic["key2"]["key2.2"] = dic["key1"]
    dic["key2"]["key2.3"] = dic

In the normal sense, nested dictionary will be a n-nary tree like data structure. But the definition doesn't exclude the possibility of a cross edge or even a back edge (thus no longer a tree). For instance, here key2.2 holds to the dictionary from key1, key2.3 points to the entire dictionary(back edge/cycle). When there is a back edge(cycle), the stack/recursion will run infinitely.

                          root<-------back edge
                        /      \           |
                     _key1   __key2__      |
                    /       /   \    \     |
               |->key1.1 key2.1 key2.2 key2.3
               |   /       |      |
               | value1  value2   |
               |                  | 
              cross edge----------|

If you print this dictionary with this implementation from Scharron

    def myprint(d):
      for k, v in d.items():
        if isinstance(v, dict):
          myprint(v)
        else:
          print "{0} : {1}".format(k, v)

You would see this error:

    RuntimeError: maximum recursion depth exceeded while calling a Python object

The same goes with the implementation from senderle.

Similarly, you get an infinite loop with this implementation from Fred Foo:

    def myprint(d):
        stack = list(d.items())
        while stack:
            k, v = stack.pop()
            if isinstance(v, dict):
                stack.extend(v.items())
            else:
                print("%s: %s" % (k, v))

However, Python actually detects cycles in nested dictionary:

    print dic
    {'key2': {'key2.1': 'value2', 'key2.3': {...}, 
       'key2.2': {'key1.1': 'value1'}}, 'key1': {'key1.1': 'value1'}}

"{...}" is where a cycle is detected.

As requested by Moondra this is a way to avoid cycles (DFS):

def myprint(d): 
    stack = list(d.items()) 
    visited = set() 
    while stack: 
        k, v = stack.pop() 
        if isinstance(v, dict): 
            if k not in visited: 
                stack.extend(v.items()) 
        else: 
            print("%s: %s" % (k, v)) 
        visited.add(k)

Answer 3

Since a dict is iterable, you can apply the classic nested container iterable formula to this problem with only a couple of minor changes. Here's a Python 2 version (see below for 3):

import collections
def nested_dict_iter(nested):
    for key, value in nested.iteritems():
        if isinstance(value, collections.Mapping):
            for inner_key, inner_value in nested_dict_iter(value):
                yield inner_key, inner_value
        else:
            yield key, value

Test:

list(nested_dict_iter({'a':{'b':{'c':1, 'd':2}, 
                            'e':{'f':3, 'g':4}}, 
                       'h':{'i':5, 'j':6}}))
# output: [('g', 4), ('f', 3), ('c', 1), ('d', 2), ('i', 5), ('j', 6)]

In Python 2, It might be possible to create a custom Mapping that qualifies as a Mapping but doesn't contain iteritems, in which case this will fail. The docs don't indicate that iteritems is required for a Mapping; on the other hand, the source gives Mapping types an iteritems method. So for custom Mappings, inherit from collections.Mapping explicitly just in case.

In Python 3, there are a number of improvements to be made. As of Python 3.3, abstract base classes live in collections.abc. They remain in collections too for backwards compatibility, but it's nicer having our abstract base classes together in one namespace. So this imports abc from collections. Python 3.3 also adds yield from, which is designed for just these sorts of situations. This is not empty syntactic sugar; it may lead to faster code and more sensible interactions with coroutines.

from collections import abc
def nested_dict_iter(nested):
    for key, value in nested.items():
        if isinstance(value, abc.Mapping):
            yield from nested_dict_iter(value)
        else:
            yield key, value

Answer 4

Alternative iterative solution:

def myprint(d):
    stack = d.items()
    while stack:
        k, v = stack.pop()
        if isinstance(v, dict):
            stack.extend(v.iteritems())
        else:
            print("%s: %s" % (k, v))

Answer 5

Slightly different version I wrote that keeps track of the keys along the way to get there

def print_dict(v, prefix=''):
    if isinstance(v, dict):
        for k, v2 in v.items():
            p2 = "{}['{}']".format(prefix, k)
            print_dict(v2, p2)
    elif isinstance(v, list):
        for i, v2 in enumerate(v):
            p2 = "{}[{}]".format(prefix, i)
            print_dict(v2, p2)
    else:
        print('{} = {}'.format(prefix, repr(v)))

On your data, it'll print

data['xml']['config']['portstatus']['status'] = u'good'
data['xml']['config']['target'] = u'1'
data['xml']['port'] = u'11'

It's also easy to modify it to track the prefix as a tuple of keys rather than a string if you need it that way.

Answer 6

Here is pythonic way to do it. This function will allow you to loop through key-value pair in all the levels. It does not save the whole thing to the memory but rather walks through the dict as you loop through it

def recursive_items(dictionary):
    for key, value in dictionary.items():
        if type(value) is dict:
            yield (key, value)
            yield from recursive_items(value)
        else:
            yield (key, value)

a = {'a': {1: {1: 2, 3: 4}, 2: {5: 6}}}

for key, value in recursive_items(a):
    print(key, value)

Prints

a {1: {1: 2, 3: 4}, 2: {5: 6}}
1 {1: 2, 3: 4}
1 2
3 4
2 {5: 6}
5 6

Answer 7

A alternative solution to work with lists based on Scharron's solution

def myprint(d):
    my_list = d.iteritems() if isinstance(d, dict) else enumerate(d)

    for k, v in my_list:
        if isinstance(v, dict) or isinstance(v, list):
            myprint(v)
        else:
            print u"{0} : {1}".format(k, v)

Answer 8

I am using the following code to print all the values of a nested dictionary, taking into account where the value could be a list containing dictionaries. This was useful to me when parsing a JSON file into a dictionary and needing to quickly check whether any of its values are None.

    d = {
            "user": 10,
            "time": "2017-03-15T14:02:49.301000",
            "metadata": [
                {"foo": "bar"},
                "some_string"
            ]
        }


    def print_nested(d):
        if isinstance(d, dict):
            for k, v in d.items():
                print_nested(v)
        elif hasattr(d, '__iter__') and not isinstance(d, str):
            for item in d:
                print_nested(item)
        elif isinstance(d, str):
            print(d)

        else:
            print(d)

    print_nested(d)

Output:

    10
    2017-03-15T14:02:49.301000
    bar
    some_string

Answer 9

Iterative solution as an alternative:

def traverse_nested_dict(d):
    iters = [d.iteritems()]

    while iters:
        it = iters.pop()
        try:
            k, v = it.next()
        except StopIteration:
            continue

        iters.append(it)

        if isinstance(v, dict):
            iters.append(v.iteritems())
        else:
            yield k, v


d = {"a": 1, "b": 2, "c": {"d": 3, "e": {"f": 4}}}
for k, v in traverse_nested_dict(d):
    print k, v

Answer 10

Your question already has been answered well, but I recommend using isinstance(d, collections.Mapping) instead of isinstance(d, dict). It works for dict(), collections.OrderedDict(), and collections.UserDict().

The generally correct version is:

def myprint(d):
    for k, v in d.items():
        if isinstance(v, collections.Mapping):
            myprint(v)
        else:
            print("{0} : {1}".format(k, v))

Answer 11

Here's a modified version of Fred Foo's answer for Python 2. In the original response, only the deepest level of nesting is output. If you output the keys as lists, you can keep the keys for all levels, although to reference them you need to reference a list of lists.

Here's the function:

def NestIter(nested):
    for key, value in nested.iteritems():
        if isinstance(value, collections.Mapping):
            for inner_key, inner_value in NestIter(value):
                yield [key, inner_key], inner_value
        else:
            yield [key],value

To reference the keys:

for keys, vals in mynested: 
    print(mynested[keys[0]][keys[1][0]][keys[1][1][0]])

for a three-level dictionary.

You need to know the number of levels before to access multiple keys and the number of levels should be constant (it may be possible to add a small bit of script to check the number of nesting levels when iterating through values, but I haven't yet looked at this).

Answer 12

I find this approach a bit more flexible, here you just providing generator function that emits key, value pairs and can be easily extended to also iterate over lists.

def traverse(value, key=None):
    if isinstance(value, dict):
        for k, v in value.items():
            yield from traverse(v, k)
    else:
        yield key, value

Then you can write your own myprint function, then would print those key value pairs.

def myprint(d):
    for k, v in traverse(d):
        print(f"{k} : {v}")

A test:

myprint({
    'xml': {
        'config': {
            'portstatus': {
                'status': 'good',
            },
            'target': '1',
        },
        'port': '11',
    },
})

Output:

status : good
target : 1
port : 11

I tested this on Python 3.6.

Answer 13

These answers work for only 2 levels of sub-dictionaries. For more try this:

nested_dict = {'dictA': {'key_1': 'value_1', 'key_1A': 'value_1A','key_1Asub1': {'Asub1': 'Asub1_val', 'sub_subA1': {'sub_subA1_key':'sub_subA1_val'}}},
                'dictB': {'key_2': 'value_2'},
                1: {'key_3': 'value_3', 'key_3A': 'value_3A'}}

def print_dict(dictionary):
    dictionary_array = [dictionary]
    for sub_dictionary in dictionary_array:
        if type(sub_dictionary) is dict:
            for key, value in sub_dictionary.items():
                print("key=", key)
                print("value", value)
                if type(value) is dict:
                    dictionary_array.append(value)



print_dict(nested_dict)