# Python – Swapping Hierarchy in Nested Dictionaries

• Last Updated : 16 Jun, 2021

Sometimes, while working with Python dictionaries, we can have a problem in which we need to perform hierarchy swapping of nested dictionaries. This problem can have application in domains which require dictionary restructuring. Let’s discuss certain ways in which this task can be performed.

Input : test_dict = {‘Gfg’: { ‘a’ : [1, 3, 7, 8], ‘b’ : [4, 9], ‘c’ : [0, 7]}}
Output : {‘c’: {‘Gfg’: [0, 7]}, ‘b’: {‘Gfg’: [4, 9]}, ‘a’: {‘Gfg’: [1, 3, 7, 8]}}
Input : test_dict = {‘Gfg’: {‘best’ : [1, 3, 4]}}
Output : {‘best’: {‘Gfg’: [1, 3, 4]}}

Method #1 : Using loop + items()
The combination of above functions can be used to solve this problem. In this, we iterate the dictionary and restructure using brute force. The items() is used to extract all the key-value pairs of dictionary.

## Python3

 `# Python3 code to demonstrate working of``# Swapping Hierarchy in Nested Dictionaries``# Using loop + items()` `# initializing dictionary``test_dict ``=` `{``'Gfg'``: { ``'a'` `: [``1``, ``3``], ``'b'` `: [``3``, ``6``], ``'c'` `: [``6``, ``7``, ``8``]},``             ``'Best'``: { ``'a'` `: [``7``, ``9``], ``'b'` `: [``5``, ``3``, ``2``], ``'d'` `: [``0``, ``1``, ``0``]}}` `# printing original dictionary``print``(``"The original dictionary : "` `+` `str``(test_dict))` `# Swapping Hierarchy in Nested Dictionaries``# Using loop + items()``res ``=` `dict``()``for` `key, val ``in` `test_dict.items():``    ``for` `key_in, val_in ``in` `val.items():``        ``if` `key_in ``not` `in` `res:``            ``temp ``=` `dict``()``        ``else``:``            ``temp ``=` `res[key_in]``        ``temp[key] ``=` `val_in``        ``res[key_in] ``=` `temp` `# printing result``print``(``"The rearranged dictionary : "` `+` `str``(res))`

Output :

The original dictionary : {‘Gfg’: {‘a’: [1, 3], ‘c’: [6, 7, 8], ‘b’: [3, 6]}, ‘Best’: {‘d’: [0, 1, 0], ‘a’: [7, 9], ‘b’: [5, 3, 2]}}
The rearranged dictionary : {‘d’: {‘Best’: [0, 1, 0]}, ‘a’: {‘Gfg’: [1, 3], ‘Best’: [7, 9]}, ‘c’: {‘Gfg’: [6, 7, 8]}, ‘b’: {‘Gfg’: [3, 6], ‘Best’: [5, 3, 2]}}

Method #2 : Using defaultdict() + loop
This is yet another brute force way to solve this problem. In this, we reduce a key test step by using defaultdict() rather than conventional dictionary.

## Python3

 `# Python3 code to demonstrate working of``# Swapping Hierarchy in Nested Dictionaries``# Using  defaultdict() + loop``from` `collections ``import` `defaultdict` `# initializing dictionary``test_dict ``=` `{``'Gfg'``: { ``'a'` `: [``1``, ``3``], ``'b'` `: [``3``, ``6``], ``'c'` `: [``6``, ``7``, ``8``]},``             ``'Best'``: { ``'a'` `: [``7``, ``9``], ``'b'` `: [``5``, ``3``, ``2``], ``'d'` `: [``0``, ``1``, ``0``]}}` `# printing original dictionary``print``(``"The original dictionary : "` `+` `str``(test_dict))` `# Swapping Hierarchy in Nested Dictionaries``# Using  defaultdict() + loop``res ``=` `defaultdict(``dict``)``for` `key, val ``in` `test_dict.items():``    ``for` `key_in, val_in ``in` `val.items():``        ``res[key_in][key] ``=` `val_in` `# printing result``print``(``"The rearranged dictionary : "` `+` `str``(``dict``(res)))`

Output :

The original dictionary : {‘Gfg’: {‘a’: [1, 3], ‘c’: [6, 7, 8], ‘b’: [3, 6]}, ‘Best’: {‘d’: [0, 1, 0], ‘a’: [7, 9], ‘b’: [5, 3, 2]}}
The rearranged dictionary : {‘d’: {‘Best’: [0, 1, 0]}, ‘a’: {‘Gfg’: [1, 3], ‘Best’: [7, 9]}, ‘c’: {‘Gfg’: [6, 7, 8]}, ‘b’: {‘Gfg’: [3, 6], ‘Best’: [5, 3, 2]}}

My Personal Notes arrow_drop_up