# Python – Convert dictionary to K sized dictionaries

• Last Updated : 02 Sep, 2020

Given a Dictionary, divide dictionary into K sized different dictionaries list.

Input : test_dict = {‘Gfg’ : 1, ‘is’ : 2, ‘best’ : 3, ‘for’ : 4, ‘geeks’ : 5, ‘CS’ : 6}, K = 3
Output : [{‘Gfg’: 1, ‘is’: 2, ‘best’: 3}, {‘for’: 4, ‘geeks’: 5, ‘CS’: 6}]
Explanation : Divided into size of 3 keys.

Input : test_dict = {‘Gfg’ : 1, ‘is’ : 2, ‘best’ : 3, ‘for’ : 4}, K = 2
Output : [{‘Gfg’: 1, ‘is’: 2}, {‘best’: 3, ‘for’: 4}]
Explanation : Divided into size of 2 keys.

Method : Using loop

In this, we iterate for all the keys in dictionary using loop and bifurcate according to size  and append to new list.

## Python3

 `# Python3 code to demonstrate working of ``# Convert dictionary to K Keys dictionaries``# Using loop`` ` `# initializing dictionary``test_dict ``=` `{``'Gfg'` `: ``1``, ``'is'` `: ``2``, ``'best'` `: ``3``, ``'for'` `: ``4``, ``'geeks'` `: ``5``, ``'CS'` `: ``6``}`` ` `# printing original dictionary``print``(``"The original dictionary is : "` `+` `str``(test_dict))`` ` `# initializing K ``K ``=` `2`` ` `res ``=` `[]``count ``=` `0``flag ``=` `0``indict ``=` `dict``()``for` `key ``in` `test_dict:``    ``indict[key] ``=` `test_dict[key]        ``    ``count ``+``=` `1``     ` `    ``# checking for K size and avoiding empty dict using flag``    ``if` `count ``%` `K ``=``=` `0` `and` `flag:``        ``res.append(indict)``         ` `        ``# reinitializing dictionary``        ``indict ``=` `dict``()``        ``count ``=` `0``    ``flag ``=` `1``     ` ` ` `# printing result ``print``(``"The converted list : "` `+` `str``(res)) `

Output

```The original dictionary is : {'Gfg': 1, 'is': 2, 'best': 3, 'for': 4, 'geeks': 5, 'CS': 6}
The converted list : [{'Gfg': 1, 'is': 2}, {'best': 3, 'for': 4}, {'geeks': 5, 'CS': 6}]
```

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