# Count dictionaries in a list in Python

• Last Updated : 29 Apr, 2020

A list in Python may have items of different types. Sometimes, while working with data, we can have a problem in which we need to find the count of dictionaries in particular list. This can have application in data domains including web development and Machine Learning. Lets discuss certain ways in which this task can be performed.

Input : test_list = [4, 5, ‘gfg’]
Output : 0

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Input : test_list = [{‘gfg’ : 1}]
Output : 1

Input : test_list = [10, {‘gfg’ : 1}, {‘ide’ : 2, ‘code’ : 3}, 20]
Output : 2

Input : test_list = [4, 5, ‘gfg’, {‘best’: 32, ‘gfg’: 1}, {‘CS’: 4}, (1, 2)]
Output : 2

Method #1 : Using list comprehension + `isinstance()`
The combination of above functionalities can be used to solve this problem. In this, we perform iteration using list comprehension and test for dictionary using isinstance().

 `# Python3 code to demonstrate working of ``# Dictionary Count in List``# Using list comprehension + isinstance()`` ` `# initializing list``test_list ``=` `[``10``, {``'gfg'` `: ``1``}, {``'ide'` `: ``2``, ``'code'` `: ``3``}, ``20``]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# Dictionary Count in List``# Using list comprehension + isinstance()``res ``=` `len``([ele ``for` `ele ``in` `test_list ``if` `isinstance``(ele, ``dict``)])`` ` `# printing result ``print``(``"The Dictionary count : "` `+` `str``(res)) `
Output:
```The original list is : [10, {'gfg': 1}, {'code': 3, 'ide': 2}, 20]
The Dictionary count : 2
```

Method #2 : Using recursion + `isinstance()` ( for nested dictionaries)
The combination of above functionalities can be used to solve this problem. In this, we also solve the problem of inner nesting using recursion.

 `# Python3 code to demonstrate working of ``# Dictionary Count in List``# Using recursion + isinstance()`` ` `# helper_func``def` `hlper_fnc(test_list):``    ``count ``=` `0``    ``if` `isinstance``(test_list, ``str``):``        ``return` `0``    ``if` `isinstance``(test_list, ``dict``):``        ``return` `hlper_fnc(test_list.values()) ``+` `hlper_fnc(test_list.keys()) ``+` `1``    ``try``:``        ``for` `idx ``in` `test_list:``            ``count ``=` `count ``+` `hlper_fnc(idx)``    ``except` `TypeError:``        ``return` `0``    ``return` `count`` ` `# initializing list``test_list ``=` `[``10``, {``'gfg'``: ``1``}, {``'code'``: ``3``, ``'ide'``: ``2``}, ``20``]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# Dictionary Count in List``# Using recursion + isinstance()``res ``=` `hlper_fnc(test_list)`` ` `# printing result ``print``(``"The Dictionary count : "` `+` `str``(res)) `
Output:
```The original list is : [10, {'gfg': 1}, {'code': 3, 'ide': 2}, 20]
The Dictionary count : 2
```

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