# Python – Frequency of unequal items in Dictionary

Sometimes, while working with Python, we can come across a problem in which we need to check for the unequal items count among two dictionaries. This has an application in cases of web development and other domains as well. Let’s discuss certain ways in which this task can be performed.

Method #1 : Using dictionary comprehension
This particular task can be performed in one line using dictionary comprehension which offers a way of compacting lengthy brute logic and just checks for unequal items and increments count.

 `# Python3 code to demonstrate working of ` `# Dissimilar items frequency in Dictionary ` `# Using dictionary comprehension ` ` `  `# initializing dictionaries ` `test_dict1 ``=` `{``'gfg'` `: ``1``, ``'is'` `: ``2``, ``'best'` `: ``3``} ` `test_dict2 ``=` `{``'gfg'` `: ``1``, ``'is'` `: ``2``, ``'good'` `: ``3``} ` ` `  `# printing original dictionaries ` `print``(``"The original dictionary 1 is : "` `+` `str``(test_dict1)) ` `print``(``"The original dictionary 2 is : "` `+` `str``(test_dict2)) ` ` `  `# Dissimilar items frequency in Dictionary ` `# Using dictionary comprehension ` `res ``=` `{key: test_dict1[key] ``for` `key ``in` `test_dict1 ``if` `key ``not` `in` `test_dict2} ` ` `  `# printing result ` `print``(``"The number of uncommon items are : "` `+` `str``(``len``(res))) `

Output :

```The original dictionary 1 is : {'best': 3, 'is': 2, 'gfg': 1}
The original dictionary 2 is : {'good': 3, 'is': 2, 'gfg': 1}
The number of uncommon items are : 1
```

Method #2 : Using` set()` + XOR operator + `items()`
The combination of above methods can be used to perform this particular task. In this, the set function removes duplicates and XOR operator computes the matched items. Result can be computed by subtracting original dict. length by new dict. length.

 `# Python3 code to demonstrate working of ` `# Dissimilar items frequency in Dictionary ` `# Using set() + XOR operator + items() ` ` `  `# initializing dictionaries ` `test_dict1 ``=` `{``'gfg'` `: ``1``, ``'is'` `: ``2``, ``'best'` `: ``3``} ` `test_dict2 ``=` `{``'gfg'` `: ``1``, ``'is'` `: ``2``, ``'good'` `: ``3``} ` ` `  `# printing original dictionaries ` `print``(``"The original dictionary 1 is : "` `+` `str``(test_dict1)) ` `print``(``"The original dictionary 2 is : "` `+` `str``(test_dict2)) ` ` `  `# Dissimilar items frequency in Dictionary ` `# Using set() + XOR operator + items() ` `res ``=` `set``(test_dict1.items()) ^ ``set``(test_dict2.items()) ` ` `  `# printing result ` `print``(``"The number of uncommon items are : "` `+` `str``(``len``(test_dict1) ``-` `len``(res))) `

Output :

```The original dictionary 1 is : {'best': 3, 'is': 2, 'gfg': 1}
The original dictionary 2 is : {'good': 3, 'is': 2, 'gfg': 1}
The number of uncommon items are : 1
```

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