# Counting the frequencies in a list using dictionary in Python

Given an unsorted list of some elements(may or may not be integers), Find the frequency of each distinct element in the list using a dictionary.

Example:

```Input : [1, 1, 1, 5, 5, 3, 1, 3, 3, 1,
4, 4, 4, 2, 2, 2, 2]
Output : 1 : 5
2 : 4
3 : 3
4 : 3
5 : 2
Explanation : Here 1 occurs 5 times, 2
occurs 4 times and so on...
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The problem can be solved in many ways. A simple approach would be to iterate over the list and use each distinct element of the list as a key of the dictionary and store the corresponding count of that key as values. Below is the Python code for this approach:

 `# Python program to count the frequency of  ` `# elements in a list using a dictionary ` ` `  `def` `CountFrequency(my_list): ` ` `  `    ``# Creating an empty dictionary  ` `    ``freq ``=` `{} ` `    ``for` `item ``in` `my_list: ` `        ``if` `(item ``in` `freq): ` `            ``freq[item] ``+``=` `1` `        ``else``: ` `            ``freq[item] ``=` `1` ` `  `    ``for` `key, value ``in` `freq.items(): ` `        ``print` `(``"% d : % d"``%``(key, value)) ` ` `  `# Driver function ` `if` `__name__ ``=``=` `"__main__"``:  ` `    ``my_list ``=``[``1``, ``1``, ``1``, ``5``, ``5``, ``3``, ``1``, ``3``, ``3``, ``1``, ``4``, ``4``, ``4``, ``2``, ``2``, ``2``, ``2``] ` ` `  `    ``CountFrequency(my_list) `

Output:

``` 1 :  5
2 :  4
3 :  3
4 :  3
5 :  2
```

Time Complexity:O(N), where N is the length of the list.

Alternative way:An alternative approach can be to use the list.count() method. This makes the program much more compact without affecting the run time.Below is the Python code for this:

 `# Python program to count the frequency of ` `# elements in a list using a dictionary ` ` `  `def` `CountFrequency(my_list): ` `     `  `    ``# Creating an empty dictionary  ` `    ``freq ``=` `{} ` `    ``for` `items ``in` `my_list: ` `        ``freq[items] ``=` `my_list.count(items) ` `     `  `    ``for` `key, value ``in` `freq.items(): ` `        ``print` `(``"% d : % d"``%``(key, value)) ` ` `  `# Driver function ` `if` `__name__ ``=``=` `"__main__"``:  ` `    ``my_list ``=``[``1``, ``1``, ``1``, ``5``, ``5``, ``3``, ``1``, ``3``, ``3``, ``1``, ``4``, ``4``, ``4``, ``2``, ``2``, ``2``, ``2``] ` `    ``CountFrequency(my_list) `

Output:

``` 1 :  5
2 :  4
3 :  3
4 :  3
5 :  2
```

Time Complexity:O(N), where N is the length of the list.

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