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Counting the frequencies in a list using dictionary in Python

• Difficulty Level : Medium
• Last Updated : 24 Mar, 2023

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

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:

Approach 1: Counting the frequencies in a list using a loop

In this method, we will use a Python loop to count the distinct element from the list and append it to the dictionary.

Python3

 `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.

Approach 2: Count the frequencies in a list using  list.count()

An alternative approach can be to use the list.count() method.

Python3

 `import` `operator`     `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(N2), where N is the length of the list. The time complexity list.count() is O(N) alone, and when used inside the loop it will become O(N2).

Approach 3: Count the frequencies in a list using dict.get()

The dict.get() method, makes the program much shorter and makes understanding how the get method is useful instead of if…else.

Python3

 `def` `CountFrequency(my_list):`   `    ``# Creating an empty dictionary` `    ``count ``=` `{}` `    ``for` `i ``in` `[``1``, ``1``, ``1``, ``5``, ``5``, ``3``, ``1``, ``3``, ``3``, ``1``, ``4``, ``4``, ``4``, ``2``, ``2``, ``2``, ``2``]:` `        ``count[i] ``=` `count.get(i, ``0``) ``+` `1` `    ``return` `count`     `# Driver function` `if` `__name__ ``=``=` `"__main__"``:` `    ``my_list ``=` `[``1``, ``1``, ``1``, ``5``, ``5``, ``3``, ``1``, ``3``, ``3``, ``1``, ``4``, ``4``, ``4``, ``2``, ``2``, ``2``, ``2``]` `    ``print``(CountFrequency(my_list))`

Output:

`{1: 5, 5: 2, 3: 3, 4: 3, 2: 4}`

Count the frequencies in a list using  operator.countOf() method

Approach 4:

1. Created a dictionary with keys as unique list values and values as unique element count in list using for loop and operator.countOf() method.
2. Displayed the keys and values of the dictionary.

Python3

 `import` `operator`   `def` `CountFrequency(my_list):`   `    ``# Creating an empty dictionary` `    ``freq ``=` `{}` `    ``for` `items ``in` `my_list:` `        ``freq[items] ``=` `operator.countOf(my_list, 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
5 :  2
3 :  3
4 :  3
2 :  4```

Time Complexity : O(N*N)
N -length of list
Auxiliary Space : O(1)

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