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# Python – Extract Similar pairs from List

Sometimes, while working with Python lists, we can have a problem in which we need to perform extraction of all the elements pairs in list. This kind of problem can have application in domains such as web development and day-day programming. Let’s discuss certain ways in which this task can be performed.

Input : test_list = [1, 2, 3, 4]
Output : []

Input : test_list = [2, 2, 2, 2, 3, 3, 4]
Output : [(2, 2), (2, 2), (3, 3)]

Method #1 : Using Counter() + list comprehension The combination of above functions can be used to solve this problem. In this, frequencies are extracted using Counter() and pairs construction is done used list comprehension.

## Python3

 `# Python3 code to demonstrate working of``# Extract Similar pairs from List``# Using Counter() + list comprehension``from` `collections ``import` `Counter` `# initializing list``test_list ``=` `[``4``, ``6``, ``7``, ``4``, ``2``, ``6``, ``2``, ``8``]` `# printing original list``print``("The original ``list` `is` `: " ``+` `str``(test_list))` `# Extract Similar pairs from List``# Using Counter() + list comprehension``res ``=` `[(key, key) ``for` `key, val ``in` `Counter(test_list).items()``                                  ``for` `_ ``in` `range``(val ``/``/` `2``)]` `# printing result``print``("The records after pairing : " ``+` `str``(res))`

Output :

```The original list is : [4, 6, 7, 4, 2, 6, 2, 8]
The records after pairing : [(4, 4), (6, 6), (2, 2)]```

Time Complexity: O(n*n) where n is the number of elements in the in the list “test_list”. The Counter() + list comprehension is used to perform the task and it takes O(n*n) time.
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the in the list “test_list”.

Method #2 : Using fromkeys() + list comprehension This is yet another way in which this task can be performed. In this, we perform the task of Counter() using fromkeys() and get() of dictionary, i.e getting frequencies.

## Python3

 `# Python3 code to demonstrate working of``# Extract Similar pairs from List``# Using fromkeys() + list comprehension` `# initializing list``test_list ``=` `[``4``, ``6``, ``7``, ``4``, ``2``, ``6``, ``2``, ``8``]` `# printing original list``print``("The original ``list` `is` `: " ``+` `str``(test_list))` `# Extract Similar pairs from List``# Using fromkeys() + list comprehension``temp ``=` `dict``.fromkeys(test_list, ``0``)``for` `key ``in` `test_list:``    ``temp[key] ``+``=` `1``res ``=` `[(key, key) ``for` `key, val ``in` `temp.items() ``for` `_ ``in` `range``(val ``/``/` `2``)]` `# printing result``print``("The records after pairing : " ``+` `str``(res))`

Output :

```The original list is : [4, 6, 7, 4, 2, 6, 2, 8]
The records after pairing : [(4, 4), (6, 6), (2, 2)]```

Method 3 :  Use a set

## Python3

 `# Python3 code to demonstrate working of``# Extract Similar pairs from List``# Using a set` `# initializing list``test_list ``=` `[``4``, ``6``, ``7``, ``4``, ``2``, ``6``, ``2``, ``8``]` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))` `# Extract Similar pairs from List``# Using a set``odd_freq ``=` `set``()``res ``=` `[]``for` `elem ``in` `test_list:``    ``if` `elem ``in` `odd_freq:``        ``res.append((elem, elem))``        ``odd_freq.remove(elem)``    ``else``:``        ``odd_freq.add(elem)` `# printing result``print``(``"The records after pairing : "` `+` `str``(res))`

Output

```The original list is : [4, 6, 7, 4, 2, 6, 2, 8]
The records after pairing : [(4, 4), (6, 6), (2, 2)]```

The time complexity of this solution is O(n), where n is the length of test_list, since we iterate over the list once. The auxiliary space is O(n) as well, since the set odd_freq can contain up to n/2 elements in the worst case

Method #4: Using dictionary

• Create an empty dictionary freq_dict to keep track of the frequency of each element in the list.
• Iterate through the elements of the input list test_list.
• If an element is already present in freq_dict, then add it to the output list res as a pair (elem, elem) and decrement its frequency in freq_dict.
• If an element is not present in freq_dict, then initialize its frequency to 1.
• Finally, return the output list res.

## Python3

 `# Python3 code to demonstrate working of``# Extract Similar pairs from List``# Using dictionary` `# initializing list``test_list ``=` `[``4``, ``6``, ``7``, ``4``, ``2``, ``6``, ``2``, ``8``]` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))` `# Extract Similar pairs from List``# Using dictionary``freq_dict ``=` `{}``res ``=` `[]``for` `elem ``in` `test_list:``    ``if` `elem ``in` `freq_dict ``and` `freq_dict[elem] > ``0``:``        ``res.append((elem, elem))``        ``freq_dict[elem] ``-``=` `1``    ``else``:``        ``freq_dict[elem] ``=` `1` `# printing result``print``(``"The records after pairing : "` `+` `str``(res))`

Output

```The original list is : [4, 6, 7, 4, 2, 6, 2, 8]
The records after pairing : [(4, 4), (6, 6), (2, 2)]```

Time complexity: O(n), where n is the length of the input list test_list.
Auxiliary space: O(n), where n is the length of the input list test_list.

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