# Python – Elements frequency count in multiple lists

• Last Updated : 20 Jan, 2023

Sometimes while working with Python lists we can have a problem in which we need to extract the frequency of elements in list. But this can be added work if we have more than 1 list we work on. Let’s discuss certain ways in which this task can be performed.

Method #1: Using dictionary comprehension + set() + count() This is one of the way in which this task can be performed. In this, we perform the task of counting using count() and set() and the extension of logic is done using dictionary comprehension.

## Python3

 `# Python3 code to demonstrate``# Elements frequency count in multiple lists``# using set() + count() + dictionary comprehension` `# Initializing lists``test_list1 ``=` `[``1``, ``3``, ``2``, ``4``, ``5``, ``1``, ``2``]``test_list2 ``=` `[``4``, ``1``, ``4``, ``3``, ``4``, ``2``, ``4``]``test_list3 ``=` `[``1``, ``4``, ``5``, ``3``, ``4``, ``5``, ``4``]` `# printing original lists``print``(``"The original list 1 is : "` `+` `str``(test_list1))``print``(``"The original list 2 is : "` `+` `str``(test_list2))``print``(``"The original list 3 is : "` `+` `str``(test_list3))` `# Elements frequency count in multiple lists``# using set() + count() + dictionary comprehension``res ``=` `{idx : [test_list1.count(idx), test_list2.count(idx), test_list3.count(idx)]``    ``for` `idx ``in` `set``(test_list1 ``+` `test_list2 ``+` `test_list3)}``            ` `# printing result``print` `(``"The frequency of each element is : "` `+` `str``(res))`

Output

```The original list 1 is : [1, 3, 2, 4, 5, 1, 2]
The original list 2 is : [4, 1, 4, 3, 4, 2, 4]
The original list 3 is : [1, 4, 5, 3, 4, 5, 4]
The frequency of each element is : {1: [2, 1, 1], 2: [2, 1, 0], 3: [1, 1, 1], 4: [1, 4, 3], 5: [1, 0, 2]}```

Time Complexity: O(n^2) and the space complexity is O(n) where n is the total number of elements in all the lists.
Auxiliary Space: O(n), where n is the number of elements in the input lists. The res dictionary and the set created for storing the unique elements of the input lists use O(n) space.

Method #2: Using Counter() + map() + dictionary comprehension The combination of above functionalities can be used to solve this problem. In this, the task of finding the frequency is done using Counter() and map().

## Python3

 `# Python3 code to demonstrate``# Elements frequency count in multiple lists``# using map() + Counter() + dictionary comprehension``from` `collections ``import` `Counter` `# Initializing lists``test_list1 ``=` `[``1``, ``3``, ``2``, ``4``, ``5``, ``1``, ``2``]``test_list2 ``=` `[``4``, ``1``, ``4``, ``3``, ``4``, ``2``, ``4``]``test_list3 ``=` `[``1``, ``4``, ``5``, ``3``, ``4``, ``5``, ``4``]` `# printing original lists``print``(``"The original list 1 is : "` `+` `str``(test_list1))``print``(``"The original list 2 is : "` `+` `str``(test_list2))``print``(``"The original list 3 is : "` `+` `str``(test_list3))` `# Elements frequency count in multiple lists``# using map() + Counter() + dictionary comprehension``freq ``=` `list``(``map``(Counter, (test_list1, test_list2, test_list3)))``res ``=` `{ele: [cnt[ele] ``for` `cnt ``in` `freq] ``for` `ele ``in` `{ele ``for` `cnt ``in` `freq ``for` `ele ``in` `cnt}}``            ` `# printing result``print` `(``"The frequency of each element is : "` `+` `str``(res))`

Output

```The original list 1 is : [1, 3, 2, 4, 5, 1, 2]
The original list 2 is : [4, 1, 4, 3, 4, 2, 4]
The original list 3 is : [1, 4, 5, 3, 4, 5, 4]
The frequency of each element is : {1: [2, 1, 1], 2: [2, 1, 0], 3: [1, 1, 1], 4: [1, 4, 3], 5: [1, 0, 2]}```

Method #3: Using extend(),list(),set() and count() methods

## Python3

 `# Python3 code to demonstrate``# Elements frequency count in multiple lists` `# Initializing lists``test_list1 ``=` `[``1``, ``3``, ``2``, ``4``, ``5``, ``1``, ``2``]``test_list2 ``=` `[``4``, ``1``, ``4``, ``3``, ``4``, ``2``, ``4``]``test_list3 ``=` `[``1``, ``4``, ``5``, ``3``, ``4``, ``5``, ``4``]` `# printing original lists``print``(``"The original list 1 is : "` `+` `str``(test_list1))``print``(``"The original list 2 is : "` `+` `str``(test_list2))``print``(``"The original list 3 is : "` `+` `str``(test_list3))` `# Elements frequency count in multiple lists``x``=``[]``res``=``dict``()``x.extend(test_list1)``x.extend(test_list2)``x.extend(test_list3)``x``=``list``(``set``(x))``for` `i ``in` `x:``    ``a``=``[]``    ``a.append(test_list1.count(i))``    ``a.append(test_list2.count(i))``    ``a.append(test_list3.count(i))``    ``res[i]``=``a` `# printing result``print` `(``"The frequency of each element is : "` `+` `str``(res))`

Output

```The original list 1 is : [1, 3, 2, 4, 5, 1, 2]
The original list 2 is : [4, 1, 4, 3, 4, 2, 4]
The original list 3 is : [1, 4, 5, 3, 4, 5, 4]
The frequency of each element is : {1: [2, 1, 1], 2: [2, 1, 0], 3: [1, 1, 1], 4: [1, 4, 3], 5: [1, 0, 2]}```

Time Complexity : O(N)
Auxiliary Space : O(N)

Method #4: Using dictionary comprehension + set() + operator.countOf() This is one of the way in which this task can be performed. In this, we perform the task of counting using operator.countOf() and set() and the extension of logic is done using dictionary comprehension.

## Python3

 `# Python3 code to demonstrate``# Elements frequency count in multiple lists``# using set() + operator.countOf() + dictionary comprehension``import` `operator as op``# Initializing lists``test_list1 ``=` `[``1``, ``3``, ``2``, ``4``, ``5``, ``1``, ``2``]``test_list2 ``=` `[``4``, ``1``, ``4``, ``3``, ``4``, ``2``, ``4``]``test_list3 ``=` `[``1``, ``4``, ``5``, ``3``, ``4``, ``5``, ``4``]` `# printing original lists``print``(``"The original list 1 is : "` `+` `str``(test_list1))``print``(``"The original list 2 is : "` `+` `str``(test_list2))``print``(``"The original list 3 is : "` `+` `str``(test_list3))` `# Elements frequency count in multiple lists``# using set() + operator.countOf() + dictionary comprehension``res ``=` `{idx : [op.countOf(test_list1,idx), op.countOf(test_list2,idx), op.countOf(test_list3,idx)]``    ``for` `idx ``in` `set``(test_list1 ``+` `test_list2 ``+` `test_list3)}``            ` `# printing result``print` `(``"The frequency of each element is : "` `+` `str``(res))`

Output

```The original list 1 is : [1, 3, 2, 4, 5, 1, 2]
The original list 2 is : [4, 1, 4, 3, 4, 2, 4]
The original list 3 is : [1, 4, 5, 3, 4, 5, 4]
The frequency of each element is : {1: [2, 1, 1], 2: [2, 1, 0], 3: [1, 1, 1], 4: [1, 4, 3], 5: [1, 0, 2]}```

Time Complexity: O(n)
Auxiliary Space: O(n)

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