# Python program for sum of consecutive numbers with overlapping in lists

• Last Updated : 01 Oct, 2020

Given a List, perform summation of consecutive elements, by overlapping.

Input : test_list = [4, 7, 3, 2]
Output : [11, 10, 5, 6]
Explanation : 4 + 7 = 11, 7 + 3 = 10, 3 + 2 = 5, and 2 + 4 = 6.

Input : test_list = [4, 7, 3]
Output : [11, 10, 7]
Explanation : 4+7=11, 7+3=10, 3+4=7.

Method 1 : Using list comprehension + zip()

In this, we zip list, with consecutive elements’ using zip() and then perform summation using + operator. The iteration part is done using list comprehension.

## Python3

 `# initializing list``test_list ``=` `[``4``, ``7``, ``3``, ``2``, ``9``, ``2``, ``1``]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# using zip() to get pairing.``# last element is joined with first for pairing``res ``=` `[a ``+` `b ``for` `a, b ``in` `zip``(test_list, test_list[``1``:] ``+` `[test_list[``0``]])]`` ` `# printing result``print``(``"The Consecutive overlapping Summation : "` `+` `str``(res))`

Output

```The original list is : [4, 7, 3, 2, 9, 2, 1]
The Consecutive overlapping Summation : [11, 10, 5, 11, 11, 3, 5]

```

Method #2 : Using sum() + list comprehension + zip()

In this, we perform the task of getting summation using sum() and rest all functionalities kept similar to the above method.

## Python3

 `# Python3 code to demonstrate working of ``# Consecutive overlapping Summation``# Using sum() + list comprehension + zip()`` ` `# initializing list``test_list ``=` `[``4``, ``7``, ``3``, ``2``, ``9``, ``2``, ``1``]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# using sum() to compute elements sum``# last element is joined with first for pairing``res ``=` `[``sum``(sub) ``for` `sub ``in` `zip``(test_list, test_list[``1``:] ``+` `[test_list[``0``]])]  `` ` `# printing result ``print``(``"The Consecutive overlapping Summation : "` `+` `str``(res))`

Output

```The original list is : [4, 7, 3, 2, 9, 2, 1]
The Consecutive overlapping Summation : [11, 10, 5, 11, 11, 3, 5]

```

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