# Python – Summation after elements removal

• Last Updated : 19 Feb, 2020

Sometimes we need to perform the operation of removing all the items from the lists that are present in other list, i.e we are given some of the invalid numbers in one list which needs to be get ridden from the original list and perform its summation. Lets discuss various ways in which this can be performed.

Method #1 : Using list comprehension + sum()
The list comprehension can be used to perform the naive method in just the one line and hence gives an easy method to perform this particular task. The task of performing summation is done using sum().

 `# Python 3 code to demonstrate ``# Summation after elements removal``# using list comprehension + sum()`` ` `# initializing list ``test_list ``=` `[``1``, ``3``, ``4``, ``6``, ``7``, ``6``]`` ` `# initializing remove list ``remove_list ``=` `[``3``, ``6``]`` ` `# printing original list ``print` `(``"The original list is : "` `+` `str``(test_list))`` ` `# printing remove list ``print` `(``"The remove list is : "` `+` `str``(remove_list))`` ` `# using list comprehension + sum() to perform task``res ``=` `sum``([i ``for` `i ``in` `test_list ``if` `i ``not` `in` `remove_list])`` ` `# printing result``print` `(``"The list after performing removal summation is : "` `+` `str``(res))`
Output :
```The original list is : [1, 3, 4, 6, 7, 6]
The remove list is : [3, 6]
The list after performing removal summation is : 12
```

Method #2 : Using `filter() + lambda + sum()`
The filter function can be used along with lambda to perform this task and creating a new filtered list of all the elements that are not present in the remove element list. The task of performing summation is done using sum().

 `# Python 3 code to demonstrate ``# Summation after elements removal``# using filter() + lambda + sum()`` ` `# initializing list ``test_list ``=` `[``1``, ``3``, ``4``, ``6``, ``7``]`` ` `# initializing remove list ``remove_list ``=` `[``3``, ``6``]`` ` `# printing original list ``print` `(``"The original list is : "` `+` `str``(test_list))`` ` `# printing remove list ``print` `(``"The remove list is : "` `+` `str``(remove_list))`` ` `# using filter() + lambda + sum() to perform task``res ``=` `sum``(``filter``(``lambda` `i: i ``not` `in` `remove_list, test_list))`` ` `# printing result``print` `(``"The list after performing removal summation is : "` `+` `str``(res))`
Output :
```The original list is : [1, 3, 4, 6, 7, 6]
The remove list is : [3, 6]
The list after performing removal summation is : 12
```

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