Python – Summation after elements removal

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().

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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))

chevron_right


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().

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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))

chevron_right


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



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.


Article Tags :

Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.