The deletion of elementary elements from list has been dealt with many times, but sometimes rather than having just a one list, we have list of list where we need to perform this particular task. Having shorthands to perform this particular task can help. Let’s discuss certain ways to perform this particular task.
Method #1 : Using list comprehension
The logic behind this kind of method is to reduce the size of code and do the task to perform using loops as a way of list comprehension itself.
# Python3 code to demonstrate # Removing element from list of lists # using list comprehension # initializing list test_list = [[ 4 , 5 , 6 ], [ 5 , 6 , 4 , 1 ], [ 4 ], [ 4 , 8 , 9 , 10 ]] # printing original list print ( "The original list : " + str (test_list)) # initializing Number to delete N = 4 # using list comprehension # Removing element from list of lists res = [[ele for ele in sub if ele ! = N] for sub in test_list] # print result print ( "The list after deletion of element : " + str (res)) |
The original list : [[4, 5, 6], [5, 6, 4, 1], [4], [4, 8, 9, 10]] The list after deletion of element : [[5, 6], [5, 6, 1], [], [8, 9, 10]]
Method 2 : Using list comprehension + list slicing
In this method, we generally do the task similar to the above method, the variation is just we use list slicing for better code readability.
# Python3 code to demonstrate # Removing element from list of lists # using list comprehension + list slicing # initializing list test_list = [[ 4 , 5 , 6 ], [ 5 , 6 , 4 , 1 ], [ 4 ], [ 4 , 8 , 9 , 10 ]] # printing original list print ( "The original list : " + str (test_list)) # initializing Number to delete N = 4 # using list comprehension + list slicing # Removing element from list of lists for sub in test_list: sub[:] = [ele for ele in sub if ele ! = N] # print result print ( "The list after deletion of element : " + str (test_list)) |
The original list : [[4, 5, 6], [5, 6, 4, 1], [4], [4, 8, 9, 10]] The list after deletion of element : [[5, 6], [5, 6, 1], [], [8, 9, 10]]
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