This particular article is indeed a very useful one for Machine Learning enthusiast as it solves a good problem for them. In Machine Learning we generally encounter this issue of getting a particular string in huge amount of data and handling that sometimes becomes a tedious task. Lets discuss certain way outs to solve this problem.
Method #1: Using remove() This particular method is quite naive and not recommended to use, but is indeed a method to perform this task. remove() generally removes the first occurrence of K string and we keep iterating this process until no K string is found in list.
Python3
# Python 3 code to demonstrate# Remove K String from String List# using remove()# initializing listtest_list = ["bad", "Lazyroar", "bad", "is", "best", "bad"]# Printing original listprint("Original list is : " + str(test_list))# initializing KK = "bad"# using remove() to# Remove K String from String Listwhile(K in test_list): test_list.remove(K)# Printing modified listprint("Modified list is : " + str(test_list)) |
Original list is : ['bad', 'Lazyroar', 'bad', 'is', 'best', 'bad'] Modified list is : ['Lazyroar', 'is', 'best']
Time Complexity: O(n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(1), constant extra space is required
Method #2: Using List Comprehension More concise and better approach to remove all the K strings, it just checks if the string is not K and re-makes the list with all strings that are not K.
Python3
# Python 3 code to demonstrate# Remove K String from String List# using list comprehension# initializing listtest_list = ["bad", "Lazyroar", "bad", "is", "best", "bad"]# Printing original listprint("Original list is : " + str(test_list))# initializing KK = "bad"# using list comprehension to# Remove K String from String Listtest_list = [i for i in test_list if i != K]# Printing modified listprint("Modified list is : " + str(test_list)) |
Original list is : ['bad', 'Lazyroar', 'bad', 'is', 'best', 'bad'] Modified list is : ['Lazyroar', 'is', 'best']
Time Complexity: O(n)
Auxiliary Space: O(n), where n is length of list.
Method #3 : Using join(),replace(),split() and remove() methods
Python3
# Python 3 code to demonstrate# Remove K String from String List# initializing listtest_list = ["bad", "Lazyroar", "bad", "is", "best", "bad"]# Printing original listprint ("Original list is : " + str(test_list))# initializing KK = "bad"x="-".join(test_list)x=x.replace(K,"")a=x.split("-")while("" in a ): a.remove("")# Printing modified listprint ("Modified list is : " + str(a)) |
Original list is : ['bad', 'Lazyroar', 'bad', 'is', 'best', 'bad'] Modified list is : ['Lazyroar', 'is', 'best']
Here’s another approach to removing a string ‘K’ from a string list, using a filter:
Python
# Python 3 code to demonstrate# Remove K String from String List# using filter() # initializing listtest_list = ["bad", "Lazyroar", "bad", "is", "best", "bad"] # Printing original listprint("Original list is : " + str(test_list)) # initializing KK = "bad" # using filter to Remove K String from String Listtest_list = list(filter(lambda x: x!=K, test_list)) # Printing modified listprint("Modified list is : " + str(test_list)) |
Original list is : ['bad', 'Lazyroar', 'bad', 'is', 'best', 'bad'] Modified list is : ['Lazyroar', 'is', 'best']
This approach uses the filter() function to create a filtered list of elements from test_list that are not equal to K. The list() function is then used to convert the filtered list into a list.
Time complexity: O(n), where n is the number of elements in the list test_list.
Space complexity: O(n), since a new list is created.
