Sometimes, in the process of data filtering we have a problem in which we need to remove words which are composite of certain letters. This kind of application is common in data science domain. Let’s discuss certain ways in which this task can be performed.
Method #1 : Using all() + list comprehension The combination of above methods can be used to perform this task. In this, we just check for all list characters using all() in each list and filters out string which has any one of characters.
Python3
# Python3 code to demonstrate # Remove words containing list characters# using list comprehension + all()from itertools import groupby# initializing list test_list = ['gfg', 'is', 'best', 'for', 'Lazyroar']# initializing char list char_list = ['g', 'o']# printing original listprint ("The original list is : " + str(test_list))# printing character listprint ("The character list is : " + str(char_list))# Remove words containing list characters# using list comprehension + all()res = [ele for ele in test_list if all(ch not in ele for ch in char_list)]# printing result print ("The filtered strings are : " + str(res)) |
The original list is : ['gfg', 'is', 'best', 'for', 'Lazyroar'] The character list is : ['g', 'o'] The filtered strings are : ['is', 'best']
Time Complexity: O(n) where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n), where n is the number of elements in the new res list
Method #2 : Using loop This is brute method in which this task can be performed. In this we use loop and conditional statements to perform this task.
Python3
# Python3 code to demonstrate # Remove words containing list characters# using loopfrom itertools import groupby# initializing list test_list = ['gfg', 'is', 'best', 'for', 'Lazyroar']# initializing char list char_list = ['g', 'o']# printing original listprint ("The original list is : " + str(test_list))# printing character listprint ("The character list is : " + str(char_list))# Remove words containing list characters# using loopres = []flag = 1for ele in test_list: for idx in char_list: if idx not in ele: flag = 1 else: flag = 0 break if(flag == 1): res.append(ele) # printing result print ("The filtered strings are : " + str(res)) |
The original list is : ['gfg', 'is', 'best', 'for', 'Lazyroar'] The character list is : ['g', 'o'] The filtered strings are : ['is', 'best']
Time Complexity: O(n*n)
Auxiliary Space: O(n), where n is length of list.
Method #3 : Using replace() and len() methods
Python3
# Python3 code to demonstrate# Remove words containing list characters# initializing listtest_list = ['gfg', 'is', 'best', 'for', 'Lazyroar']# initializing char listchar_list = ['g', 'o']# printing original listprint ("The original list is : " + str(test_list))# printing character listprint ("The character list is : " + str(char_list))# Remove words containing list charactersres=[]for i in test_list: x=i for j in char_list: i=i.replace(j,"") if(len(i)==len(x)): res.append(i)# printing resultprint ("The filtered strings are : " + str(res)) |
The original list is : ['gfg', 'is', 'best', 'for', 'Lazyroar'] The character list is : ['g', 'o'] The filtered strings are : ['is', 'best']
Time Complexity : O(N*N)
Auxiliary space : O(1)
Method #4 : Using Counter() function
Python3
# Python3 code to demonstrate# Remove words containing list charactersfrom collections import Countertest_list = ['gfg', 'is', 'best', 'for', 'Lazyroar']# initializing char listchar_list = ['g', 'o']# printing original listprint("The original list is : " + str(test_list))# printing character listprint("The character list is : " + str(char_list))# Remove words containing list charactersres = []freqCharList = Counter(char_list)for i in test_list: unique = True for char in i: if char in freqCharList.keys(): unique =False break if(unique): res.append(i) # printing resultprint ("The filtered strings are : " + str(res)) |
The original list is : ['gfg', 'is', 'best', 'for', 'Lazyroar'] The character list is : ['g', 'o'] The filtered strings are : ['is', 'best']
Time Complexity : O(N*N)
Auxiliary space : O(N)
Method #5: Using operator.countOf() method
Python3
# Python3 code to demonstrate# Remove words containing list charactersfrom itertools import groupbyimport operator as op# initializing listtest_list = ['gfg', 'is', 'best', 'for', 'Lazyroar']# initializing char listchar_list = ['g', 'o']# printing original listprint("The original list is : " + str(test_list))# printing character listprint("The character list is : " + str(char_list))# Remove words containing list characters# using loopres = []flag = 1for ele in test_list: for idx in char_list: if op.countOf(ele, idx) == 0: flag = 1 else: flag = 0 break if(flag == 1): res.append(ele)# printing resultprint("The filtered strings are : " + str(res)) |
The original list is : ['gfg', 'is', 'best', 'for', 'Lazyroar'] The character list is : ['g', 'o'] The filtered strings are : ['is', 'best']
Time Complexity : O(N*N)
Auxiliary space : O(1)
Method #6: Using the recursive method.
Python3
# Python3 code to demonstrate# Remove words containing list charactersdef remove_words(start,lst,charlst,newlist=[]): if start==len(lst): return newlist flag=0 for i in charlst: if i in lst[start]: flag=1 if flag==0: newlist.append(lst[start]) return remove_words(start+1,lst,charlst,newlist)# initializing listtest_list = ['gfg', 'is', 'best', 'for', 'Lazyroar']# initializing char listchar_list = ['g', 'o']# printing original listprint("The original list is : " + str(test_list))# printing character listprint("The character list is : " + str(char_list))# Remove words containing list characters# using recursive functionres = remove_words(0,test_list,char_list)# printing resultprint("The filtered strings are : " + str(res))#this code contributed by tvsk |
The original list is : ['gfg', 'is', 'best', 'for', 'Lazyroar'] The character list is : ['g', 'o'] The filtered strings are : ['is', 'best']
Time Complexity : O(N*N)
Auxiliary space : O(N)
Method #7: Using filter() and lambda function:
Python3
test_list = ['gfg', 'is', 'best', 'for', 'Lazyroar']char_list = ['g', 'o']# printing original listprint("The original list is : " + str(test_list)) # printing character listprint("The character list is : " + str(char_list))res = list(filter(lambda x: not any(y in x for y in char_list), test_list))print(res) |
The original list is : ['gfg', 'is', 'best', 'for', 'Lazyroar'] The character list is : ['g', 'o'] ['is', 'best']
Time Complexity : O(N*N)
Auxiliary space : O(N)
Method #8 : Using set intersection
Python3
# Python3 code to demonstrate# Remove words containing list characters# initializing listtest_list = ['gfg', 'is', 'best', 'for', 'Lazyroar']# initializing char listchar_list = set(['g', 'o'])# printing original listprint("The original list is : " + str(test_list))# printing character listprint("The character list is : " + str(char_list))# Remove words containing list charactersres = list(filter(lambda x: not set(x).intersection(char_list), test_list))# printing resultprint("The filtered strings are : " + str(res)) |
The original list is : ['gfg', 'is', 'best', 'for', 'Lazyroar']
The character list is : {'o', 'g'}
The filtered strings are : ['is', 'best']
Time complexity: O(N), where N is the number of elements in test_list
Auxiliary Space: O(N), in the worst case, all elements in test_list are kept in the resulting list
Explanation: This method uses filter and set to achieve the desired result. set is used to convert the list of characters into a set, which allows for constant time look ups. filter is used to keep only those elements from test_list for which the intersection with char_list is empty.
