Python – Filter rows with only Alphabets from List of Lists
Last Updated :
26 Apr, 2023
Given Matrix, write a Python program to extract rows which only contains alphabets in its strings.
Examples:
Input : test_list = [[“gfg”, “is”, “best”], [“Geeks4Geeks”, “good”], [“Gfg is good”], [“love”, “gfg”]]
Output : [[‘gfg’, ‘is’, ‘best’], [‘love’, ‘gfg’]]
Explanation : All strings with just alphabets are extracted.
Input : test_list = [[“gfg”, “is”, “!best”], [“Geeks4Geeks”, “good”], [“Gfg is good”], [“love”, “gfg”]]
Output : [[‘love’, ‘gfg’]]
Explanation : All strings with just alphabets are extracted.
Method #1: Using isalpha() + all() + list comprehension
In this, we check for all the alphabets using isalpha() and all() is used to ensure all the strings contain just the alphabets. The list comprehension is used to iterate through rows.
Python3
test_list = [[ "gfg" , "is" , "best" ], [ "Geeks4Geeks" , "good" ],
[ "Gfg is good" ], [ "love" , "gfg" ]]
print ( "The original list is : " + str (test_list))
res = [sub for sub in test_list if all (ele.isalpha() for ele in sub)]
print ( "Filtered Rows : " + str (res))
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Output
The original list is : [['gfg', 'is', 'best'], ['Geeks4Geeks', 'good'], ['Gfg is good'], ['love', 'gfg']]
Filtered Rows : [['gfg', 'is', 'best'], ['love', 'gfg']]
Time Complexity: O(n*n)
Auxiliary Space: O(n)
Method #2 : Using filter() + lambda + join() + isalpha()
In this, we concatenate each string using join() and test if its all alphabets using isalpha(), and add if the verdict returns true.
Python3
test_list = [[ "gfg" , "is" , "best" ], [ "Geeks4Geeks" , "good" ],
[ "Gfg is good" ], [ "love" , "gfg" ]]
print ( "The original list is : " + str (test_list))
res = list ( filter ( lambda sub: ''.join(
[ele for ele in sub]).isalpha(), test_list))
print ( "Filtered Rows : " + str (res))
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Output
The original list is : [['gfg', 'is', 'best'], ['Geeks4Geeks', 'good'], ['Gfg is good'], ['love', 'gfg']]
Filtered Rows : [['gfg', 'is', 'best'], ['love', 'gfg']]
The time and space complexity for all the methods are the same:
Time Complexity: O(n2)
Auxiliary Space : O(n)
Method #3 : Using itertools.filterfalse() method
Python3
import itertools
test_list = [[ "gfg" , "is" , "best" ], [ "Geeks4Geeks" , "good" ],
[ "Gfg is good" ], [ "love" , "gfg" ]]
print ( "The original list is : " + str (test_list))
res = list (itertools.filterfalse( lambda sub: not ''.join(
[ele for ele in sub]).isalpha(), test_list))
print ( "Filtered Rows : " + str (res))
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Output
The original list is : [['gfg', 'is', 'best'], ['Geeks4Geeks', 'good'], ['Gfg is good'], ['love', 'gfg']]
Filtered Rows : [['gfg', 'is', 'best'], ['love', 'gfg']]
Time Complexity: O(n2)
Auxiliary Space: O(n)
Method #4 : Using re
Here’s another approach that you can use to extract the rows that contain only alphabets in its strings.
This approach uses the built-in re module to match the string against a regular expression pattern for alphabets and returns the filtered rows.
Python3
import re
test_list = [[ "gfg" , "is" , "best" ], [ "Geeks4Geeks" , "good" ], [ "Gfg is good" ], [ "love" , "gfg" ]]
result = [row for row in test_list if all (re.match(r '^[a-zA-Z]+$' , ele) for ele in row)]
print ( "Filtered Rows:" , result)
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Output
Filtered Rows: [['gfg', 'is', 'best'], ['love', 'gfg']]
Time Complexity: O(n^2)
Auxiliary Space: O(n)
Note: In the above code, re.match returns a match object if there is a match, and None if there is no match. The all function returns True if all the elements in the given iterable are True, and False otherwise. In this case, we use all to check if all the elements in the row list match the regular expression pattern for alphabets.
Method #5: Using a nested loop to iterate over each element and character of the list and check if it is an alphabet.
Step-by-step approach:
- Initialize an empty list to store the filtered rows.
- Use a nested for loop to iterate over each sublist and each character in the sublist.
- Use the isalpha() method to check if the character is an alphabet.
- If all characters in the sublist are alphabets, append the sublist to the filtered list.
- Return the filtered list.
Python3
test_list = [[ "gfg" , "is" , "best" ], [ "Geeks4Geeks" , "good" ],
[ "Gfg is good" ], [ "love" , "gfg" ]]
print ( "The original list is : " + str (test_list))
filtered_list = []
for sublist in test_list:
is_alphabetic = True
for char in sublist:
if not char.isalpha():
is_alphabetic = False
break
if is_alphabetic:
filtered_list.append(sublist)
print ( "Filtered Rows : " + str (filtered_list))
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Output
The original list is : [['gfg', 'is', 'best'], ['Geeks4Geeks', 'good'], ['Gfg is good'], ['love', 'gfg']]
Filtered Rows : [['gfg', 'is', 'best'], ['love', 'gfg']]
Time complexity: O(n*m), where n is the number of sublists and m is the length of the longest sublist.
Auxiliary space: O(k), where k is the number of alphabetic sublists.
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