Python – Filter rows without Space Strings

Last Updated : 12 Nov, 2020

Given Matrix, extract rows in which Strings don’t have spaces.

Examples:

Input : test_list = [[“gfg is”, “best”], [“gfg”, “good”], [“gfg is cool”], [“love”, “gfg”]]
Output : [[‘gfg’, ‘good’], [‘love’, ‘gfg’]]
Explanation : Both the lists have strings don’t have spaces.

Input : test_list = [[“gfg is”, “best”], [“gfg “, “good”], [“gfg is cool”], [“love”, “gfg”]]
Output : [[‘love’, ‘gfg’]]
Explanation : List has strings don’t have spaces.

Method #1: Using list comprehension + any() + regex

In this, we check for no space in each string using regex, any() is used to check this for any string found with spaces, that row is not added.

Python3

# Python3 code to demonstrate working of
# Filter rows without Space Strings
# Using list comprehension + any() + regex
import re
  
# initializing list
test_list = [["gfg is", "best"], ["gfg", "good"],
             ["gfg is cool"], ["love", "gfg"]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# checking for spaces using regex
# not including row if any string has space
res = [row for row in test_list if not any(
    bool(re.search(r"\s", ele)) for ele in row)]
  
# printing result
print("Filtered Rows : " + str(res))

Output:

The original list is : [[‘gfg is’, ‘best’], [‘gfg’, ‘good’], [‘gfg is cool’], [‘love’, ‘gfg’]]
Filtered Rows : [[‘gfg’, ‘good’], [‘love’, ‘gfg’]]

Method #2 : Using filter() + lambda + any() + regex

In this, we perform task of filtering using filter() and lambda function, rest all the functionalities are performed alike the above method.

Python3

# Python3 code to demonstrate working of
# Filter rows without Space Strings
# Using filter() + lambda + any() + regex
import re
  
# initializing list
test_list = [["gfg is", "best"], ["gfg", "good"],
             ["gfg is cool"], ["love", "gfg"]]
  
# printing original list
print("The original list is : " + str(test_list))
  
# checking for spaces using regex
# not including row if any string has space
res = list(filter(lambda row: not any(bool(re.search(r"\s", ele))
                                      for ele in row), test_list))
  
# printing result
print("Filtered Rows : " + str(res))

Output:

The original list is : [[‘gfg is’, ‘best’], [‘gfg’, ‘good’], [‘gfg is cool’], [‘love’, ‘gfg’]]
Filtered Rows : [[‘gfg’, ‘good’], [‘love’, ‘gfg’]]

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