Python - Filter rows without Space Strings

Last Updated : 26 Apr, 2023

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 that don't have spaces.

Input: test_list = [["gfg is", "best"], ["gfg ", "good"], ["gfg is cool"], ["love", "gfg"]]
Output: [['love', 'gfg']]
Explanation: The list has strings that 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 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))

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

Complexity Analysis:

Time Complexity: O(N²), (loop * re.search())
Auxiliary Space: O(N)

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 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))

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

Complexity Analysis:

Time Complexity: O(N²), for loop takes the time complexity of O(n) and the filter also takes O(n) so together the final complexity is O(n²),
Auxiliary Space: O(N), the size of array, so O(n)

Method #3 : Using join() and find() methods

In this method, we perform task of joining all the strings using join() method and then checking if there is a space between the string using find() method.

# Python3 code to demonstrate working of
# Filter rows without Space Strings

# 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 = []
for i in test_list:
    a = "".join(i)
    if(a.find(" ") == -1):
        res.append(i)
# printing result
print("Filtered Rows : " + str(res))

 
# Python3 code to demonstrate working of
# Filter rows without Space Strings​# initializing listtest_list = [["gfg is", "best"], ["gfg", "good"],             ["gfg is cool"], ["love", "gfg"]]​# printing original listprint("The original list is : " + str(test_list))​# checking for spaces using regex# not including row if any string has spaceres = []for i in test_list:    a = "".join(i)    if(a.find(" ") == -1):        res.append(i)# printing resultprint("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']]

Time Complexity: O(n*n)
Auxiliary Space: O(n)

Method #4:Using itertools.filterfalse() method

# Python3 code to demonstrate working of
# Filter rows without Space Strings
import itertools
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(itertools.filterfalse(lambda row:  any(bool(re.search(r"\s", ele))
                                      for ele in row), test_list))

# printing result
print("Filtered Rows : " + str(res))

 
# Python3 code to demonstrate working of
# Filter rows without Space Stringsimport itertoolsimport re​# initializing listtest_list = [["gfg is", "best"], ["gfg", "good"],             ["gfg is cool"], ["love", "gfg"]]​# printing original listprint("The original list is : " + str(test_list))​# checking for spaces using regex# not including row if any string has spaceres = list(itertools.filterfalse(lambda row:  any(bool(re.search(r"\s", ele))                                      for ele in row), test_list))​# printing resultprint("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']]

Time Complexity: O(N2)

Auxiliary Space: O(N)

Method #5: Here is a new approach using a list comprehension and the split and all() method:

# Python3 code to demonstrate working of
# Filter rows without Space Strings
# Using list comprehension + split() method
 
# 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 the split() method
# not including row if any string has space
res = [row for row in test_list if all(ele.split() == [ele] for ele in row)]
 
# printing result
print("Filtered Rows : " + str(res))

 
# Python3 code to demonstrate working of
# Filter rows without Space Strings# Using list comprehension + split() method # initializing listtest_list = [["gfg is", "best"], ["gfg", "good"],             ["gfg is cool"], ["love", "gfg"]] # printing original listprint("The original list is : " + str(test_list)) # checking for spaces using the split() method# not including row if any string has spaceres = [row for row in test_list if all(ele.split() == [ele] for ele in row)] # printing resultprint("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']]

Time Complexity: O(N2), loop * split() method
Auxiliary Space: O(N)

Method #6: Using nested loops and flag variable

Step by step approach:

Initialize a list of lists called test_list with some test data.
Print the original list using the print() function and string concatenation.
Initialize an empty list called res to store the filtered rows.
For each row in test_list, do the following:
Initialize a flag variable called flag to True.
For each element (ele) in the current row, do the following:
Check if the element contains any spaces using the in keyword and the string " " as a parameter. If it does, set the flag variable to False and break out of the loop using the break keyword.
If the flag variable is still True after checking all elements in the current row, append the current row to the res list.
Print the filtered rows using the print() function and string concatenation.

# Python3 code to demonstrate working of
# Filter rows without Space Strings
# Using nested loops and flag variable
 
# 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))
 
# removing rows containing space strings
res = []
for row in test_list:
    flag = True
    for ele in row:
        if " " in ele:
            flag = False
            break
    if flag:
        res.append(row)
 
# printing result
print("Filtered Rows : " + str(res))

 
# Python3 code to demonstrate working of
# Filter rows without Space Strings# Using nested loops and flag variable # initializing listtest_list = [["gfg is", "best"], ["gfg", "good"],             ["gfg is cool"], ["love", "gfg"]] # printing original listprint("The original list is : " + str(test_list)) # removing rows containing space stringsres = []for row in test_list:    flag = True    for ele in row:        if " " in ele:            flag = False            break    if flag:        res.append(row) # printing resultprint("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']]

Time complexity: O(n^2) (nested loop)
Auxiliary space: O(k) (where k is the length of the longest row)

Python - Filter rows without Space Strings

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Python - Filter rows without Space Strings

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