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
import re
test_list = [["gfg is", "best"], ["gfg", "good"],
["gfg is cool"], ["love", "gfg"]]
print("The original list is : " + str(test_list))
res = [row for row in test_list if not any(
bool(re.search(r"\s", ele)) for ele in row)]
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']]
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
import re
test_list = [["gfg is", "best"], ["gfg", "good"],
["gfg is cool"], ["love", "gfg"]]
print("The original list is : " + str(test_list))
res = list(filter(lambda row: not any(bool(re.search(r"\s", ele))
for ele in row), test_list))
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']]
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
test_list = [["gfg is", "best"], ["gfg", "good"],
["gfg is cool"], ["love", "gfg"]]
print("The original list is : " + str(test_list))
res = []
for i in test_list:
a = "".join(i)
if(a.find(" ") == -1):
res.append(i)
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']]
Time Complexity: O(n*n)
Auxiliary Space: O(n)
Method #4:Using itertools.filterfalse() method
Python3
import itertools
import re
test_list = [["gfg is", "best"], ["gfg", "good"],
["gfg is cool"], ["love", "gfg"]]
print("The original list is : " + str(test_list))
res = list(itertools.filterfalse(lambda row: any(bool(re.search(r"\s", ele))
for ele in row), test_list))
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']]
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
test_list = [["gfg is", "best"], ["gfg", "good"],
["gfg is cool"], ["love", "gfg"]]
print("The original list is : " + str(test_list))
res = [row for row in test_list if all(ele.split() == [ele] for ele in row)]
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']]
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
test_list = [["gfg is", "best"], ["gfg", "good"],
["gfg is cool"], ["love", "gfg"]]
print("The original list is : " + str(test_list))
res = []
for row in test_list:
flag = True
for ele in row:
if " " in ele:
flag = False
break
if flag:
res.append(row)
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']]
Time complexity: O(n^2) (nested loop)
Auxiliary space: O(k) (where k is the length of the longest row)
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