Python | Check for Whitespace in List (original) (raw)

Last Updated : 13 Apr, 2023

Sometimes, we might have a problem in which we need to check if the List of strings has any of blank spaces. This kind of problem can be in Machine Learning domain to get specific type of data set. Let’s discuss certain ways in which this kind of problem can be solved.

Method #1: Using regex + any() This kind of problem can be solved using the regex utility offered by python. By feeding the appropriate regex string in search(), we can check presence of space in a string and iterate through entire list using any().

Python3

import re

test_list = [ "Geeks forGeeks" , "is" , "best" ]

print ( "The original list is : " + str (test_list))

res = any ( bool (re.search(r "\s" , ele)) for ele in test_list)

print ( "Does any string contain spaces ? " + str (res))

Output

The original list is : ['Geeks forGeeks', 'is', 'best'] Does any string contain spaces ? True

Time Complexity: O(n) where n is the number of elements in the string list. The regex + any() is used to perform the task and it takes O(n) time.
Auxiliary Space: O(1) constant additional space is required

Method #2: Using in operator + any() This task can also be performed using in operator. Just required to check for a space in the string. The verdict returned is true even if a single space is found in any of string of list and false otherwise.

Python3

import re

test_list = [ "Geeks forGeeks" , "is" , "best" ]

print ( "The original list is : " + str (test_list))

res = any ('' in ele for ele in test_list)

print ( "Does any string contain spaces ? " + str (res))

Output

The original list is : ['Geeks forGeeks', 'is', 'best'] Does any string contain spaces ? True

Time Complexity: O(n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(1), constant additional space required

Method #3 : Using count() method

Python3

test_list = [ "Geeks forGeeks" , "is" , "best" ]

print ( "The original list is : " + str (test_list))

res = False

for i in test_list:

`` if (i.count( " " )> = 1 ):

`` res = True

`` break

print ( "Does any string contain spaces ? " + str (res))

Output

The original list is : ['Geeks forGeeks', 'is', 'best'] Does any string contain spaces ? True

Method #4: Using list comprehension and join()

Python3

test_list = [ "Geeks forGeeks" , "is" , "best" ]

print ( "The original list is : " + str (test_list))

res = " " in "".join(test_list)

print ( "Does any string contain spaces ? " + str (res))

Output

The original list is : ['Geeks forGeeks', 'is', 'best'] Does any string contain spaces ? True

Time complexity: O(n)
Auxiliary space: O(1)

Method #5: Using loop and if condition:

Python3

test_list = [ "Geeks forGeeks" , "is" , "best" ]

print ( "The original list is : " + str (test_list))

flag = False

for i in test_list:

`` if " " in i:

`` print ( "Does any string contain spaces ? " , True )

`` flag = True

`` break

if not flag:

`` print ( "Does any string contain spaces ? " , False )

Output

The original list is : ['Geeks forGeeks', 'is', 'best'] Does any string contain spaces ? True

Time complexity: O(n)
Auxiliary space: O(1)

Method #5: Using isspace():

Python3

test_list = [ "Geeks forGeeks" , "is" , "best" ]

print ( "The original list is : " + str (test_list))

res = False

for ele in test_list:

`` if any (c.isspace() for c in ele):

`` res = True

`` break

print ( "Does any string contain spaces ? " + str (res))

Output

The original list is : ['Geeks forGeeks', 'is', 'best'] Does any string contain spaces ? True

Time complexity: O(n*m)
Auxiliary space: O(1)