Python | Splitting list on empty string

Sometimes, we may face an issue in which we require to split a list to list of list on the blank character sent as deliminator. This kind of problem can be used to send messages or can be used in cases where it is desired to have list of list of native list. Let’s discuss certain ways in which this can be done. Method #1 : Using index() and list slicing The list slicing can be used to get the sublists from the native list and index function can be used to check for the empty string which can potentially act as a separator. The drawback of this is that it only works for a single split i.e can only divide a list to 2 sublists. 

Python3

# Python3 code to demonstrate
# divide list to siblist on deliminator
# using index() + list slicing

# initializing list
test_list = ['Geeks', 'for', '', 'Geeks', 1, 2]

# printing original list
print("The original list : " + str(test_list))

# using index() + list slicing
# divide list to siblist on deliminator
temp_idx = test_list.index('')
res = [test_list[: temp_idx], test_list[temp_idx + 1: ]]

# print result
print("The list of sublist after separation : " + str(res))

Output : 

The original list : [‘Geeks’, ‘for’, ”, ‘Geeks’, 1, 2] The list of sublist after separation : [[‘Geeks’, ‘for’], [‘Geeks’, 1, 2]]

Time Complexity : O(N), where N is the length of the input string.

Auxiliary Space: O(N)

Method #2 : Using itertools.groupby() + list comprehension The problem of the above proposed method can be solved using the groupby function which could divide on all the list breaks given by the empty strings. 

Python3

# Python3 code to demonstrate
# divide list to siblist on deliminator
# using itertools.groupby() + list comprehension
from itertools import groupby

# initializing list
test_list = ['Geeks', '', 'for', '', 4, 5, '',
                  'Geeks', 'CS', '', 'Portal']

# printing original list
print("The original list : " + str(test_list))

# using itertools.groupby() + list comprehension
# divide list to siblist on deliminator
res = [list(sub) for ele, sub in groupby(test_list, key = bool) if ele]

# print result
print("The list of sublist after separation : " + str(res))

Output : 

The original list : [‘Geeks’, ”, ‘for’, ”, 4, 5, ”, ‘Geeks’, ‘CS’, ”, ‘Portal’] The list of sublist after separation : [[‘Geeks’], [‘for’], [4, 5], [‘Geeks’, ‘CS’], [‘Portal’]]

Approach#3: Using groupby(): This code uses the groupby() function from the itertools module to group the elements in the list based on the empty string. The lambda function passed to the key parameter of the groupby() function returns True for empty strings and False for all other elements. Then, list comprehension is used to create a list of sublists by iterating over the groups of elements and creating a sublist for each group of elements that is not an empty string.

1. Import the groupby() function from the itertools module.
2. Define the input list last.
3. Use the groupby() function to group the elements in lst based on the empty string. The key parameter of the groupby() function takes a lambda function that returns True for empty string and False for all other elements.
4. Use a list comprehension to create a list of sublists by iterating over the groups of elements and creating a sublist for each group of elements that is not an empty string.
5. Print the list of sublists after separation.

Python3

from itertools import groupby

lst = ['Geeks', 'for', '', 'Geeks', 1, 2]

sublists = [list(g) for k, g in groupby(lst, key=lambda x: x == '') if not k]

print("The list of sublists after separation:", sublists)

Output

The list of sublists after separation: [['Geeks', 'for'], ['Geeks', 1, 2]]

Time Complexity: O(n), where n is the length of the input list.

Space Complexity: O(n), where n is the length of the input list,