Given a List and a sublist, count occurrence of sublist in list.
Input : test_list = [4, 5, 3, 5, 7, 8, 3, 5, 7, 2, 3, 5, 7], sublist = [3, 5, 7] Output : 3 Explanation : 3, 5, 7 occurs 3 times. Input : test_list = [4, 5, 3, 5, 8, 8, 3, 2, 7, 2, 3, 6, 7], sublist = [3, 5, 7] Output : 0 Explanation : No occurrence found.
Method #1 : Using list comprehension + slicing
In this, we test for each sublist of list extracted using slicing, if found, the element is added to list, at last length of list is computed using len().
Python3
# Python3 code to demonstrate working of # Sublist Frequency# Using list comprehension + slicing # initializing listtest_list = [4, 5, 3, 5, 7, 8, 3, 5, 7, 2, 7, 3, 2]# printing original listprint("The original list is : " + str(test_list))# initializing Sublistsublist = [3, 5, 7]# slicing is used to extract chunks and compareres = len([sublist for idx in range(len(test_list)) if test_list[idx : idx + len(sublist)] == sublist])# printing result print("The sublist count : " + str(res)) |
The original list is : [4, 5, 3, 5, 7, 8, 3, 5, 7, 2, 7, 3, 2] The sublist count : 2
Time Complexity: O(n), where n is the length of the input list.
Auxiliary Space: O(1) additional space is not required
Method #2 : Using zip_longest() + islice() + all() + loop
In this, we perform task of slicing using islice() and check if each element is matching using all(), zip_longest() helps in mapping elements to check for equality from sublist. Loop is used to get to index match with first element of sublist in list, to make it more efficient.
Python3
# Python3 code to demonstrate working of # Sublist Frequency# Using zip_longest() + islice() + all() + loop from itertools import zip_longest, islice# initializing listtest_list = [4, 5, 3, 5, 7, 8, 3, 5, 7, 2, 7, 3, 2]# printing original listprint("The original list is : " + str(test_list))# initializing Sublistsublist = [3, 5, 7]# slicing is used to extract chunks and compareres = []idx = 0 while True: try: # getting to the index idx = test_list.index(sublist[0], idx) except ValueError: break # using all() to check for all elements equivalence if all(x == y for (x, y) in zip_longest(sublist, islice(test_list, idx, idx + len(sublist)))): res.append(sublist) idx += len(sublist) idx += 1res = len(res)# printing result print("The sublist count : " + str(res)) |
The original list is : [4, 5, 3, 5, 7, 8, 3, 5, 7, 2, 7, 3, 2] The sublist count : 2
