Python – Count frequency of Sublist in given list

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 list
test_list = [4, 5, 3, 5, 7, 8, 3, 5, 7, 2, 7, 3, 2]

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

# initializing Sublist
sublist = [3, 5, 7]

# slicing is used to extract chunks and compare
res = 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))

Output

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 list
test_list = [4, 5, 3, 5, 7, 8, 3, 5, 7, 2, 7, 3, 2]

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

# initializing Sublist
sublist = [3, 5, 7]

# slicing is used to extract chunks and compare
res = []
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 += 1

res = len(res)

# printing result
print("The sublist count : " + str(res))

Output

The original list is : [4, 5, 3, 5, 7, 8, 3, 5, 7, 2, 7, 3, 2]
The sublist count : 2