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Python | Consecutive Custom Chunked elements Product

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There are numerous ways to perform product in list the list with the elements, but sometimes, its required to find product of the lists with the numbers in a sliced way. This can be custom and hence knowledge of this can come handy. Let’s discuss certain ways in which this can be done. 

Method #1 : Using list comprehension + enumerate() + loop The list comprehension can do the possible iteration part and enumerate can help in the part of logic and checking for the valid elements required in the list. The loop is used to perform product. 

Python3




# Python3 code to demonstrate
# Consecutive Custom Chunked elements Product
# using list comprehension + enumerate() + loop
 
# getting Product
def prod(val) :
    res = 1
    for ele in val:
        res *= ele
    return res 
 
# initializing lists
test_list = list(range(1, 50))
 
# printing original list
print ("The original list is : " + str(test_list))
 
# interval elements
N = 5
 
# interval difference
K = 15
 
# using list comprehension + enumerate() + loop
# Consecutive Custom Chunked elements Product
res = prod([i for j, i in enumerate(test_list) if j % K < N ])
     
# printing result
print ("The modified range product list : " + str(res))


Output : 

The original list is : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49] The modified range product list : 44225406378688905216000

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

  Method #2 : Using itertools.compress() + itertools.cycle() + loop The above two function can combine to facilitate the solution of the discussed problem. The cycle function can to the task of repetition and the compress function can be beneficial when it comes to clubbing the segments together. The loop is used to perform product. 

Python3




# Python3 code to demonstrate
# Consecutive Custom Chunked elements Product
# using itertools.compress() + itertools.cycle() + loop
from itertools import compress, cycle
 
# getting Product
def prod(val) :
    res = 1
    for ele in val:
        res *= ele
    return res 
 
# initializing lists
test_list = list(range(1, 50))
 
# printing original list
print ("The original list is : " + str(test_list))
 
# interval elements
N = 5
 
# interval difference
K = 15
 
# using itertools.compress() + itertools.cycle() + loop
# Consecutive Custom Chunked elements Product
func = cycle([True] * N + [False] * (K - N))
res = prod(list(compress(test_list, func)))
     
# printing result
print ("The modified range product list : " + str(res))


Output : 

The original list is : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49] The modified range product list : 44225406378688905216000

Time Complexity: O(n*n), where n is the length of the input list. This is because we’re using itertools.compress() + itertools.cycle() + loop which has a time complexity of O(n*n) in the worst case.
Auxiliary Space: O(1), as we’re using constant additional space 



Last Updated : 13 Apr, 2023
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