GeeksforGeeks App
Open App
Browser
Continue

# Python | Consecutive Custom Chunked elements Product

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

My Personal Notes arrow_drop_up