# Python | Interval Initialization in list

• Difficulty Level : Easy
• Last Updated : 20 Feb, 2019

There are numerous ways to initialize the list with the elements, but sometimes, its required to initialize 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()`
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.

 `# Python3 code to demonstrate ``# interval initializing in list ``# using list comprehension + enumerate()`` ` `# initializing lists``test_list ``=` `list``(``range``(``50``))`` ` `# printing original list``print` `(``"The original list is : "` `+` `str``(test_list))`` ` `# interval elements``N ``=` `5`` ` `# interval difference``K ``=` `15`` ` `# using list comprehension + enumerate()``# interval initializing in list ``res ``=`  `[i ``for` `j, i ``in` `enumerate``(test_list) ``if` `j ``%` `K < N ]``     ` `# printing result ``print` `(``"The modified initialized list : "` `+`  `str``(res))`

Output :

The original list is : [0, 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 initialized list : [0, 1, 2, 3, 4, 15, 16, 17, 18, 19, 30, 31, 32, 33, 34, 45, 46, 47, 48, 49]

Method #2 : Using `itertools.compress() + itertools.cycle()`
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.

 `# Python3 code to demonstrate ``# interval initializing in list ``# using itertools.compress() + itertools.cycle()``from` `itertools ``import` `compress, cycle `` ` `# initializing lists``test_list ``=` `list``(``range``(``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()``# interval initializing in list ``func ``=` `cycle([``True``] ``*` `N ``+` `[``False``] ``*` `(K ``-` `N))``res ``=` `list``(compress(test_list, func))``     ` `# printing result ``print` `(``"The modified initialized list : "` `+`  `str``(res))`

Output :

The original list is : [0, 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 initialized list : [0, 1, 2, 3, 4, 15, 16, 17, 18, 19, 30, 31, 32, 33, 34, 45, 46, 47, 48, 49]

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