Python | Interval Initialization in list
Last Updated :
20 Apr, 2023
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
test_list = list ( range ( 50 ))
print ("The original list is : " + str (test_list))
N = 5
K = 15
res = [i for j, i in enumerate (test_list) if j % K < N ]
print ("The modified initialized list : " + str (res))
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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]
The time complexity of this approach is O(N), where N is the length of the input list.
The auxiliary space required is also O(N), where N is the length of the input list.
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
from itertools import compress, cycle
test_list = list ( range ( 50 ))
print ("The original list is : " + str (test_list))
N = 5
K = 15
func = cycle([ True ] * N + [ False ] * (K - N))
res = list (compress(test_list, func))
print ("The modified initialized list : " + str (res))
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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]
Time Complexity: O(n), where n is the length of the list test_list
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list
Method #3 : Using slicing and looping
The function takes a list test_list, an integer N, and an integer K as input. It initializes an empty list res and an index i to zero. Then, it enters a while loop that will continue until i becomes greater than or equal to the length of test_list. Inside the loop, it adds the elements from test_list to res, starting at index i and ending at index i + N. It then increases i by `K
Python3
def interval_init(test_list, N, K):
res = []
i = 0
while i < len (test_list):
res + = test_list[i:i + N]
i + = K
return res
test_list = list ( range ( 50 ))
print (interval_init(test_list, 5 , 15 ))
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Output
[0, 1, 2, 3, 4, 15, 16, 17, 18, 19, 30, 31, 32, 33, 34, 45, 46, 47, 48, 49]
This approach has a time complexity of O(n), where n is the length of the original list. The space complexity is also O(n), as we are creating a new list to store the result.
This approach is similar to the first method using list comprehension, but it avoids the use of enumerate and modulo operator. It simply uses a while loop to iterate through the original list and add the desired elements to the result list.
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