Python | Triplet iteration in List
Last Updated :
22 Apr, 2023
List iteration is common in programming, but sometimes one requires to print the elements in consecutive triplets. This particular problem is quite common and having a solution to it always turns out to be handy. Lets discuss certain way in which this problem can be solved.
Method #1 : Using list comprehension List comprehension can be used to print the triplets by accessing current, next and next to next element in the list and then printing the same. Care has to be taken while pairing the last elements with the first ones to form a cyclic triplet pairs.
Python3
from itertools import compress
test_list = [0, 1, 2, 3, 4, 5]
print ("The original list is : " + str(test_list))
res = [((i), (i + 1) % len(test_list), (i + 2) % len(test_list))
for i in range(len(test_list))]
print ("The triplet list is : " + str(res))
|
Output
The original list is : [0, 1, 2, 3, 4, 5]
The triplet list is : [(0, 1, 2), (1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 0), (5, 0, 1)]
Time Complexity: O(n)
Auxiliary space: O(n)
Method#2 : Using zip() function zip function can be used to pair the element of the list provided. We generate the list whose starting element is next element and next to next element. Take care while generating list while merging the last element to first element to form the cyclic triplet.
Python3
import itertools
l = [0, 1, 2, 3, 4, 5]
print("The original list is : " + str(l))
def gen(l, n):
k1 = itertools.cycle(l);
k2 = itertools.dropwhile(lambda x: x!=n, k1)
k3 = itertools.islice(k2, None, len(l))
return list(k3)
ans = []
for i in zip(l, gen(l, 1), gen(l, 2)):
ans.append(tuple(i))
print ("The triplet list is : " + str(ans))
|
Output
The original list is : [0, 1, 2, 3, 4, 5]
The triplet list is : [(0, 1, 2), (1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 0), (5, 0, 1)]
Time Complexity: O(n)
Auxiliary space: O(n)
Method#3: Using For loop For loop can be used to print the triplets by accessing current, next and next to next in the list then print the same. Care has to been taken while pairing the last element with first element when forming cyclic triplet.
Python3
l = [0, 1, 2, 3, 4, 5]
k = len(l)
print("The original list is : " + str(l))
ans = []
for i in range(k):
x = (l[i], l[(i+1)%k], l[(i+2)%k]);
ans.append(x)
print ("The triplet list is : " + str(ans))
|
Output
The original list is : [0, 1, 2, 3, 4, 5]
The triplet list is : [(0, 1, 2), (1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 0), (5, 0, 1)]
Time Complexity: O(n)
Auxiliary space: O(n)
Method#4: Using zip() and slicing:
Python3
test_list = [0, 1, 2, 3, 4, 5]
result = list(zip(test_list, test_list[1:] + test_list[:1], test_list[2:] + test_list[:2]))
print(result)
|
Output
[(0, 1, 2), (1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 0), (5, 0, 1)]
Time Complexity: O(n)
Auxiliary space: O(n)
Method #5: Using recursion
Here’s a solution using recursion in Python:
Python3
def get_triplets_recursive(lst, triplet_lst, index):
if index == len(lst):
return
triplet_lst.append((lst[index], lst[(index + 1) % len(lst)], lst[(index + 2) % len(lst)]))
get_triplets_recursive(lst, triplet_lst, index + 1)
def get_triplets(lst):
triplet_lst = []
get_triplets_recursive(lst, triplet_lst, 0)
return triplet_lst
original_list = [0, 1, 2, 3, 4, 5]
triplet_list = get_triplets(original_list)
print(triplet_list)
|
Output
[(0, 1, 2), (1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 0), (5, 0, 1)]
Time complexity: O(n)
Auxiliary Space: O(n)
Method 6 : using the itertools module
step by step:
- Import the itertools module.
- Initialize the list.
- Use the itertools.cycle() function to cycle through the list.
- Use the itertools.islice() function to slice the list into triplets.
- Convert the sliced triplets into a list.
- Print the resulting list of triplets.
Python3
import itertools
l = [0, 1, 2, 3, 4, 5]
print("The original list is : " + str(l))
ans = [list(itertools.islice(itertools.cycle(l), i, i+3)) for i in range(len(l))]
print ("The triplet list is : " + str(ans))
|
Output
The original list is : [0, 1, 2, 3, 4, 5]
The triplet list is : [[0, 1, 2], [1, 2, 3], [2, 3, 4], [3, 4, 5], [4, 5, 0], [5, 0, 1]]
Time complexity: O(n), where n is the length of the list.
Auxiliary space: O(n), where n is the length of the list.
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