Python | Ways to create triplets from given list

To create triplets from a given list, we aim to group consecutive elements into sublists of three. This process involves iterating through the list and extracting segments of three adjacent elements, resulting in a new list where each item is a triplet.

Using list comprehension

List Comprehension in Python provides a clean and efficient approach to achieve this. It iterates through the list, slicing the elements into smaller groups of three, and stores these sublists in the result list.

li = ['I', 'am', 'Paras', 'Jain', 'I', 'Study', 'DS', 'Algo']
res = [li[i:i + 3] for i in range(len(li) - 2)]
print(res)

[['I', 'am', 'Paras'], ['am', 'Paras', 'Jain'], ['Paras', 'Jain', 'I'], ['Jain', 'I', 'Study'], ['I', 'Study', 'DS'], ['Study', 'DS', 'Algo']]

Explanation:

• List comprehension: This iterates through the list li and slices it into sublists of size 3, starting from index i.
• range(len(li) - 2): This ensures that the loop runs enough times to create triplets without going out of bounds when slicing the list.
• li[i:i + 3]: This creates a sublist starting at index i and ending at index i + 3.

Using tee()

tee() function from the itertools module in Python allows the creation of multiple independent iterators from a single iterable, making it possible to generate triplets without re-iterating over the list. By using tee() with the islice() function,we can create three iterators that simulate a sliding window of triplets.

import itertools

li= ['I', 'am', 'Paras', 'Jain', 'I', 'Study', 'DS', 'Algo']

# Create 3 iterators
it1, it2, it3 = itertools.tee(li, 3)

# Advance iterators
it2 = itertools.islice(it2, 1, None)
it3 = itertools.islice(it3, 2, None)

# Create triplets
res = list(zip(it1, it2, it3))

print(res)

[('I', 'am', 'Paras'), ('am', 'Paras', 'Jain'), ('Paras', 'Jain', 'I'), ('Jain', 'I', 'Study'), ('I', 'Study', 'DS'), ('Study', 'DS', 'Algo')]

Explanation:

• tee() : This creates 3 independent iterators from the original list li .
• islice():it1 starts from the first element, it2 starts from the second and it3 starts from the third effectively simulating a sliding window.
• zip(): This combines the three iterators to form triplets, iterating through the list once.

Using loop

simple loop is a straightforward approach to manually collect triplets from a list. This method iterates through the list and slices it to create triplets, providing more control over the process.

li= ['I', 'am', 'Paras', 'Jain', 'I', 'Study', 'DS', 'Algo']

res = []    # An empty list res is initialized 
for i in range(len(li) - 2):
    res.append(li[i:i + 3])

print(res)

[['I', 'am', 'Paras'], ['am', 'Paras', 'Jain'], ['Paras', 'Jain', 'I'], ['Jain', 'I', 'Study'], ['I', 'Study', 'DS'], ['Study', 'DS', 'Algo']]

Explanation:

• len(li) - 2: As we need 3 elements to form each triplet that's why we need to stop loop before reaching the last 2 elements to avoid slicing beyond the list length.
• li[i:i+3]: In each iteration this slices 3 consecutive elements starting from index i.
• res.append(...): Each triplet formed by the slice is appended to the res list .

Using deque

deque from collections module can be used to create triplets by maintaining a sliding window. This method uses a deque to store the last three elements. When the deque size reaches 3 it appends the current window to the list of triplets.

from collections import deque

li = ['I', 'am', 'Paras', 'Jain', 'I', 'Study', 'DS', 'Algo']
window = deque(maxlen=3)

# An empty list res is initialized
res = []
for word in li:
    window.append(word)
    if len(window) == 3:
        res.append(list(window))

print(res)

[['I', 'am', 'Paras'], ['am', 'Paras', 'Jain'], ['Paras', 'Jain', 'I'], ['Jain', 'I', 'Study'], ['I', 'Study', 'DS'], ['Study', 'DS', 'Algo']]

Explanation:

• deque(maxlen=3): This creates a deque that holds up to 3 elements, automatically discarding the oldest when a new element is added.
• for word in li: This Loops through each word in li and adds it to the window deque.
• window.append(word): This adds the word to the deque, removing the oldest if it exceeds 3 elements.
• if len(window) == 3: This checks if the deque has 3 elements before forming a triplet.
• res.append(list(window)): This converts the deque to a list and appends it to res as a triplet.