Program to find replicated list by replicating each element n times

Suppose we have a list of n elements; we have to repeat each element in the list n number of times.

So, if the input is like nums = [1,5,8,3], then the output will be [1, 1, 1, 1, 5, 5, 5, 5, 8, 8, 8, 8, 3, 3, 3, 3]

To solve this, we will follow these steps −

• n := size of nums
• ret := a new list
• for each num in nums, do
  • ret := ret concatenate a list with n number of nums
• return ret

Example

Let us see the following implementation to get better understanding ?

def solve(nums):
    n = len(nums)
    ret = []
    for num in nums:
        ret += [num] * n
    return ret

nums = [1, 5, 8, 3]
print(solve(nums))

The output of the above code is ?

[1, 1, 1, 1, 5, 5, 5, 5, 8, 8, 8, 8, 3, 3, 3, 3]

Using List Comprehension

We can also solve this problem using list comprehension for a more concise solution ?

def solve_compact(nums):
    n = len(nums)
    return [num for num in nums for _ in range(n)]

nums = [1, 5, 8, 3]
result = solve_compact(nums)
print(result)

[1, 1, 1, 1, 5, 5, 5, 5, 8, 8, 8, 8, 3, 3, 3, 3]

Using itertools.repeat()

Another approach uses itertools.repeat() with itertools.chain() ?

import itertools

def solve_itertools(nums):
    n = len(nums)
    return list(itertools.chain.from_iterable(itertools.repeat(num, n) for num in nums))

nums = [1, 5, 8, 3]
result = solve_itertools(nums)
print(result)

[1, 1, 1, 1, 5, 5, 5, 5, 8, 8, 8, 8, 3, 3, 3, 3]

Comparison

Conclusion

Use the basic loop approach for clarity and readability. List comprehension offers a more Pythonic solution, while itertools provides the best performance for large datasets.