如何将列表扩展到一定大小而不重复每个单独的列表元素 n 次？

Question

I'm looking to keep the individual elements of a list repeating for x number of times, but can only see how to repeat the full list x number of times.

For example, I want to repeat the list [3, 5, 1, 9, 8] such that if x=12 , then I want to produce tthe following list (ie the list continues to repeat in order until there are 12 individual elements in the list:

[3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5]

I can do the below but this is obviously not what I want and I'm unsure how to proceed from here.

my_list = [3, 5, 1, 9, 8]
x = 12

print(my_list * 12)

[3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5, 1, 9, 8]

Answer 1

Your code repeats list 12 times. You need to repeat list until length is matched. This can achieved using Itertools - Functions creating iterators for efficient looping

from itertools import cycle, islice

lis = [3, 5, 1, 9, 8]
out = list(islice(cycle(lis), 12))
print(out)

Gives #

[3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5]

More pythonic #

Use a for loop to access each element in list and iterate over 'length' times. Repeat Ith element you access through loop in same list until length matches.

lis = [3, 5, 1, 9, 8]
length = 12

out = [lis[i%len(lis)] for i in range(length)]
print(out)

Gives ##

[3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5]

Answer 2

There are multiple ways to go about it. If x is the final length desired and lst is the list (please do not use list as a variable name because it overwrites the builtin list function), then you can do:

lst = (lst * (1 + x // len(lst)))[:x]

This multiplies the list by the smallest number needed to get at least N elements, and then it slice the list to keep only the first N. For your example:

>>> lst = [3, 5, 1, 9, 8]
>>> x = 12
>>> (lst * (1 + x // len(lst)))[:x]
[3, 5, 1, 9, 8, 3, 5, 1, 9, 8, 3, 5]

You could also use a loop, for example:

index = 0
while len(lst) < x:
    lst.append(lst[index])
    index += 1