What is an efficient way to create the list [0, 2, 3, ..., 234] in python?

Question

I want to create the list [0, 2, 3, ..., 234] in python, possibly within a row and without listing all numbers explicitly

I know that following code works:

list1 = [0]
for i in range(2,235):
    list1.append(i)

But are there any possibilities to finish that within a row?

You can literally write [0, 2, 3, ..., 234], that's valid Python :-) — no comment
– no comment, Commented Sep 10, 2024 at 0:10

President James K. Polk · Accepted Answer · 2024-09-12 15:04:07Z

4

Another efficient way, combining simplicity and performance, is to use unpacking:

list1 = [0, *range(2, 235)]

edited Sep 12, 2024 at 15:04

President James K. Polk

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answered Sep 9, 2024 at 23:19

Aldair Huamani

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1 Comment

jdunlop Over a year ago

I would suggest that this be accepted as the correct answer rather than mine (though I agree with @nocomment about overloading the keyword), since it's not only a one-liner per the OP but empirically performs better than mine.

jdunlop · Accepted Answer · 2024-09-09 23:07:47Z

2

list1 = [0] + list(range(2,235))

list will expand a lazy enumerator.

answered Sep 9, 2024 at 23:07

jdunlop

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Comments

Di Nguyen · Accepted Answer · 2024-09-12 15:31:22Z

0

This is probably not as efficient as the other two answers but is a direct translation of your code to a 1 liner.

list1 = [0] + [i for i in range(2,235)]

answered Sep 12, 2024 at 15:31

Di Nguyen

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