Create a list with n numbers with definite intervals in python

Question

I need to create a list from 1 to n numbers in a uniform interval.For example, the length of the list is 6 and I need with an interval of 2,it needs to create 6 numbers in uniform intervals. the interval and size of list are dynamic values.

l=[1,3,5,7,9,11]

I tried np.arange() and np.linespace() but they don't accomplish what I need.I am wondering whether there is any function to do this

The usage is `range(start, end, step)`. The step size is 2, so if the range has 6 elements, the end of the range needs to be 2*6. — Samwise, Jul 21 '20 at 17:45
@SriTest see docs: https://docs.python.org/3/library/functions.html#func-range — Mark, Jul 21 '20 at 17:46
Does this answer your question? [How to create a range of numbers with a given increment](https://stackoverflow.com/questions/18325312/how-to-create-a-range-of-numbers-with-a-given-increment) — CPak, Jul 21 '20 at 17:47
yes it does. `range(1, 2*13, 2) -> [1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25]` — Samwise, Jul 21 '20 at 18:00
How did you use `arange` and `linspace`? What was wrong with the result? Or `range`? — hpaulj, Jul 21 '20 at 18:11

Suryansu Dash · Accepted Answer · 2020-07-27T14:27:59.433

I believe you can do that with np.arange(), but I am unable to get the specifics of your question, but for now, given a dynamic start number, interval value and number of elements for the list, you can do something like

import numpy as np
start = 1
n = 6
interval = 2
l = np.arange(start, interval * n , interval)

Else if your start and final number are given, you can edit the above for the same too. The same can be accomplished with range() function too, with the same parameters as above, and

l = list(range(start, interval*n, interval))

edit:- you can also do interval*n + 1, if the list has to be inclusive of the number at the end.

Create a list with n numbers with definite intervals in python

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