How to write a function to calculate the frequency of occurrence of a digit in a series?

Question

Programming

There is a random series of zeros where the numbers 1 interspersed with random intervals. For example [0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0 , 0, 0, 0, 1, 0, 0]

Tell me how to write a function for calculating the frequency of occurrence of the number 1. That is, for example, a unit occurs on average every 2 positions, every 5 positions, every 1.5 positions, etc.
For the [0, 1, 0, 1, 0, 1, 0, 1] series, the frequency is 2 (one in every second position), but you need to calculate the average frequency for a random series with different intervals between 1.

Perhaps there is some formula, or tell me an example code, no matter what PL, you can even use pseudocode

Answer 1

In python, this is very easy to do.
len(x)/x.count(1)

Answer 2

If understood correctly:

from typing import Optional

def freq(arr: list, digit: int = 1) -> Optional[float]:
    count = 1
    res = []
    for x in arr:
        if x == digit:
            res.append(count)
            count = 0
        count += 1
    try:
        return sum(res) / len(res)
    except ZeroDivisionError:
        return None

a = [0, 0, 0, 0, 1, 0, 0, 0, 0, 1]
print(a, ' --> ', freq(a))

a = [0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0]
print(a, ' --> ', freq(a))

# [0, 0, 0, 0, 1, 0, 0, 0, 0, 1] --> 5.0
# [0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0] --> 3.142857142857143

Answer 3

let arr = [0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0];

let p1 = arr.reduce((acc,val)=>acc+=val,0)/arr.length; // вероятность встретить 1
let p0 = 1 - p1; // вероятность встретить 0

let ratio10 = p1/p0; // Соотношение единиц к нулям
let ratio01 = p0/p1; // Соотношение нулей к единицам

Answer 4

here in js

const getPeriodicity=(arr)=>
{
  let counter=0;
  let summ=0;
  let len=0
  for(let i in   arr)
  {
    let e=arr[i];
    
    if(e===1)
    {
      counter+=2;
    }
    else if(+i!==0){
      summ+=counter;
      len++;
      counter=0;
    }
  }
  return summ/len;
}