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Franqo Balsamo
Asked: 2020-09-11 09:01:46 +0800 CST 2020-09-11 09:01:46 +0800 CST

Percentage of numbers greater than, equal to, and less than 0 in an array

  • 772

I am trying to generate a function that is parameterized and returns an array with results in percentage following an order. The final result should be an array showing the percentage of values ​​greater than zero, equal to zero, and less than zero.

Example:

masMenos(1,2,0,-1)

should result in:

0.5, 0.25, 0.25

Where 0.5 is the percentage of numbers greater than 0, 0.25 equal to zero, and 0.25 less than zero.

Any suggestion?

    var array=[1,2,0,-1];
    function masMenos(array){
      var cantidad=1;
      for(var i =0;i<array.length;i++){
        if(array[i]>0){
          cantidad=(array[i]*cantidad)/100;
        }else if(array[i]==0){
          cantidad=(array[i]*cantidad)/100;
        }else {
          cantidad=(array[i]*cantidad)/100;
        }
      }return cantidad;
    }
    console.log(masMenos(1,2,0,-1));

javascript

3 Answers

  1. Best Answer

    You can do it simply with the function filter():

        function masMenos(array){
      may=array.filter(num => num>0);
      igu=array.filter(num => num==0);
      men=array.filter(num => num<0);
      return {
        mayores: may.length/array.length,
        iguales: igu.length/array.length,
        menores: men.length/array.length
      };
    }
    
    console.log(masMenos([1,2,-1,0,1,0,1,0]));

    The function filter()filters the values ​​of the array according to a given function (in this case if they are positive, negative or zero).

    Although this method is simple and readable, it is not the most optimal, since each time that function is used, the array is traversed once.

    A more optimal and recommended option that uses a single loop is the following:

    function masMenos(array, res=[0,0,0], n=array.length) {
    for (i in array)
      array[i]>0 ? res[0]++ : array[i]<0 ? res[2]++ : res[1]++
    return res.map(a =>a/n);
  }

  console.log(masMenos([1,2,-1,0,1,0,1,0]));

    Of course the most optimal answers lose legibility.

    I hope that the solution has been clearer, greetings.

    • 6

  2. If you want it to show you three results, you need to declare and return an array, not a simple variable. You can do it like this:

    var array=[1,2,0,-1];
    function masMenos(array){      
      var cantidad = [0,0,0]
      for(var i =0;i<array.length;i++){
        if(array[i]>0){
          cantidad[0]+=1;
        }else if(array[i]==0){
          cantidad[1]+=1;
        }else {
          cantidad[2]+=1;
        }
      }
      cantidad[0] = cantidad[0]/array.length;
      cantidad[1] = cantidad[1]/array.length;
      cantidad[2] = cantidad[2]/array.length;
      return cantidad;
    }
    console.log(masMenos(array));

    • 3

  3. The most optimal thing would be to solve it in a single iteration, and it can be using for, as they have already answered, or forEach:

    const arr = [1, 2, 0, -1];

let p = [0, 0, 0];
const inv = 1 / arr.length;

arr.forEach(e => e > 0 ? p[0] += inv : e < 0 ? p[2] += inv : p[1] += inv);

console.log(p);

    dividing the quantity by the whole can be avoided if one does the sum of the parts 1/largo, because we apply the distributive property (1+2)/2 = (1/2)+(2/2) .

    Using reduce()is almost the same:

    const arr = [1, 2, 0, -1];
const inv = 1 / arr.length;

const res = arr.reduce((p, c) => {
  c > 0 ? p[0] += inv : c < 0 ? p[2] += inv : p[1] = inv;
  return p;
}, [0, 0, 0]);

console.log(res);

    • 3