'Counting negative, positive and zero element in an array using plain javascript

I have an array:

 var ar1=[];
 var ar=[-1,-2,-3,0,0,5,12,0,-10];

I am trying to categorize numbers as "zeros", "negative numbers" and "positive numbers" and counting them.

This is my code:

function counter(ar)  { 
   var num,array1=[0,0,0];
   for (i=0;i<ar.length;i++)    {
      switch (ar[i]<0)      {
         case true : array1[0]++;break;
         case false : 
                        if (ar[i]=0) array1[1]++;
                        else array1[2]++;
                        break;
         default : break;
      }
   }
  return(array1);
}

Full code:

<script type="text/javascript">
function counter(ar)  { 
   var num,array1=[0,0,0];
   for (i=0;i<ar.length;i++)    {
      switch (ar[i]<0)      {
         case true : array1[0]++;break;
         case false : 
                         if (ar[i]=0) array1[1]++;
                        else array1[2]++;
                        break;
         default : break;
      }
   }
  return(array1);
}
</script></head>
<body>
<script type="text/javascript">
var ar1=[];
 var ar=[-1,-2,-3,0,0,5,12,0,-10];
 ar1=counter(ar);
 alert("No of Negative, Zero and Positive Elements are : "+ar1);
</script>

The logic seems correct to me, but somehow it's not working. Can someone help me please.

Update

I am new in coding and just learning with javascript basics. I forgot to mention this earlier and now I see hate all over the comments.

"Is this the best way to do it?" I should have added this to my question.

I see in comments that switch case is not a good option. So what would be an alternative to that?

javascripthtml


Solution 1:[1]

To compare you need to use ===.

= is used to assign a value to a variable.

Don't use a switch in your case. You only need if else 

var ar = [-1, -2, -3, 0, 0, 5, 12, 0, -10];

function counter(ar) {
  var counter = [0, 0, 0];
  ar.forEach(function(a) {
    if (a < 0)
      counter[0]++;
    else if (a > 0)
      counter[2]++;
    else
      counter[1]++;
  });
  return counter;
}

var result = counter(ar);
alert("No of Negative, Zero and Positive Elements are : " + result);

If you want get all the values separately, one may use object destructuring:

 let [negative,zero,positive] = counter([0,0,1,1,2,2]);

 if(! negative) alert("no negative ones but "+zero+" nulls");

Solution 2:[2]

You can use the filter method to count with more ease. The following code worked for me.

var arr=[-1,-2,-3,0,0,5,12,0,-10]
function myFunction(arr){

   let negatives=arr.filter((el) => el < 0).length;
   let zeros=arr.filter((el) => el == 0).length;
   let positives =arr.filter((el) => el > 0).length;
   return{
       neg:negatives,
       zer:zeros,
       pos:positives
   }
}
console.log(myFunction(arr))

Solution 3:[3]

The answer is if (ar[i] == 0) array1[1]++;.

Solution 4:[4]

For counting positive, zero and negative number use below logic, just paste inside console you will get the results.

var ar = [-1, -2, -3, 0, 0, 5, 12, 0, -10];
var zeroCount = 0,
  posCount = 0,
  negativeCount = 0;
ar.forEach((item) => {
  if (item === 0) {
    zeroCount++
  } else if (item < 0) {
    negativeCount++
  } else if (item > 0) {
    posCount++
  }
})

console.log("ZeroCount:: " + zeroCount);
console.log("PositiveCount :: " + posCount);
console.log("NegativeCount:: " + negativeCount);

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Rajesh
Solution 2 Suraj Rao
Solution 3 Mustansir Zia
Solution 4 Barmar

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