counting how many times an item appears in a multidimensional array in javascript

Question

Given a multidimensional array like this:

 var arr = [
\"apple\",
[\"banana\", \"strawberry\",\"dsffsd\", \"apple\"],
\"banana\",
[\"sdfdsf\",\"apple\",[\"apple\         


        

Answer 1

i think if you continue to return the count value and use it in your calculation you will be able to to get a final number without an outside variable:

function countItems(arr, item) {
  let totc = 0; // totc is now local
  for (let i = 0; i < arr.length; i++ ){ // iterate on arr
        let isarray = arr[i].constructor === Array; // if the element is a nested array
        if(isarray){  totc += countItems(arr[i], item) } // recursion, using the return value



        let isin = arr[i] === item; // ELSE if the item is in the arr
        if(isin) { totc ++;  };



    }
    return totc; // the length of the sum array show how many items founded
  }
}

note that the only changes i made were to instantiate totc inside the function and to take the result of recursive calls and add them to the local total.

Nina's answer is more elegant, but perhaps this will be easier to understand.

Answer 2

You could iterate the flat version of arr using arr.toString().split(",") so you don't need recursion.

var arr =  [
    "apple",
    ["banana", "strawberry","dsffsd", "apple"],
    "banana",
    ["sdfdsf","apple",["apple",["nonapple", "apple",["apple"]]]]
    ,"apple"];


var counts = {};


 arr.toString().split(",").forEach(e=>{
    counts[e] = (counts[e] || 0) +1
 })   

 console.log(counts.apple)

Doesn't work if elements have "," inside, but works with.flat() instead of .toString().split(",")

Answer 3

You could take a recursive approach for arrays or check the value and add the result of the boolean value.

function count(array, value) {
    return array.reduce((s, a) => s + (Array.isArray(a) ? count(a, value) : a === value), 0);
}

var array = ["apple", ["banana", "strawberry", "dsffsd", "apple"], "banana", ["sdfdsf", "apple", ["apple", ["nonapple", "apple", ["apple"]]]], "apple"];

console.log(count(array, 'apple'));

Version with a for loop.

function count(array, value) {
    var i,
        sum = 0;

    for (i = 0; i < array.length; i++) {
        if (Array.isArray(array[i])) {
            sum += count(array[i], value);
            continue;
        }
        sum += array[i] === value;
    }
    return sum;
}

var array = ["apple", ["banana", "strawberry", "dsffsd", "apple"], "banana", ["sdfdsf", "apple", ["apple", ["nonapple", "apple", ["apple"]]]], "apple"];

console.log(count(array, 'apple'));

Answer 4

I think this is a simple approach but you can get tangled with it:

 function countItems(arr, item, count = 0){
     if(!arr.length) return count; //if the array is empty then there's nothing else to count
     let cTemp;
     if(Array.isArray(arr[0])){ //if the next item is an array
         cTemp = countItems(arr[0], item); //count the items in that array
     } else {
         cTemp = arr[0] === item ? 1 : 0; //if it's a string the compare it with item
         //1 if we found it
         //0 if we didn't
     }
     return countItems(arr.slice(1), item, count+cTemp);
     //count the items of the rest of the array and add what we found
     //arr.slice(1) is the rest of the array
     //cTemp is the count for the first item in the array
 }

Which of course can be rewritten into a single line:

 let countItems = ([first, ...rest], item, count = 0) => !first ? count : countItems(rest, item, count + (Array.isArray(first) ? countItems(first, item) : +(first === item)))