Question 217715 in Javascript Question posted in
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Javascript – Find the count of values in a given Object

const obj1 = {
  1: 'a',
  2: 'b',
  3: 'a',
  4: 'c',
  5: 'a',
  6: 'd',
  7: 'c',
};

For the above code I need to get output as below

output = { a: 3, b:  1, c: 2, d: 1 }
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2

Answers


  1. 0

    this function works

    function countOccurrences(obj) {
  const result = {};

  for (const key in obj) {
    const value = obj[key];
    result[value] = (result[value] || 0) + 1;
  }
  return result;
}
// Input object
const obj1 = {
  1: 'a',
  2: 'b',
  3: 'a',
  4: 'c',
  5: 'a',
  6: 'd',
  7: 'c'
};

// Call the function
const output = countOccurrences(obj1);
console.log(output); // Output: { a: 3, b: 1, c: 2, d: 1 }
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  2. 0

    A concise / advanced solution would look like this:

    Object.values(obj1).reduce((o,n)=>{
    if (!o[n]) o[n]=0;
    o[n]+=1;
    return o;
}, {})

    Object.values() gives you the values of an object (see documentation).

    Array.protoype.reduce() is a function which takes a function as its argument. Because we want to start on an empty object, we give it {} as a second input parameter. The reduce function iterates over all given array values and applying the given function to it. The result is a single value opposed to what Array.prototype.map does.
    (see documentation)

    For read-/memorability reasons I choose the parameter names o for old and n for new for the reduce function.
    What we are doing here is simple:

    • look whether on the given "old" value (which is in the first step {}) exists a key of n (n is a in the first step). If not give it a count of 0.
    • anyways increase the count at o[n].
    • return o

    As you see the o is passed in and back the function and it accumulates the count. So this part is often called accumulator. But I use o because old is easier for me to remember than accumulator.

    Functions like Array.prototype.map and Array.protoype.reduce ( and don’t forget Array.protoype.reduceRight) are hard to understand at first. But once understood they will be the best tools on your belt.

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