Determine the frequency of a specific key in an array of objects

Question

Determine the frequency of a specific key in an array of objects

original array: ................

[
   {
    from: {_id: "60dd7c7950d9e01088e438e0"}
   },
   {
    from: {_id: "60dd7c7950d9e01088e438e0"}
   },
   {
    from: {_id: "60dd7e19e6b26621247a35cd"}
   }
]

A new array is created to count the instances of each _id and store it in messageCount .................

[
 {
  from: {_id: "60dd7c7950d9e01088e438e0"},
  messageCount: 2
 },
 {
  from: {_id: "60dd7e19e6b26621247a35cd"},
  messageCount: 1
 }
]

javascript arrays

Answer 1

Answer №1

To tackle this issue, as suggested in the feedback, you can employ the reduce method.

const items = [
   {
    from: {_id: "60dd7c7950d9e01088e438e0"}
   },
   {
    from: {_id: "60dd7c7950d9e01088e438e0"}
   },
   {
    from: {_id: "60dd7e19e6b26621247a35cd"}
   }
];

/**
 * Function to extract the id from an item
 */
function getId(item){
    return item['from']['_id'];
}

// The initial value is an empty array passed as the second argument of the reduce function
// @see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce
const result = items.reduce(function (previous, next) {
    // Retrieve the id of the next item (or current item in the iteration)
    const id = getId(next);
    // Search for an existing item in the array
    // @see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/findIndex
    const index = previous.findIndex(item => id === getId(item));

    if (index !== -1) {
        // If there is already an item with the same id, increment the message count
        previous[index]['messageCount'] += 1;
    } else {
        // If there isn't an item, add a new one with a message count of 1
        previous.push({
            ...next,
            messageCount: 1,
        });
    }

    return previous;
}, []);

console.log(result);

References:

Answer 2

To tackle this issue, as suggested in the feedback, you can employ the reduce method.

const items = [
   {
    from: {_id: "60dd7c7950d9e01088e438e0"}
   },
   {
    from: {_id: "60dd7c7950d9e01088e438e0"}
   },
   {
    from: {_id: "60dd7e19e6b26621247a35cd"}
   }
];

/**
 * Function to extract the id from an item
 */
function getId(item){
    return item['from']['_id'];
}

// The initial value is an empty array passed as the second argument of the reduce function
// @see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce
const result = items.reduce(function (previous, next) {
    // Retrieve the id of the next item (or current item in the iteration)
    const id = getId(next);
    // Search for an existing item in the array
    // @see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/findIndex
    const index = previous.findIndex(item => id === getId(item));

    if (index !== -1) {
        // If there is already an item with the same id, increment the message count
        previous[index]['messageCount'] += 1;
    } else {
        // If there isn't an item, add a new one with a message count of 1
        previous.push({
            ...next,
            messageCount: 1,
        });
    }

    return previous;
}, []);

console.log(result);

References:

Answer 3

Answer №2

Utilizing array methods, I incorporated a combination of .forEach loop along with the .reduce method ... Additionally, to determine the frequency of occurrences, I devised a straightforward array method ...





const array = [
   {
    from: {_id: "60dd7c7950d9e01088e438e0"}
   },
   {
    from: {_id: "60dd7c7950d9e01088e438e0"}
   },
   {
    from: {_id: "60dd7e19e6b26621247a35cd"}
   }
];

/*
  Formulating an array method capable of identifying the number of times the specified item appears within the array 
*/

Array.prototype.getItemCount = function(item) {
  let counts = {};
  for (let i = 0; i < this.length; i++) {
    let num = this[i];
    counts[num] = counts[num] ? counts[num]+1: 1;
  }
  return counts[item] || 0;
};


let result = [];

// Collating all ids and storing them in a constant using the array.reduce method
const allIds = array.reduce((acc,item)=> acc.concat(item.from._id),[]);

// Employing a forEach loop coupled with a ternary operator to filter out unique ids (conditions)
let filtered_id = [];
allIds.forEach((id)=> {
   !filtered_id.includes(id) ? filtered_id.push(id) : null;
});

// Eventually, compiling all the pertinent data into the designated result!

filtered_id.forEach(id =>{
    result.push({
    from: { _id: id },
    messageCount :  allIds.getItemCount(id)
    });
});

console.log(result);

Answer 4

Utilizing array methods, I incorporated a combination of .forEach loop along with the .reduce method ... Additionally, to determine the frequency of occurrences, I devised a straightforward array method ...





const array = [
   {
    from: {_id: "60dd7c7950d9e01088e438e0"}
   },
   {
    from: {_id: "60dd7c7950d9e01088e438e0"}
   },
   {
    from: {_id: "60dd7e19e6b26621247a35cd"}
   }
];

/*
  Formulating an array method capable of identifying the number of times the specified item appears within the array 
*/

Array.prototype.getItemCount = function(item) {
  let counts = {};
  for (let i = 0; i < this.length; i++) {
    let num = this[i];
    counts[num] = counts[num] ? counts[num]+1: 1;
  }
  return counts[item] || 0;
};


let result = [];

// Collating all ids and storing them in a constant using the array.reduce method
const allIds = array.reduce((acc,item)=> acc.concat(item.from._id),[]);

// Employing a forEach loop coupled with a ternary operator to filter out unique ids (conditions)
let filtered_id = [];
allIds.forEach((id)=> {
   !filtered_id.includes(id) ? filtered_id.push(id) : null;
});

// Eventually, compiling all the pertinent data into the designated result!

filtered_id.forEach(id =>{
    result.push({
    from: { _id: id },
    messageCount :  allIds.getItemCount(id)
    });
});

console.log(result);

Determine the frequency of a specific key in an array of objects

Answer №1

Answer №2

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