Identify duplicate values in an array by comparing pairs of elements

Question

Identify duplicate values in an array by comparing pairs of elements

If the array contains the data shown below:

let array = [{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 },
        { name: "Suresh", SalseVersion: 12, MarketingCode: 13 },
        { name: "Siva", SalseVersion: 10, MarketingCode: 14 },
        { name: "Sakthi", SalseVersion: 10, MarketingCode: 11 },...]

Then I expect the following result:

[{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 },
 { name: "Sakthi", SalseVersion: 10, MarketingCode: 11 }]

If the array contains the data shown below:

let array = [{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 },
        { name: "Suresh", SalseVersion: 12, MarketingCode: 14},
        { name: "Siva", SalseVersion: 12, MarketingCode: 14 },
        { name: "Sakthi", SalseVersion: 10, MarketingCode: 11 },...]

Then I expect the following result:

        [{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 },
        { name: "Sakthi", SalseVersion: 10, MarketingCode: 11 }
        { name: "Suresh", SalseVersion: 12, MarketingCode: 14},
        { name: "Siva", SalseVersion: 12, MarketingCode: 14 }]

I have attempted to solve this by utilizing the following method:

let arr = [{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 },
        { name: "Suresh", SalseVersion: 12, MarketingCode: 13 },
        { name: "Siva", SalseVersion: 10, MarketingCode: 14 },
        { name: "Sakthi", SalseVersion: 10, MarketingCode: 11 }]


var sorted_arr = arr.slice().sort();
var results = [];
for (var i = 0; i < sorted_arr.length - 1; i++) {
    if (sorted_arr[i + 1].SalesVersion == sorted_arr[i].SalesVersion && sorted_arr[i + 1].MarketingCode == sorted_arr[i].MarketingCode) {
        results.push(sorted_arr[i]);
    }
}

console.log(results);

Unfortunately, I was unable to retrieve the duplicated values. Can you assist me in resolving this issue?

Note: Please note that this solution should work regardless of whether the SalseVersion and MarketingCode values are strings, numbers, or Booleans.

I have tried some of the solutions provided, but I encountered the following error:

https://i.sstatic.net/ylOzt.jpg

javascript arrays typescript ecmascript-6

Answer 1

Answer №1

Utilize the reduce method for this task.

Initially, the first reduce function will group the values by combining the values of SalseVersion and MarketingCode.

Subsequently, the second reduce function will verify if the group contains more than one element. If so, it will concatenate the values into a single array.

let array = [{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 },{ name: "Suresh", SalseVersion: 12, MarketingCode: 14},{ name: "Siva", SalseVersion: 12, MarketingCode: 14 },{ name: "Sakthi", SalseVersion: 10, MarketingCode: 11 }];

let result = Object.values(array.reduce((c, v) => {
  let k = v.SalseVersion + '-' + v.MarketingCode;
  c[k] = c[k] || [];
  c[k].push(v);
  return c;
}, {})).reduce((c, v) => v.length > 1 ? c.concat(v) : c, []);

console.log( result );

Answer 2

Utilize the reduce method for this task.

Initially, the first reduce function will group the values by combining the values of SalseVersion and MarketingCode.

Subsequently, the second reduce function will verify if the group contains more than one element. If so, it will concatenate the values into a single array.

let array = [{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 },{ name: "Suresh", SalseVersion: 12, MarketingCode: 14},{ name: "Siva", SalseVersion: 12, MarketingCode: 14 },{ name: "Sakthi", SalseVersion: 10, MarketingCode: 11 }];

let result = Object.values(array.reduce((c, v) => {
  let k = v.SalseVersion + '-' + v.MarketingCode;
  c[k] = c[k] || [];
  c[k].push(v);
  return c;
}, {})).reduce((c, v) => v.length > 1 ? c.concat(v) : c, []);

console.log( result );

Answer 3

Answer №2

If you're looking for elements in an array that have at least one other element with similar properties, you can achieve this by using the .filter method and then sorting the results:

const findDuplicates = array => {
  const filteredDuplicates = array.filter(({ name, SalseVersion, MarketingCode }, i, arr) => (
    arr.find((item, findI) => (
      findI !== i &&
      SalseVersion === item.SalseVersion &&
      MarketingCode === item.MarketingCode
    ))
  ));
  return filteredDuplicates.sort((a, b) =>
    String(a.SalseVersion).localeCompare(String(b.SalseVersion), 'kn') ||
    String(a.MarketingCode).localeCompare(String(b.MarketingCode), 'kn')
  );
};

console.log(findDuplicates([{name:"Ramesh",SalseVersion:10,MarketingCode:11},{name:"Suresh",SalseVersion:12,MarketingCode:13},{name:"Siva",SalseVersion:10,MarketingCode:14},{name:"Sakthi",SalseVersion:10,MarketingCode:11}]))
console.log(findDuplicates([{name:"Ramesh",SalseVersion:10,MarketingCode:11},{name:"Suresh",SalseVersion:12,MarketingCode:14},{name:"Siva",SalseVersion:12,MarketingCode:14},{name:"Sakthi",SalseVersion:10,MarketingCode:11}]))

Answer 4

If you're looking for elements in an array that have at least one other element with similar properties, you can achieve this by using the .filter method and then sorting the results:

const findDuplicates = array => {
  const filteredDuplicates = array.filter(({ name, SalseVersion, MarketingCode }, i, arr) => (
    arr.find((item, findI) => (
      findI !== i &&
      SalseVersion === item.SalseVersion &&
      MarketingCode === item.MarketingCode
    ))
  ));
  return filteredDuplicates.sort((a, b) =>
    String(a.SalseVersion).localeCompare(String(b.SalseVersion), 'kn') ||
    String(a.MarketingCode).localeCompare(String(b.MarketingCode), 'kn')
  );
};

console.log(findDuplicates([{name:"Ramesh",SalseVersion:10,MarketingCode:11},{name:"Suresh",SalseVersion:12,MarketingCode:13},{name:"Siva",SalseVersion:10,MarketingCode:14},{name:"Sakthi",SalseVersion:10,MarketingCode:11}]))
console.log(findDuplicates([{name:"Ramesh",SalseVersion:10,MarketingCode:11},{name:"Suresh",SalseVersion:12,MarketingCode:14},{name:"Siva",SalseVersion:12,MarketingCode:14},{name:"Sakthi",SalseVersion:10,MarketingCode:11}]))

Answer 5

Answer №3

Make sure to update your code with the following changes:

let list = [{ name: "John", Version: 10, Code: 11 },
    { name: "Jane", Version: 12, Code: 14},
    { name: "Jill", Version: 12, Code: 14 },
    { name: "Jack", Version: 10, Code: 11 },
    { name: "Jenny", Version: 10, Code: 11 }]

// Implement a custom sorting function
var sorted_list = list.slice().sort((a,b) => {
     if(a.Version === b.Version) {
          return a.Code - b.Code;
      }
      return a.Version - b.Version;
});
var results = [];
for (var i = 0; i < sorted_list.length - 1; i++) {
    if (sorted_list[i + 1].Version == sorted_list[i].Version && sorted_list[i + 1].Code == sorted_list[i].Code) {
        /* push both elements when there is a match, unless it has already been pushed for 2 matches */
       if (i >= 1 && sorted_list[i - 1].Version == sorted_list[i].Version && sorted_list[i - 1].Code == sorted_list[i].Code) {
         results.push(sorted_list[i+1]);
       } else {
         results.push(sorted_list[i], sorted_list[i+1]);
       }            
    }
}

console.log(results);

Answer 6

Make sure to update your code with the following changes:

let list = [{ name: "John", Version: 10, Code: 11 },
    { name: "Jane", Version: 12, Code: 14},
    { name: "Jill", Version: 12, Code: 14 },
    { name: "Jack", Version: 10, Code: 11 },
    { name: "Jenny", Version: 10, Code: 11 }]

// Implement a custom sorting function
var sorted_list = list.slice().sort((a,b) => {
     if(a.Version === b.Version) {
          return a.Code - b.Code;
      }
      return a.Version - b.Version;
});
var results = [];
for (var i = 0; i < sorted_list.length - 1; i++) {
    if (sorted_list[i + 1].Version == sorted_list[i].Version && sorted_list[i + 1].Code == sorted_list[i].Code) {
        /* push both elements when there is a match, unless it has already been pushed for 2 matches */
       if (i >= 1 && sorted_list[i - 1].Version == sorted_list[i].Version && sorted_list[i - 1].Code == sorted_list[i].Code) {
         results.push(sorted_list[i+1]);
       } else {
         results.push(sorted_list[i], sorted_list[i+1]);
       }            
    }
}

console.log(results);

Answer 7

Answer №4

In a nutshell, you can achieve grouping by sorting the array based on the desired key. This will result in an array where all keys and their corresponding values are grouped together.

Subsequently, you can filter the array by comparing each key with its adjacent values to identify the desired groups.

var array = [{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 }, { name: "Suresh", SalseVersion: 12, MarketingCode: 13 }, { name: "Siva", SalseVersion: 10, MarketingCode: 14 }, { name: "Sakthi", SalseVersion: 10, MarketingCode: 11 }],
    keys = ['SalseVersion', 'MarketingCode'],
    result = array
        .sort((a, b) => {
            var diff;
            keys.some(k => diff = a[k] - b[k]);
            return diff;
        })
        .filter((o, i, a) => 
            keys.every(k => a[i - 1] && a[i - 1][k] === o[k]) ||
            keys.every(k => a[i + 1] && o[k] === a[i + 1][k])
        );

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

If the data is unsorted, an alternative method involves using a hash table to temporarily store and identify the first object in each group or detect duplicates within the group.

var array = [{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 }, { name: "Suresh", SalseVersion: 12, MarketingCode: 13 }, { name: "Siva", SalseVersion: 10, MarketingCode: 14 }, { name: "Sakthi", SalseVersion: 10, MarketingCode: 11 }],
    keys = ['SalseVersion', 'MarketingCode'],
    result = array.reduce((hash => (res, obj) => {
        var key = keys.map(k => obj[k]).join('|');
        if (key in hash) {
            if (hash[key]) {
                res.push(hash[key]);
                hash[key] = false;
            }
            res.push(obj);
        } else {
             hash[key] = obj;
        }
        return res;
    })(Object.create(null)), []);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 8

In a nutshell, you can achieve grouping by sorting the array based on the desired key. This will result in an array where all keys and their corresponding values are grouped together.

Subsequently, you can filter the array by comparing each key with its adjacent values to identify the desired groups.

var array = [{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 }, { name: "Suresh", SalseVersion: 12, MarketingCode: 13 }, { name: "Siva", SalseVersion: 10, MarketingCode: 14 }, { name: "Sakthi", SalseVersion: 10, MarketingCode: 11 }],
    keys = ['SalseVersion', 'MarketingCode'],
    result = array
        .sort((a, b) => {
            var diff;
            keys.some(k => diff = a[k] - b[k]);
            return diff;
        })
        .filter((o, i, a) => 
            keys.every(k => a[i - 1] && a[i - 1][k] === o[k]) ||
            keys.every(k => a[i + 1] && o[k] === a[i + 1][k])
        );

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

If the data is unsorted, an alternative method involves using a hash table to temporarily store and identify the first object in each group or detect duplicates within the group.

var array = [{ name: "Ramesh", SalseVersion: 10, MarketingCode: 11 }, { name: "Suresh", SalseVersion: 12, MarketingCode: 13 }, { name: "Siva", SalseVersion: 10, MarketingCode: 14 }, { name: "Sakthi", SalseVersion: 10, MarketingCode: 11 }],
    keys = ['SalseVersion', 'MarketingCode'],
    result = array.reduce((hash => (res, obj) => {
        var key = keys.map(k => obj[k]).join('|');
        if (key in hash) {
            if (hash[key]) {
                res.push(hash[key]);
                hash[key] = false;
            }
            res.push(obj);
        } else {
             hash[key] = obj;
        }
        return res;
    })(Object.create(null)), []);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Identify duplicate values in an array by comparing pairs of elements

Answer №1

Answer №2

Answer №3

Answer №4

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