What is the best way to identify and delete duplicate rows in a JSON array?

Question

What is the best way to identify and delete duplicate rows in a JSON array?

Suppose we have two variables, x and y. The task at hand is to compare these variables using JavaScript code in order to generate the output by eliminating any duplicate data.

Duplicate data here refers to instances where both x and y contain values: 202, 203, 204, 205.

var x = [{'pID': 200},{'pID': 201},{'pID': 202},{'pID': 203},{'pID': 204},{'pID': 205}];
var y = [{'fID': 202},{'fID': 203},{'fID': 204},{'fID': 205}];
output = [{'ID': 200}, {'ID': 201}];

How can this be achieved? The two variables share some common data points but with different key names. Please advise if there are any errors in my naming conventions.

The JSON array x functions as the primary data source, meaning it contains more data than the JSON array y.

Thank you.

javascript json

Answer 1

Answer №1

If you're looking for a simple solution, using filter and map can do the trick. First, apply the filter and then map the result type. Since all items in b are present in a, this approach should work smoothly.

a.filter(aIt => !b.find(bIt => bIt.fID === aIt.pID)).map(res => ({'ID': res.pID}))

Check out this working example:

var a = [{'pID': 200},{'pID': 201},{'pID': 202},{'pID': 203},{'pID': 204},{'pID': 205}];
var b = [{'fID': 202},{'fID': 203},{'fID': 204},{'fID': 205}];

var res = a.filter(aIt => !b.find(bIt => bIt.fID === aIt.pID)).map(res => ({'ID': res.pID}));

console.log(res);

Alternatively, you can use reduce to skip the additional mapping step:

a.reduce((res, aIt) => (!b.find(bIt => bIt.fID === aIt.pID) && !res.push({ID: aIt.pID}) || res), []);

Here's another working example with reduce:

var a = [{'pID': 200},{'pID': 201},{'pID': 202},{'pID': 203},{'pID': 204},{'pID': 205}];
var b = [{'fID': 202},{'fID': 203},{'fID': 204},{'fID': 205}];

var res = a.reduce((res, aIt) => (!b.find(bIt => bIt.fID === aIt.pID) && !res.push({ID: aIt.pID}) || res), []);

console.log(res)

Tip: You can also consider using the some method to filter effectively, as it returns true or false more directly compared to find which returns an object or undefined.

Answer 2

If you're looking for a simple solution, using filter and map can do the trick. First, apply the filter and then map the result type. Since all items in b are present in a, this approach should work smoothly.

a.filter(aIt => !b.find(bIt => bIt.fID === aIt.pID)).map(res => ({'ID': res.pID}))

Check out this working example:

var a = [{'pID': 200},{'pID': 201},{'pID': 202},{'pID': 203},{'pID': 204},{'pID': 205}];
var b = [{'fID': 202},{'fID': 203},{'fID': 204},{'fID': 205}];

var res = a.filter(aIt => !b.find(bIt => bIt.fID === aIt.pID)).map(res => ({'ID': res.pID}));

console.log(res);

Alternatively, you can use reduce to skip the additional mapping step:

a.reduce((res, aIt) => (!b.find(bIt => bIt.fID === aIt.pID) && !res.push({ID: aIt.pID}) || res), []);

Here's another working example with reduce:

var a = [{'pID': 200},{'pID': 201},{'pID': 202},{'pID': 203},{'pID': 204},{'pID': 205}];
var b = [{'fID': 202},{'fID': 203},{'fID': 204},{'fID': 205}];

var res = a.reduce((res, aIt) => (!b.find(bIt => bIt.fID === aIt.pID) && !res.push({ID: aIt.pID}) || res), []);

console.log(res)

Tip: You can also consider using the some method to filter effectively, as it returns true or false more directly compared to find which returns an object or undefined.

Answer 3

Answer №2

To find unique values, you can combine two arrays and calculate the occurrences of each value using the array#reduce method.

var array1 = [{'id': 100}, {'id': 101}, {'id': 102}, {'id': 103},{'id': 104}, {'id': 105}];    
var array2 = [{'id': 102}, {'id': 103}, {'id': 104}, {'id': 105}];

var mergedArray = array1.concat(array2).reduce((result, value) => {
  var val = Object.values(value)[0];
  result[val] = (result[val] || 0) + 1;
  return result;
},{});

var uniqueValues = Object.keys(mergedArray).reduce((resultArray, key) => {
  if(mergedArray[key] == 1)
    resultArray.push({ID : key});
  return resultArray;
},[]);
console.log(uniqueValues);

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Answer 4

To find unique values, you can combine two arrays and calculate the occurrences of each value using the array#reduce method.

var array1 = [{'id': 100}, {'id': 101}, {'id': 102}, {'id': 103},{'id': 104}, {'id': 105}];    
var array2 = [{'id': 102}, {'id': 103}, {'id': 104}, {'id': 105}];

var mergedArray = array1.concat(array2).reduce((result, value) => {
  var val = Object.values(value)[0];
  result[val] = (result[val] || 0) + 1;
  return result;
},{});

var uniqueValues = Object.keys(mergedArray).reduce((resultArray, key) => {
  if(mergedArray[key] == 1)
    resultArray.push({ID : key});
  return resultArray;
},[]);
console.log(uniqueValues);

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Answer 5

Answer №3

To simplify the process, start by extracting the values from the object and then utilize the filter method to eliminate duplicates.

var getValues = ( array ) => array.reduce( (a,b) => a.concat( Object.values( b ) ), [] );
var getCommonValues = ( arr1, arr2 ) => arr1.filter( item => arr2.indexOf( item ) == -1 );

var arrayA = getValues(a); // retrieve values from object A
var arrayB = getValues(b);

var result = arrayA.length > arrayB.length ? 
         getCommonValues(arrayA, arrayB) : getCommonValues(arrayB, arrayA) ; // use the larger array as the first parameter

Example

var a = [{
  'pID': 200
}, {
  'pID': 201
}, {
  'pID': 202
}, {
  'pID': 203
}, {
  'pID': 204
}, {
  'pID': 205
}];
var b = [{
  'fID': 202
}, {
  'fID': 203
}, {
  'fID': 204
}, {
  'fID': 205
}];
var getValues = (arr) => arr.reduce((a, b) => a.concat(Object.values(b)), []);
var getIntersection = (a, b) => a.filter(item => b.indexOf(item) == -1);

var arrayA = getValues(a); //get the values
var arrayB = getValues(b);
var output = arrayA.length > arrayB.length ?
  getIntersection(arrayA, arrayB) : getIntersection(arrayB, arrayA);

console.log(output);

Answer 6

To simplify the process, start by extracting the values from the object and then utilize the filter method to eliminate duplicates.

var getValues = ( array ) => array.reduce( (a,b) => a.concat( Object.values( b ) ), [] );
var getCommonValues = ( arr1, arr2 ) => arr1.filter( item => arr2.indexOf( item ) == -1 );

var arrayA = getValues(a); // retrieve values from object A
var arrayB = getValues(b);

var result = arrayA.length > arrayB.length ? 
         getCommonValues(arrayA, arrayB) : getCommonValues(arrayB, arrayA) ; // use the larger array as the first parameter

Example

var a = [{
  'pID': 200
}, {
  'pID': 201
}, {
  'pID': 202
}, {
  'pID': 203
}, {
  'pID': 204
}, {
  'pID': 205
}];
var b = [{
  'fID': 202
}, {
  'fID': 203
}, {
  'fID': 204
}, {
  'fID': 205
}];
var getValues = (arr) => arr.reduce((a, b) => a.concat(Object.values(b)), []);
var getIntersection = (a, b) => a.filter(item => b.indexOf(item) == -1);

var arrayA = getValues(a); //get the values
var arrayB = getValues(b);
var output = arrayA.length > arrayB.length ?
  getIntersection(arrayA, arrayB) : getIntersection(arrayB, arrayA);

console.log(output);

Answer 7

Answer №4

To eliminate repeated values that are unwanted, you can utilize the Array Filter method in JavaScript. The approach involves fetching all values from variable b and then removing objects from variable a that share the same values.

// sample data
var a = [{'pID': 200},{'pID': 201},{'pID': 202},{'pID': 203},{'pID': 204},{'pID': 205}];
var b = [{'fID': 202},{'fID': 203},{'fID': 204},{'fID': 205}];

// extract values from variable b
var bValues = b.map((element) => (element.fID));

// filter out elements in variable a based on values in bValues
var result = a.filter((element) => {
    return (bValues.indexOf(element.pID) === -1);
});
console.log(result);

Answer 8

To eliminate repeated values that are unwanted, you can utilize the Array Filter method in JavaScript. The approach involves fetching all values from variable b and then removing objects from variable a that share the same values.

// sample data
var a = [{'pID': 200},{'pID': 201},{'pID': 202},{'pID': 203},{'pID': 204},{'pID': 205}];
var b = [{'fID': 202},{'fID': 203},{'fID': 204},{'fID': 205}];

// extract values from variable b
var bValues = b.map((element) => (element.fID));

// filter out elements in variable a based on values in bValues
var result = a.filter((element) => {
    return (bValues.indexOf(element.pID) === -1);
});
console.log(result);

Answer 9

Answer №5

To only check the keys ending with ID, you can utilize a Set. Remove the values from the second array and create a new array with those remaining.

var firstArray = [{ pID: 200 }, { pID: 201 }, { pID: 202 }, { pID: 203 }, { pID: 204 }, { pID: 205 }],
    secondArray = [{ fID: 202 }, { fID: 203 }, { fID: 204 }, { fID: 205 }],
    getValue = obj => obj[Object.keys(obj).find(key => key.endsWith('ID'))],
    set = new Set(firstArray.map(obj => getValue(obj))),
    result;

secondArray.forEach(obj => set.delete(getValue(obj)));

result = Array.from(set, ID => ({ ID }));

console.log(result);

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Answer 10

To only check the keys ending with ID, you can utilize a Set. Remove the values from the second array and create a new array with those remaining.

var firstArray = [{ pID: 200 }, { pID: 201 }, { pID: 202 }, { pID: 203 }, { pID: 204 }, { pID: 205 }],
    secondArray = [{ fID: 202 }, { fID: 203 }, { fID: 204 }, { fID: 205 }],
    getValue = obj => obj[Object.keys(obj).find(key => key.endsWith('ID'))],
    set = new Set(firstArray.map(obj => getValue(obj))),
    result;

secondArray.forEach(obj => set.delete(getValue(obj)));

result = Array.from(set, ID => ({ ID }));

console.log(result);

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What is the best way to identify and delete duplicate rows in a JSON array?

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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