How to get the difference between two arrays of objects in Typescript/Javascript

Question

I have two arrays of objects like:

[{id: "1", name: "item1", checked: true}, {id: "2", name: "item2", checked: true}, {id: "3", name: "item3", checked: true}]

and

[{id: "1", name: "item1", checked: true}, {id: "2", name: "item2", checked: true}, {id: "3", name: "item3", checked: false}]

The first array holds the default states, the second one hold the current states. I'd like to filter out the objects that "checked" is changed. The expected result would be

[{id: "3", name: "item3", checked: false}]

I tried with loadash differentBy, but it returns empty array:

const diff=_.differenceBy(array1, array2, "checked")

Can anyone help out? Thanks.

tokland · Accepted Answer · 2022-06-15 23:10:01Z

1

A very compact solution -but O(n^2)- with _.differenceWith:

const xs = [
  {id: "1", name: "item1", checked: true}, 
  {id: "2", name: "item2", checked: true}, 
  {id: "3", name: "item3", checked: true}
]

const ys = [
  {id: "1", name: "item1", checked: true}, 
  {id: "2", name: "item2", checked: true}, 
  {id: "3", name: "item3", checked: false}
]

const zs = _.differenceWith(xs, ys, _.isEqual)
// [{ id: "3", name: "item3", checked: true }]

A bit more verbose, but this one is O(n):

const xsById = _.keyBy(xs, x => x.id)
const zs = ys.filter(y => xsById[y.id]?.checked !== y.checked)

edited Jun 15, 2022 at 23:10

answered Jun 15, 2022 at 18:00

tokland

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Comments

M.Fazil · Accepted Answer · 2022-06-15 17:48:54Z

You could use a method like this, if an item in the array compared by keyToCompare and identified by it's identifier (In your situation checked and id).

const differenceBy = (arr1, arr2, keyToCompare, identifier) => {
    const diff = []
    arr1.forEach(arr => {
        const arr2Item = arr2.find(i => i[identifier] === arr[identifier])
        if(arr[keyToCompare] !== arr2Item[keyToCompare]) {
            diff.push(arr2Item)
        }
    })

    return diff
}

Example:

const arr1 = [{id: "1", name: "item1", checked: true}, {id: "2", name: "item2", checked: true}, {id: "3", name: "item3", checked: true}]
const arr2 = [{id: "1", name: "item1", checked: true}, {id: "2", name: "item2", checked: true}, {id: "3", name: "item3", checked: false}]

const differenceBy = (arr1, arr2, keyToCompare, identifier) => {
    const diff = []
    arr1.forEach(arr => {
        const arr2Item = arr2.find(i => i[identifier] === arr[identifier])
        if(arr[keyToCompare] !== arr2Item[keyToCompare]) {
            diff.push(arr2Item)
        }
    })

    return diff
}

console.log(differenceBy(arr1, arr2, 'checked', 'id'))

José Lourenço · Accepted Answer · 2022-06-15 17:24:06Z

0

I guess you could use something like that

function compareArrays(arr1, arr2, keyCompared) {
    let d = []
    arr1.forEach((e,i)=>{
    if(e[keyCompared] !== arr2[i][keyCompared])
        d.push(e)
    })
    return d
}

Example:

let array1 = [{id: "1", name: "item1", checked: true}, {id: "2", name: "item2", checked: true}, {id: "3", name: "item3", checked: true}]
let array2 = [{id: "1", name: "item1", checked: true}, {id: "2", name: "item2", checked: true}, {id: "3", name: "item3", checked: false}] 
function compareArrays(arr1, arr2, keyCompared) {
    let d = []
    arr1.forEach((e,i)=>{
        if(e[keyCompared] !== arr2[i][keyCompared])
            d.push(e)
    })
    return d
} 
console.log(compareArrays(array1,array2,'checked'))

answered Jun 15, 2022 at 17:24

José Lourenço

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1 Comment

GLP Over a year ago

Thanks for your quick response. It does not seem working if the orders of the objects are not the same.

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