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Compare two arrays of objects, and remove if object value is equal

I've tried modifying some of the similar solutions on here but I keep getting stuck, I believe I have part of this figured out however, the main caveat is that:
Some of the objects have extra keys, which renders my object comparison logic useless.
I am trying to compare two arrays of objects. One array is the original array, and the other array contains the items I want deleted from the original array. However there's one extra issue in that the second array contains extra keys, so my comparison logic doesn't work.
An example would make this easier, let's say I have the following two arrays:
const originalArray = [{id: 1, name: "darnell"}, {id: 2, name: "funboi"},
{id: 3, name: "jackson5"}, {id: 4, name: "zelensky"}];
const itemsToBeRemoved = [{id: 2, name: "funboi", extraProperty: "something"},
{id: 4, name: "zelensky", extraProperty: "somethingelse"}];
after running the logic, my final output should be this array:
[{id: 1, name: "darnell"}, {id: 3, name: "jackson5"}]
And here's the current code / logic that I have, which compares but doesn't handle the extra keys. How should I handle this? Thank you in advance.
const prepareArray = (arr) => {
return arr.map((el) => {
if (typeof el === "object" && el !== null) {
return JSON.stringify(el);
} else {
return el;
}
});
};
const convertJSON = (arr) => {
return arr.map((el) => {
return JSON.parse(el);
});
};
const compareArrays = (arr1, arr2) => {
const currentArray = [...prepareArray(arr1)];
const deletedItems = [...prepareArray(arr2)];
const compared = currentArray.filter((el) => deletedItems.indexOf(el) === -1);
return convertJSON(compared);
};

How about using filter and some? You can extend the filter condition on select properties using &&.
const originalArray = [
{ id: 1, name: 'darnell' },
{ id: 2, name: 'funboi' },
{ id: 3, name: 'jackson5' },
{ id: 4, name: 'zelensky' },
];
const itemsToBeRemoved = [
{ id: 2, name: 'funboi', extraProperty: 'something' },
{ id: 4, name: 'zelensky', extraProperty: 'somethingelse' },
];
console.log(
originalArray.filter(item => !itemsToBeRemoved.some(itemToBeRemoved => itemToBeRemoved.id === item.id))
)
Or you can generalise it as well.
const originalArray = [
{ id: 1, name: 'darnell' },
{ id: 2, name: 'funboi' },
{ id: 3, name: 'jackson5' },
{ id: 4, name: 'zelensky' },
];
const itemsToBeRemoved = [
{ id: 2, name: 'funboi', extraProperty: 'something' },
{ id: 4, name: 'zelensky', extraProperty: 'somethingelse' },
];
function filterIfSubset(originalArray, itemsToBeRemoved) {
const filteredArray = [];
for (let i = 0; i < originalArray.length; i++) {
let isSubset = false;
for (let j = 0; j < itemsToBeRemoved.length; j++) {
// check if whole object is a subset of the object in itemsToBeRemoved
if (Object.keys(originalArray[i]).every(key => originalArray[i][key] === itemsToBeRemoved[j][key])) {
isSubset = true;
}
}
if (!isSubset) {
filteredArray.push(originalArray[i]);
}
}
return filteredArray;
}
console.log(filterIfSubset(originalArray, itemsToBeRemoved));
Another simpler variation of the second approach:
const originalArray = [
{ id: 1, name: 'darnell' },
{ id: 2, name: 'funboi' },
{ id: 3, name: 'jackson5' },
{ id: 4, name: 'zelensky' },
];
const itemsToBeRemoved = [
{ id: 2, name: 'funboi', extraProperty: 'something' },
{ id: 4, name: 'zelensky', extraProperty: 'somethingelse' },
];
const removeSubsetObjectsIfExists = (originalArray, itemsToBeRemoved) => {
return originalArray.filter(item => {
const isSubset = itemsToBeRemoved.some(itemToBeRemoved => {
return Object.keys(item).every(key => {
return item[key] === itemToBeRemoved[key];
});
});
return !isSubset;
});
}
console.log(removeSubsetObjectsIfExists(originalArray, itemsToBeRemoved));

The example below is a reusable function, the third parameter is the key to which you compare values from both arrays.
Details are commented in example
const arr=[{id:1,name:"darnell"},{id:2,name:"funboi"},{id:3,name:"jackson5"},{id:4,name:"zelensky"}],del=[{id:2,name:"funboi",extraProperty:"something"},{id:4,name:"zelensky",extraProperty:"somethingelse"}];
/** Compare arrayA vs. delArray by a given key's value.
--- ex. key = 'id'
**/
function deleteByKey(arrayA, delArray, key) {
/* Get an array of only the values of the given key from delArray
--- ex. delList = [1, 2, 3, 4]
*/
const delList = delArray.map(obj => obj[key]);
/* On every object of arrayA compare delList values vs
current object's key's value
--- ex. current obj[id] = 2
--- [1, 2, 3, 4].includes(obj[id])
Any match returns an empty array and non-matches are returned
in it's own array.
--- ex. ? [] : [obj]
The final return is a flattened array of the non-matching objects
*/
return arrayA.flatMap(obj => delList.includes(obj[key]) ? [] : [obj]);
};
console.log(deleteByKey(arr, del, 'id'));

let ff = [{ id: 1, name: 'darnell' }, { id: 2, name: 'funboi' },
{ id: 3, name: 'jackson5' },
{ id: 4, name: 'zelensky' }]
let cc = [{ id: 2, name: 'funboi', extraProperty: 'something' },
{ id: 4, name: 'zelensky', extraProperty: 'somethingelse' }]
let ar = []
let out = []
const result = ff.filter(function(i){
ar.push(i.id)
cc.forEach(function(k){
out.push(k.id)
})
if(!out.includes(i.id)){
// console.log(i.id, i)
return i
}
})
console.log(result)

I need to delete the object list by using matched key. (example the key have 1 and 2 , the result will only show 3)

let selectedRow = ["1","2","3"];
let arr = [
{ id:1, name:"eddie" },
{ id:2, name:"jake" },
{ id:3, name:"susan" },
];

Updation on the answer provided by Andy, If you don't want to update the exiting array and want to result in a new array
let selectedRow = ["1", "2"];
let arr = [
{ id: 1, name: "eddie" },
{ id: 2, name: "jake" },
{ id: 3, name: "susan" },
];
const result = arr.filter(item => !selectedRow.includes(item.id.toString()))
console.log(result)
If you want changes in a current array and don't want to store results in a new array (Not the most efficient solution though)
let selectedRow = ["1", "2"];
let arr = [
{ id: 1, name: "eddie" },
{ id: 2, name: "jake" },
{ id: 3, name: "susan" },
];
for (const row of selectedRow) {
const index = arr.findIndex(item => item.id.toString() === row)
if (index !== -1)
arr.splice(index, 1)
}
console.log(arr)

Make sure your selectedRow array is an array of numbers (because your object ids are numbers).
filter over the array of objects and only keep the ones that selectedRow doesn't include.
const arr = [{ id: 1, name: 'eddie' }, { id: 2, name: 'jake' }, { id: 3, name: 'susan' }];
const selectedRow = ['1', '2'].map(Number);
const result = arr.filter(obj => {
return !selectedRow.includes(obj.id);
});
console.log(result);

How to filter an array of object with an array of numbers

Given an array of objects arr1 how can I filter out to a new array the objects that do not have a property equal to any value in the array of numbers arr2
const arr1 = [
{
key: 1,
name: 'Al'
},
{
key: 2,
name: 'Lo'
},
{
key: 3,
name: 'Ye'
}
];
const arr2 = [2, 3]
// Failed attempt
const newArr = arr1.filter(obj1 => arr2.some(num1 => num1 !== obj1.key))
console.log(newArr)
// Expected: [{ key: 1, name: 'Al' }]
// Received: [
// { key: 1, name: 'Al' },
// { key: 2, name: 'Lo' },
// { key: 3, name: 'Ye' }
// ]

Using your syntax:
You have to match on the somein case it's the same and not different. Then if it matches, do not keep the value.
const arr1 = [
{
key: 1,
name: 'Al',
},
{
key: 2,
name: 'Lo',
},
{
key: 3,
name: 'Ye',
},
];
const arr2 = [2, 3];
const newArr= arr1.filter(x => !arr2.some(y => y === x.key));
console.log(newArr);
Alternative syntax below :
const arr1 = [{
key: 1,
name: 'Al',
},
{
key: 2,
name: 'Lo',
},
{
key: 3,
name: 'Ye',
},
];
const arr2 = [2, 3];
const newArr = arr1.filter(({
key,
}) => !arr2.some(y => y === key));
console.log(newArr);
That said, you should be using Array.includes() like some ppl answered. It's simplier for the same outcome
const arr1 = [{
key: 1,
name: 'Al',
},
{
key: 2,
name: 'Lo',
},
{
key: 3,
name: 'Ye',
},
];
const arr2 = [2, 3];
const newArr = arr1.filter(({
key,
}) => !arr2.includes(key));
console.log(newArr);

You can do this
const newArr = arr1.filter(obj => !arr2.includes(obj.key));

This will work for you:
const arr1 = [
{
key: 1,
name: 'Al'
},
{
key: 2,
name: 'Lo'
},
{
key: 3,
name: 'Ye'
}
];
const arr2 = [2, 3]
const filtered = arr1.filter(val => !arr2.includes(val.key))
console.log(filtered)
:)

For situations like this Set is also very cool (and for big arrays more performant):
const arr1 = [
{
key: 1,
name: 'Al'
},
{
key: 2,
name: 'Lo'
},
{
key: 3,
name: 'Ye'
}
];
const arr2 = [2, 3]
const arr2Set = new Set(arr2);
const newArr = arr1.filter(obj1 => !arr2Set.has(obj1.key))
console.log(newArr)

You can use indexOf like this:
const newArr = arr1.filter(obj => arr2.indexOf(obj.key) > -1);

You need to filter the arr1 when arr1 element does not exist in arr2, so I think it could be better to use indexOf() like this
const newArr = arr1.filter(obj1 => arr2.indexOf(obj1.key) === -1)
if the element does not exist in arr2 it will return -1 which what you need.

group array by groups and sort by position

I have an array. I need to group this array by groups and sort by position. I tied to create a new array with group names as keys and values as sorted array grouped by group, but didn't work well. How can I do this?
a = [
{id:1,name:'qw'group:'C',name:'hite',position:'1'},
{id:2,name:'qwe'group:'B',name:'ite',position:'2'},
{id:3,name:'qwer'group:'A',name:'ite',position:'3'},
{id:4,name:'qer'group:'D',name:'te',position:'4'},
{id:5,name:'wer'group:'C',name:'whit',position:'5'},
{id:6,name:'er'group:'B',name:'whi',position:'6'},
]
function groupDo(array){
var groups = [];
for (var i in array){
groups[array[i].group] = array[i].group;
}
for (var i in array){
if (groups[array[i].group] == array[i].group){
groups[array[i].group] = array[i];
}
}
}

Here's a simple straight forward answer:
var sortByPosition = function(obj1, obj2) {
return obj1.position - obj2.position;
};
var arr = [
{ id: 1, name: 'qw', group: 'C', name: 'hite', position: '1' },
{ id: 2, name: 'qwe', group: 'B', name: 'ite', position: '2' },
{ id: 3, name: 'qwer', group: 'A', name: 'ite', position: '3' },
{ id: 4, name: 'qer', group: 'D', name: 'te', position: '4' },
{ id: 5, name: 'wer', group: 'C', name: 'whit', position: '5' },
{ id: 6, name: 'er', group: 'B', name: 'whi', position: '6' },
];
var grouped = {};
for (var i = 0; i < arr.length; i += 1) {
if(!grouped[arr[i].group]) {
grouped[arr[i].group] = [];
}
grouped[arr[i].group].push(arr[i]);
}
for (var group in grouped) {
grouped[group] = grouped[group].sort(sortByPosition);
}
console.log(grouped);
When you want to do stuff like this though, it's usually recommended to use a utility library like lodash or underscore.js, so that you don't have to "reinvent the wheel". Here's how it would look like using one of these libraries:
var arr = [
{ id: 1, name: 'qw', group: 'C', name: 'hite', position: '1' },
{ id: 2, name: 'qwe', group: 'B', name: 'ite', position: '2' },
{ id: 3, name: 'qwer', group: 'A', name: 'ite', position: '3' },
{ id: 4, name: 'qer', group: 'D', name: 'te', position: '4' },
{ id: 5, name: 'wer', group: 'C', name: 'whit', position: '5' },
{ id: 6, name: 'er', group: 'B', name: 'whi', position: '6' },
];
var grouped = _.groupBy(arr, 'group');
for (var group in grouped) {
_.sortBy(grouped[group], 'position');
}
console.log(grouped);

Here ya go!
a = [
{id:1,name:'qw',group:'C',name:'hite',position:'1'},
{id:2,name:'qwe',group:'B',name:'ite',position:'2'},
{id:3,name:'qwer',group:'A',name:'ite',position:'3'},
{id:4,name:'qer',group:'D',name:'te',position:'4'},
{id:5,name:'wer',group:'C',name:'whit',position:'5'},
{id:6,name:'er',group:'B',name:'whi',position:'6'},
]
function groupAndSort(array, groupField, sortField) {
var groups = {}; // This object will end being keyed by groups, and elements will be arrays of the rows within the given array, which have been sorted by the sortField
// Put all the rows into groups
for (var i = 0; i < array.length; i++) {
var row = array[i];
var groupValue = row[groupField];
groups[groupValue] = groups[groupValue] || [];
groups[groupValue].push(row);
}
// Sort each group
for (var groupValue in groups) {
groups[groupValue] = groups[groupValue].sort(function(a, b) {
return a[sortField] - b[sortField];
});
}
// Return the results
return groups;
}
var groupedAndSorted = groupAndSort(a, "group", "position");

If you want to group objects, first think about what the resulting data would look like. Maybe something like this?
var grouped = {
A : [
{id:3,name:'qwer', group:'A',name:'ite',position:'3'}
],
B : [],
C : [],
D : []
};
And so on. To transform a list into an object, consider using .reduce().
.reduce() takes a function as its first argument, and a resulting object as the second. The function iterates through each element of the array and reduces it into the given object.
var data = [
{id:1,name:'qw', group:'C',name:'hite',position:'1'},
{id:2,name:'qwe', group:'B',name:'ite',position:'2'},
{id:3,name:'qwer', group:'A',name:'ite',position:'3'},
{id:4,name:'qer', group:'D',name:'te',position:'4'},
{id:5,name:'wer', group:'C',name:'whit',position:'5'},
{id:6,name:'er', group:'B',name:'whi',position:'6'},
]
// acc is the accumulated object, x is each element of the array
data.reduce(function(acc, x) {
// first check if the given group is in the object
acc[x.group] = acc[x.group] ? acc[x.group].concat(x) : [x];
return acc;
}, {}); // this is the resulting object
Now all you need to do is use the built in sort to order the resulting arrays. You could do this by iterating through the keys of the resulting object and applying .sort() to each array. .sort() takes a function as an argument which accesses the data and provides a comparison function.
// a and b are elements of the array
array.sort(function(a, b) {
if (a.position > b.position) {
return -1;
} else if (b.position > a.position) {
return 1;
} else {
return 0;
}
});
And you would implement it like so
var result = Object.keys(data).map(function(d){
return d.sort(f); // f is the function above
});

JavaScript filter array by data from another

I have an array object:
[
{ id:1, name: 'Pedro'},
{ id:2, name: 'Miko'},
{ id:3, name: 'Bear'},
{ id:4, name: 'Teddy'},
{ id:5, name: 'Mouse'}
]
And I have an array with ids [1, 3, 5],
How can I filter the array object to leave records only with id's from the second one?

If Array.includes() is supported, you can use it with Array.filter() to get the items:
const array = [
{ id: 1, name: 'Pedro'},
{ id: 2, name: 'Miko'},
{ id: 3, name: 'Bear'},
{ id: 4, name: 'Teddy'},
{ id: 5, name: 'Mouse'}
];
const filterArray = [1,3,5];
const result = array.filter(({ id }) => filterArray.includes(id));
console.log(result);
If includes is not supported, you can use Array.indexOf() instead:
var array = [
{ id: 1, name: 'Pedro'},
{ id: 2, name: 'Miko'},
{ id: 3, name: 'Bear'},
{ id: 4, name: 'Teddy'},
{ id: 5, name: 'Mouse'}
];
var filterArray = [1,3,5];
var result = array.filter(function(item) {
return filterArray.indexOf(item.id) !== -1;
});
console.log(result);

Maybe take a Array.prototype.reduce in combination with an Array.prototype.some. This keeps the order of the given array need.
var data = [
{ id: 3, name: 'Bear' },
{ id: 4, name: 'Teddy' },
{ id: 5, name: 'Mouse' },
{ id: 1, name: 'Pedro' },
{ id: 2, name: 'Miko' },
],
need = [1, 3, 5],
filtered = need.reduce(function (r, a) {
data.some(function (el) {
return a === el.id && r.push(el);
});
return r;
}, []);
document.write('<pre>' + JSON.stringify(filtered, 0, 4) + '</pre>');
To keep the order of data you can use Array.prototype.filter:
var data = [
{ id: 3, name: 'Bear' },
{ id: 4, name: 'Teddy' },
{ id: 5, name: 'Mouse' },
{ id: 1, name: 'Pedro' },
{ id: 2, name: 'Miko' },
],
need = [1, 3, 5],
filtered = data.filter(function (a) {
return ~need.indexOf(a.id);
});
document.write('<pre>' + JSON.stringify(filtered, 0, 4) + '</pre>');

In case the data set is small, you are ok with any of the offered solution (ones that use indexOf).
However, these solutions are O(n^2) ones, therefore, given the data set big enough, the lag can become noticeable. In this case, you should build an index prior to selecting elements.
Example:
function filterFast(data, ids) {
var index = ids.reduce(function(a,b) {a[b] = 1; return a;}, {});
return data.filter(function(item) {
return index[item.id] === 1;
});
}
And some benchmarking can be tested here.

You can use the filter method on your Array:
var data = [
{ id:1, name: 'Pedro'},
{ id:2, name: 'Miko'},
{ id:3, name: 'Bear'},
{ id:4, name: 'Teddy'},
{ id:5, name: 'Mouse'}
];
var ids = [1, 3, 5];
var filteredData = filterData(data, 'id', ids[1]);
function filterData(data, prop, values) {
return data.filter(function(item) {
return ~values.indexOf(item[prop]); // ~ returns 0 if indexOf returns -1
});
}
See it in action in this JSFiddle.

Or if you are using jQuery, another option may be:
var arr1 = [1, 3, 5],
arr2 = [{ id: 1, name: 'Pedro' },
{ id: 2, name: 'Miko' },
{ id: 3, name: 'Bear' },
{ id: 4, name: 'Teddy' },
{ id: 5, name: 'Mouse' }],
filtered = $.grep(arr2, function (item) {
if (arr1.indexOf(item.id) > -1) {
return true;
}
});

You can use a for loop on the object array and check hasOwnProperty in another for loop for each ids in [1,3,5] (break out of the loop once an id found). (And break out of the bigger for-loop once all ids are found) If your array object is ordered (e.g. elements sorted from smallest id to biggest id) and so are your list, this solution should be quite efficient.
var c = 0;
for(var i =0; i< objects.length; i++){
for(var v =0; v< list.length; v++)
if(objects[i].hasOwnProperty(list[v])){
delete objects[i]; c++; break;
}
if(c===list.length) break;
}
or use array.splice( i, 1 ); if you don't want an empty slot.

Using filter and indexOf will do the trick:
var filteredArray = dataArray.filter(function(obj) {
return idsArray.indexOf(obj.id) > -1;
});
However, indexOf has linear performance, and it will be called lots of times.
In ES6 you can use a set instead, whose has call has sublinear performance (on average):
var idsSet = new Set(idsArray),
filteredArray = dataArray.filter(obj => idsSet.has(obj.id));
Assuming the toString method of your ids is injective, you can achieve something similar in ES5:
var idsHash = Object.create(null);
idsArray.forEach(function(id) {
idsHash[id] = true;
});
var filteredArray = dataArray.filter(function(obj) {
return idsHash[obj.id];
});