How to join objects - javascript

Given the following two arrays of objects :
var items = [
{colorId:'2',name:'qqq'},
{colorId:'5',name:'www'},
{colorId:'2',name:'eee'},
{colorId:'4',name:'rrr'}
];
var colors = [
{id:'5',name:'blue'},
{id:'2',name:'red'}
];
I need to make a join between the items and colors based on the colorId.
The desired result :
var arr3 = [
{id:'2', name:'qqq', name:'red'},
{id:'5', name:'www', name:'blue'},
{id:'2', name:'eee', name:'red'}
];
What is a elegant way to do this ?

this a pure JavaScript solution, which will change the items array itself.
if you don't want to modify the same array, you can create new array and push cloned items into it.
var items= [{colorId:'2',name:'qqq'},
{colorId:'5',name:'www'},
{colorId:'2',name:'eee'},
{colorId:'4',name:'rrr'}]
var colors= [{id:'5',name:'blue'},
{id:'2',name:'red'}, ]
items.forEach(function(d) {
var matchColor = colors.filter(function(item){ return item.id === d.colorId});
if(matchColor.length){
d.color = matchColor[0].name;
}
});
console.log(items);

Using _.map you can add the color value to each element and return a new array.
It looks like this:
var newItems = _.map(items, function(item) {
// `black` will be the default fallback color
// We use _.result in case we can't find the colorId in the colors array
var color = _.result(_.find(colors, { id: item.colorId }), 'name', 'black');
// Append the color to the original item in the array
item.color = color
// Return the modified item
return item;
});
// newItems:
// [
// { "colorId": "2", "name": "qqq", "color": "red" },
// { "colorId": "5", "name": "www", "color": "blue" },
// { "colorId": "2", "name": "eee", "color": "red" },
// { "colorId": "4", "name": "rrr", "color": "black" }
// ]

A nice way is to put the names together in an array where the id is the same. The id is the key and is a number.
Structure :
var data = {
id: { // id is a number
names: [],
color: ""
}
};
Example :
var items= [{colorId:'2',name:'qqq'},
{colorId:'5',name:'www'},
{colorId:'2',name:'eee'},
{colorId:'4',name:'rrr'}];
var colors= [{id:'5',name:'blue'},
{id:'2',name:'red'},
{id: '1', name: 'violett'}];
// Result
var data = {};
items.map(function(obj) {
if(!(obj.colorId in data)) {
data[obj.colorId] = {};
data[obj.colorId].names = [];
}
data[obj.colorId].names.push(obj.name);
});
colors.map(function(obj) {
if(!(obj.id in data)) {
data[obj.id] = {};
data[obj.id].names = [];
}
data[obj.id].color = obj.name;
});
console.log(data);

You could use a hash table and loop every array only once.
var items = [{ colorId: '2', name: 'qqq' }, { colorId: '5', name: 'www' }, { colorId: '2', name: 'eee' }, { colorId: '4', name: 'rrr' }],
colors = [{ id: '5', name: 'blue' }, { id: '2', name: 'red' }],
arr3 = [];
items.forEach(function (hash) {
colors.forEach(function (a) {
hash[a.id] = a.name;
});
return function (a) {
hash[a.colorId] && arr3.push({ id: a.colorId, name: a.name, color: hash[a.colorId] });
};
}(Object.create(null)), []);
console.log(arr3);
ES6
var items = [{ colorId: '2', name: 'qqq' }, { colorId: '5', name: 'www' }, { colorId: '2', name: 'eee' }, { colorId: '4', name: 'rrr' }],
colors = [{ id: '5', name: 'blue' }, { id: '2', name: 'red' }],
arr3 = [];
items.forEach((hash => {
colors.forEach(a => hash[a.id] = a.name);
return a => hash[a.colorId] && arr3.push({ id: a.colorId, name: a.name, color: hash[a.colorId] });
})(Object.create(null)), []);
console.log(arr3);

Related

Filter out array using another array of objects

I am having two arrays
const selected = [];
const current = [
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
const result = []
I need to compare these two arrays and the result should only have the single entry instead of duplicates. In the above example result should have the following output.
Also items in the selected should be taken into consideration and should be in the beginning of the result
result = [
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
Also when the input is following
const selected = [ {id:5, name: "xyz" }];
const current = [
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
result = [[
{ id: 5, name: "xyz" },
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
Also when the input is following
const selected = [ {id:1, name: "abc" }, {id:4, name: "lmn" }];
const current = [
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
result = [[
{ id: 1, name: "abc" },
{ id: 4, name: "lmn" }
{ id: 2, name: "def" }
];
Note the comparison should be made using name field
Code that I tried
const res = [...(selected || [])].filter((s) =>
current.find((c) => s.name === c.name)
);
Sandbox: https://codesandbox.io/s/nervous-shannon-j1vn5k?file=/src/index.js:115-206
You could get all items and filter the array by checking the name with a Set.
const
filterBy = (key, s = new Set) => o => !s.has(o[key]) && s.add(o[key]),
selected = [{ id: 1, name: "abc" }, { id: 1, name: "lmn" }],
current = [{ id: 1, name: "abc" }, { id: 2, name: "def" }],
result = [...selected, ...current].filter(filterBy('name'));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Loop through selected, and if there is no object in current with a name that matches the name of the object in the current iteration push it into current.
const selected=[{id:1,name:"abc"},{id:6,name:"def"},{id:4,name:"lmn"}];
const current=[{id:1,name:"abc"},{id:2,name:"def"}];
for (const sel of selected) {
const found = current.find(cur => cur.name === sel.name);
if (!found) current.push(sel);
}
console.log(current);
This is a good use for .reduce, avoids multiple loops/finds and doesn't need filtering with side-effects.
const selected = [ {id:1, name: "abc" }, {id:4, name: "lmn" }];
const current = [
{ id: 1, name: "abc" },
{ id: 2, name: "def" }
];
const result = Object.values(
[...selected, ...current].reduce((obj, item) => {
obj[item.name] = obj[item.name] || item;
return obj;
}, {})
)
console.log(result);

Get the branch of collection without the sibling elements, searching by property

I have the object with the next structure:
let array = [
{
name: 'Name1',
items: [
{
name: 'Name1.1',
items: [
{ id: '1', name: 'Name1.1.1' },
{ id: '2', name: 'Name1.1.2' },
{ id: '3', name: 'Name1.1.3' },
...
],
},
{
name: 'Name1.2',
items: [
{ id: '4', name: 'Name1.2.1' },
{ id: '5', name: 'Name1.2.2' },
],
},
],
},
{
name: 'Name2',
items: [
{
name: 'Name2.1',
items: [
{ id: '6', name: 'Name2.1.1' },
{ id: '7', name: 'Name2.1.2' },
],
},
],
},
];
I want to get the branch without the sibling elements, searching by id. The desired result is the next structure by id = '4':
let array = [
{
name: 'Name1',
items: [
{
name: 'Name1.2',
items: [
{ id: '4', name: 'Name1.2.1' },
],
},
],
}
];
I could find only the end element of the tree ({ id: '4', name: 'Name1.2.1' }). But I don't understand how to get intermediate structures of the tree.
const test = (data, id) => {
if (!data || !data.length) return null;
for (var j = 0; j < data.length; j++) {
var result = data[j].items
? test(data[j].items, id)
: data[j].id
? data[j].id === id
? data[j]
: undefined
: undefined;
if (result !== undefined) {
return result;
}
}
return undefined;
};
test(array, '4');
You should indeed take a recursive approach, but your function currently can only return an id value (a string) or null or undefined. It never returns an array, yet that is what you expect to get.
When a solution is found as a base case, you need to wrap that solution in an array and plain object, each time you get out of the recursion tree.
Here is a working solution:
function getPath(forest, targetid) {
for (let root of forest) {
if (root.id === targetid) return [root]; // base case
let items = root.items && getPath(root.items, targetid);
if (items) return [{ ...root, items }]; // wrap!
}
}
// Example run:
let array = [{name: 'Name1',items: [{name: 'Name1.1',items: [{ id: '1', name: 'Name1.1.1' },{ id: '2', name: 'Name1.1.2' },{ id: '3', name: 'Name1.1.3' },],},{name: 'Name1.2',items: [{ id: '4', name: 'Name1.2.1' },{ id: '5', name: 'Name1.2.2' },],},],},{name: 'Name2',items: [{name: 'Name2.1',items: [{ id: '6', name: 'Name2.1.1' },{ id: '7', name: 'Name2.1.2' },],},],},];
console.log(getPath(array, '4'));

How to merge two array of objects by key AND keep unique keys in a single array of objects?

I want to merge two array of objects where objects with the same ID will merge properties and objects with unique IDs will be its own object in the merged array. The following code does the first part where similar IDs will merge but how do I keep objects with unique ids from arr2 in the merged array and have it work with arrays of varying lengths?
Expected output:
[
{
"id": "1",
"date": "2017-01-24",
"name": "test"
},
{
"id": "2",
"date": "2017-01-22",
"bar": "foo"
}
{ "id": "3",
"foo": "bar",
}
]
The code:
let arr1 = [{
id: '1',
createdDate: '2017-01-24'
},
{
id: '2',
createdDate: '2017-01-22'
},
];
let arr2 = [{
id: '1',
name: 'test'
},
{
id: '3',
foo: 'bar'
},
{
id: '2',
bar: 'foo'
},
];
let merged = [];
for (let i = 0; i < arr1.length; i++) {
merged.push({
...arr1[i],
...arr2.find((itmInner) => itmInner.id === arr1[i].id),
},
);
}
console.log(merged);
Iterate over the larger array, the one that contains the smaller array, instead:
let arr1=[{id:"1",createdDate:"2017-01-24"},{id:"2",createdDate:"2017-01-22"}],arr2=[{id:"1",name:"test"},{id:"3",foo:"bar"},{id:"2",bar:"foo"}];
const merged = arr2.map(item => ({
...arr1.find(({ id }) => id === item.id),
...item
}));
console.log(merged);
(if order matters, you can sort if afterwards too)
If you don't know in advance which one / if one will contain the other, then use an object to index the merged objects by IDs first:
let arr1=[{id:"1",createdDate:"2017-01-24"},{id:"2",createdDate:"2017-01-22"}],arr2=[{id:"1",name:"test"},{id:"3",foo:"bar"},{id:"2",bar:"foo"}];
const resultObj = Object.fromEntries(
arr1.map(
item => [item.id, { ...item }]
)
);
for (const item of arr2) {
if (!resultObj[item.id]) {
resultObj[item.id] = item;
} else {
Object.assign(resultObj[item.id], item);
}
}
const merged = Object.values(resultObj);
console.log(merged);
You can create new object that contains the elems of arr1 and arr2 group by id key as follows and the merged array will be stored on object values.
You can get object values using Object.values func.
let arr1 = [{
id: '1',
createdDate: '2017-01-24'
},
{
id: '2',
createdDate: '2017-01-22'
},
];
let arr2 = [{
id: '1',
name: 'test'
},
{
id: '3',
foo: 'bar'
},
{
id: '2',
bar: 'foo'
},
];
const groupById = {};
for (let i = 0; i < Math.min(arr1.length, arr2.length); i ++) {
if (arr1[i]) {
groupById[arr1[i].id] = { ...groupById[arr1[i].id], ...arr1[i] };
}
if (arr2[i]) {
groupById[arr2[i].id] = { ...groupById[arr2[i].id], ...arr2[i] };
}
}
const merged = Object.values(groupById);
console.log(merged);
You could take a single loop approach by storing the objects in a hash table, sorted by id.
const
mergeTo = (target, objects = {}) => o => {
if (!objects[o.id]) target.push(objects[o.id] = {});
Object.assign(objects[o.id], o);
},
array1 = [{ id: '1', createdDate: '2017-01-24' }, { id: '2', createdDate: '2017-01-22' }],
array2 = [{ id: '1', name: 'test' }, { id: '3', foo: 'bar' }, { id: '2', bar: 'foo' }],
merged = [],
merge = mergeTo(merged);
array1.forEach(merge);
array2.forEach(merge);
console.log(merged);
.as-console-wrapper { max-height: 100% !important; top: 0; }
A different approach could be merged the two array as is, and then "squash" it:
let arr1 = [{
id: '1',
createdDate: '2017-01-24'
},
{
id: '2',
createdDate: '2017-01-22'
},
];
let arr2 = [{
id: '1',
name: 'test'
},
{
id: '3',
foo: 'bar'
},
{
id: '2',
bar: 'foo'
},
];
let merged = [...arr1, ...arr2].reduce(
(acc, {id, ...props}) =>
(acc.set(id, {...(acc.get(id) || {}), ...props}), acc), new Map());
console.log([...merged].map( ([id, props]) => ({id, ...props}) ))
Notice that you might not need the last line, it used just to obtain the format you want to, since the above reduce is using a Map as accumulator, you can already access to everything with just merged.get("1").createdDate for example (where "1" is the id).
Since you're operating on one array by merging them at the beginning, you don't care about the length of them or even which one contains more elements. You can also have several arrays instead of just two, it doesn't matter.
What it matters is the order: if more than one array contains the same property for the same "id", the value you'll get is the value from the most recent array added (in the example above, would be arr2).
You can write a function to reduce the arrays to an object and then extract the value from that object which will return the values that you want. You can see the code below:
let arr1 = [
{
id: '1',
createdDate: '2017-01-24',
},
{
id: '2',
createdDate: '2017-01-22',
},
];
let arr2 = [
{
id: '1',
name: 'test',
},
{
id: '3',
foo: 'bar',
},
{
id: '2',
bar: 'foo',
},
];
function merge(arr1 = [], arr2 = []) {
return Object.values(
arr1.concat(arr2).reduce(
(acc, curr) => ({
...acc,
[curr.id]: { ...(acc[curr.id] ?? {}), ...curr },
}),
{}
)
);
}
const merged = merge(arr1, arr2);
Output:
[
{
"id": "1",
"createdDate": "2017-01-24",
"name": "test"
},
{
"id": "2",
"createdDate": "2017-01-22",
"bar": "foo"
},
{
"id": "3",
"foo": "bar"
}
]

What is an alternative way to update an array of objects?

I have a array of objects. I want to update an object using id.
I am able to do using the map function. Is there an alternative way or more efficient way to update the array?
Here is my code:
https://stackblitz.com/edit/js-xgfwdw?file=index.js
var id = 3
var obj = {
name: "test"
}
let arr = [{
name: "dd",
id: 1
}, {
name: "dzxcd",
id: 3
}, {
name: "nav",
id: 5
}, {
name: "hhh",
id: 4
}]
function getUpdated(obj, id) {
var item = [...arr];
const t = item.map((i) => {
if(i.id==id){
return {
...obj,
id
}
}else {
return i;
}
})
return t
}
console.log(getUpdated(obj,id))
The expected output is correct but I want to achieve the same functionality using an alternative way.
[{
name: "dd",
id: 1
}, {
name: "test",
id: 3
}, {
name: "nav",
id: 5
}, {
name: "hhh",
id: 4
}]
you are in the correct way, basically the bad thing that you are doing is creating new arrays [...arr], when map already gives you a new array.
other things to use, may be the ternary operator and return directly the result of the map function
check here the improvedGetUpdate:
var id = 3;
var obj = {
name: "test"
};
let arr = [{
name: "dd",
id: 1
}, {
name: "dzxcd",
id: 3
}, {
name: "nav",
id: 5
}, {
name: "hhh",
id: 4
}]
function getUpdated(obj, id) {
var item = [...arr];
const t = item.map((i) => {
if (i.id == id) {
return {
...obj,
id
}
} else {
return i;
}
})
return t
}
improvedGetUpdate = (obj, id) => arr.map(i => {
return i.id !== id ? i : {
...obj,
id
}
})
console.log(getUpdated(obj, id))
console.log(improvedGetUpdate(obj, id))
var id = 3
var obj = {
name: "test"
}
let arr = [{
name: "dd",
id: 1
}, {
name: "dzxcd",
id: 3
}, {
name: "nav",
id: 5
}, {
name: "hhh",
id: 4
}]
const result = arr.map((el) => el.id === id ? {...obj, id} : el)
console.log(result);
Use splice method which can be used to update the array too:
var obj = {
id: 3,
name: "test"
}
let arr = [{
name: "dd",
id: 1
}, {
name: "dzxcd",
id: 3
}, {
name: "nav",
id: 5
}, {
name: "hhh",
id: 4
}]
arr.splice(arr.findIndex(({id}) => id === obj.id), 0, obj);
console.log(arr);
#quirimmo suggested short code.
I suggest fast code.
var id = 3;
var obj = {
id: 3,
name: "test"
}
let arr = [{
name: "dd",
id: 1
}, {
name: "dzxcd",
id: 3
}, {
name: "nav",
id: 5
}, {
name: "hhh",
id: 4
}]
var arr2 = [...arr];
console.time('⏱');
arr.splice(arr.findIndex(({id}) => id === obj.id), 0, obj);
console.timeEnd('⏱');
console.time('⏱');
for (let item of arr2) {
if (item.id === id) {
item.name = obj.name;
break;
}
}
console.timeEnd('⏱');
console.log(arr2);

deleting an element in nested array using filter() function

I have been trying to delete an element with an ID in nested array.
I am not sure how to use filter() with nested arrays.
I want to delete the {id: 111,name: "A"} object only.
Here is my code:
var array = [{
id: 1,
list: [{
id: 123,
name: "Dartanan"
}, {
id: 456,
name: "Athos"
}, {
id: 789,
name: "Porthos"
}]
}, {
id: 2,
list: [{
id: 111,
name: "A"
}, {
id: 222,
name: "B"
}]
}]
var temp = array
for (let i = 0; i < array.length; i++) {
for (let j = 0; j < array[i].list.length; j++) {
temp = temp.filter(function(item) {
return item.list[j].id !== 123
})
}
}
array = temp
You can use the function forEach and execute the function filter for every array list.
var array = [{ id: 1, list: [{ id: 123, name: "Dartanan" }, { id: 456, name: "Athos" }, { id: 789, name: "Porthos" }] }, { id: 2, list: [{ id: 111, name: "A" }, { id: 222, name: "B" }] }];
array.forEach(o => (o.list = o.list.filter(l => l.id != 111)));
console.log(array);
.as-console-wrapper { max-height: 100% !important; top: 0; }
To remain the data immutable, use the function map:
var array = [{ id: 1, list: [{ id: 123, name: "Dartanan" }, { id: 456, name: "Athos" }, { id: 789, name: "Porthos" }] }, { id: 2, list: [{ id: 111, name: "A" }, { id: 222, name: "B" }] }],
result = array.map(o => ({...o, list: o.list.filter(l => l.id != 111)}));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
You could create a new array which contains elements with filtered list property.
const result = array.map(element => (
{
...element,
list: element.list.filter(l => l.id !== 111)
}
));
You can use Object.assign if the runtime you are running this code on does not support spread operator.
Array.filter acts on elements:
var myArray = [{something: 1, list: [1,2,3]}, {something: 2, list: [3,4,5]}]
var filtered = myArray.filter(function(element) {
return element.something === 1;
// true = keep element, false = discard it
})
console.log(filtered); // logs [{something: 1, list: [1,2,3]}]
You can use it like this:
var array = [{
id: 1,
list: [{
id: 123,
name: "Dartanan"
}, {
id: 456,
name: "Athos"
}, {
id: 789,
name: "Porthos"
}]
}, {
id: 2,
list: [{
id: 111,
name: "A"
}, {
id: 222,
name: "B"
}]
}]
for (var i = 0; i < array.length; ++i) {
var element = array[i]
// Filter the list
element.list = element.list.filter(function(listItem) {
return listItem.id !== 111 && listItem.name !== 'A';
})
}
console.log(array)

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