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In JavaScript, how do I check if an array has duplicate multiple values?

I'm sorry I can't speak English well.
These are my simple code with a few parameters array:
if (link.indexOf({"x" : "1" , "y":"2" , "z": "3"}) === -1) {
link.push({
"x": "1",
"y": "2",
"z": "3"
});
} else {
alert("Duplicate");
}
Used in "for" loop but not alert Duplicate.

You can create a separate function for this to check if element exists in list or not.
Try this:
function doesExistInList(list, obj) {
for (let i = 0; i < list.length; i++) {
if (list[i].x === obj.x && list[i].y === obj.y && list[i].z === obj.z) {
return true;
}
}
return false;
}
let link = [];
let obj = { "x": "1", "y": "2", "z": "3" };
if (doesExistInList(link, obj) == false) {
link.push(obj);//insert same object to list
} else {
alert("Duplicate");
}
console.log(link);

JavaScript compares objects by reference -- that is, do the two objects point to the exact same spot in memory?
If so, then not only are they equal -- but they are the exact same object.
What you want is a compare-by-value -- and you'll have to do that yourself.
(You could also use the find method, but that isn't quite as supported.)
const link = [];
const testVal = {
"x": "1",
"y": "2",
"z": "3"
};
const results = link.filter(k =>
k.x === testVal.x &&
k.y === testVal.y &&
k.z === testVal.z);
if (results.length === 0) {
link.push(testVal);
} else {
alert("Duplicate");
}
console.log(link);

If two objects have same keys and same values in javascript they are different objects. To compare two objects you need to create separate function. Then use some() to compare each object in array with given object
Note: The below doesn't work for nested object. But it will work for your given case
function compareObjs(a, b){
if(Object.keys(a).length === Object.keys(b).length){
for(let k in a){
if(a[k] !== b[k]){
return false;
}
}
}
else{
return false;
}
return true;
}
const obj = {
"x": "1",
"y": "2",
"z": "3"
}
if(!link.some(a => compareObjs(obj, a))) {
link.push(obj);
} else {
alert("Duplicate");
}

There are couple of options to do that. The issue is with your code even if it looks like you are trying to check with indexOf if the array contains the targeted object. The main issue is it does not check the values of the properties but the reference of the original object.
So the first solution is, checking each property of the object in a hard coded way if you have the same structure for your object:
const links = [
{x: '2', y: '3', z: '1'},
{x: '11', y: '32', z: '73'},
{x: '1', y: '2', z: '3'},
{x: '93', y: '6', z: '76'},
];
const aim = {x: '1', y: '2', z: '3'};
links.forEach(link => {
const result = link.x === aim.x && link.y === aim.y && link.z === aim.z;
console.log(link, `This object has same property values as aim: ${result}`);
});
There are smarter solutions, just like getting the keys of the object what it has dynamically and comparing them by using some():
const links = [
{x: '2', y: '3', z: '1'},
{x: '11', y: '32', z: '73'},
{x: '1', y: '2', z: '3'},
{x: '93', y: '6', z: '76'},
];
const aim = {x: '1', y: '2', z: '3'};
links.forEach(link => {
const keysOfLink = Object.keys(link);
const keysOfAim = Object.keys(aim);
const result = keysOfLink.length === keysOfAim.length &&
keysOfLink.some(key => link[key] === aim[key]);
console.log(link, `This object has same property values as aim: ${result}`);
});
I would go with the second option, that's definitely smartest way to check.
From Array.prototype.some()'s documentation:
The some() method tests whether at least one element in the array passes the test implemented by the provided function. It returns a Boolean value.
Once in the result variable you have true value, you can push as you wanted originally.
I hope this helps!

Javascript - Compare array of objects and sum coincident values

I have this array of objects:
arr=[
{a: 1,b: 0,x: 100},
{a: 0,b: 0,x: 100},
{a: 1,b: 1,x: 100},
{a: 1,b: 0,x: 200},
{a: 1,b: 1,x: 200},
....
]
Now, what I need to do is to compare "x" values and if they coincide, tranfer summed "a" and "b" values in another array. For example:
arr2=[{a=2,b=1,x=100},{a=2,b=1,x=200}....]
Second thing to do, is to count how many objects are been joined with the same "x" value. For example in the first "arr2" object are joined 3 "arr" objects and in the second "arr2" object are joined 2 "arr" objects and so on.

This question doesn't seem to make sense. First of all, the word "coincident" doesn't have a technical definition that I'm aware of. Can you be more clear about what you mean?
Secondly its not clear what your expected results are. Perhaps what I would do if I was you would be to start with some simple example inputs and then come up with what you would expect the corresponding output to be, then use a unit testing tool to develop the code to do what you want.
For example, I have to guess what you are wanting but it might look like this (in javascript) using the libraries chai and mocha:
import { expect } from 'chai'
function doWork(input) {
// code goes here
}
const tests = [
{
name: 'Same x values coalesce',
data: [
{a=1,b=0,x=100},
{a=0,b=0,x=100}
],
expected: {
100: [1, 0]
}
}
]
describe('work', () => {
tests.forEach(test => {
it(t.name, () => {
const result = doWork(test.data)
expect(result).to.deep.equal(test.expected)
})
})
})
This technique may help you come to to an answer on your own.
Try to be a little more precise with your terms and give an example of what you are expecting to get as output.

Array reduce and map are fun methods. Object.keys() will give you an array of keys and that lets you do more array reducing and mapping. Fun times.
let arr = [
{ a: 1, b: 0, x: 100 },
{ a: 0, b: 0, x: 100 },
{ a: 1, b: 1, x: 100 },
{ a: 1, b: 0, x: 200 },
{ a: 1, b: 1, x: 200 }
];
let lists = arr.reduce((prev, curr) => {
let list = prev[curr.x] || [];
list.push(curr);
prev[curr.x] = list;
return prev;
}, {});
console.log(lists);
let flat = Object.keys(lists).reduce((prev, curr) => {
prev.push(
lists[curr].reduce(
(prev, curr) => {
prev.a += curr.a;
prev.b += curr.b;
prev.x = curr.x;
return prev;
},
{ a: 0, b: 0 }
)
);
return prev;
}, []);
console.log(flat);
Object.keys(lists).forEach(el => console.log(el + ': ' + lists[el].length));

Is it possible to change the order of parents/childs in an object with items?

I have an object that looks like this:
sum = {
value:{
a: 10,
b: 11
}
value2:{
a: 33,
c: 12
}
..
}
My object is more complex, the letters really contains objects with different values, but the key thing is that, I wish to loop through all letters that exist in every value and return objects like:
a:{
value: 10,
value2: 33
}
I could loop through the entire thing build a new object, but is there a more effient way of "flipping an objects order?
I only want to use the keys that are present in all value-objects, I currently get them like this, but that is not a requirement.
value = ['value', 'value2']
tags.forEach( (tag) =>
keys.push(Object.keys(sum[tag]))
)
matches = _.intersection.apply(_, keys);
matches.forEach( (match) => {
...
}

This solution features a reorganisation of the order of the properties.
grouped[kk][k] = sum[k][kk] for all elements
var sum = {
value: {
a: 10,
b: 11
},
value2: {
a: 33,
c: 12
}
},
grouped = {};
Object.keys(sum).forEach(function (k) {
Object.keys(sum[k]).forEach(function (kk) {
grouped[kk] = grouped[kk] || {};
grouped[kk][k] = sum[k][kk];
});
});
document.write('<pre>' + JSON.stringify(grouped, 0, 4) + '</pre>');

How do I merge an array of objects in Javascript?

Example:
var array1 = [ {'key':1, 'property1': 'x'}, {'key':2, 'property1': 'y'} ]
var array2 = [ {'key':2, 'property2': 'a'}, {'key':1, 'property2': 'b'} ]
I want merge(array1, array2) to give me:
[
{'key':1, 'property1': 'x', 'property2' : 'b'},
{'key':2, 'property1': 'y', 'property2' : 'a'}
]
Is there an easy way to do this?
EDIT: several people have answered without looking too closely at my problem, please be note that I want to match similar objects in each array and combine their properties into my final array. Keys are unique and there will only ever be at most one object with a particular key in each array.

I wrote a quick not-so-quick solution. The one problem you might want to consider is whether a property from an object in the second array should override the same property, if present, in the second object it's being compared to.
Solution 1
This solution is of complexity O(n²). Solution 2 is much faster; this solution is only for those who don't want to be Sanic the Hedgehog fast.
JavaScript
var mergeByKey = function (arr1, arr2, key) {
// key is the key that the function merges based on
arr1.forEach(function (d, i) {
var prop = d[key];
// since keys are unique, compare based on this key's value
arr2.forEach(function (f) {
if (prop == f[key]) { // if true, the objects share keys
for (var x in f) { // loop through each key in the 2nd object
if (!(x in d)) // if the key is not in the 1st object
arr1[i][x] = f[x]; // add it to the first object
// this is the part you might want to change for matching properties
// which object overrides the other?
}
}
})
})
return arr1;
}
Test Case
var arr = [ {'key':1, 'property1': 'x'},
{'key':2, 'property1': 'y'} ],
arr2= [ {'key':2, 'property2': 'a'},
{'key':1, 'property2': 'b'} ];
console.log(mergeByKey(arr, arr2, "key"));
Results
/* returns:
Object
key: 1
property1: "x"
property2: "b"
__proto__: Object
and
Object
key: 2
property1: "y"
property2: "a"
__proto__: Object
*/
fiddle
Solution 2
As Vivin Paliath pointed out in the comments below, my first solution was of O(n²) complexity (read: bad). His answer is very good and provides a solution with a complexity of O(m + n), where m is the size of the first array and n of the second array. In other words, of complexity O(2n).
However, his solution does not address objects within objects. To solve this, I used recursion—read: the devil, just like O(n²).
JavaScript
var mergeByKey = function (arr1, arr2, key) {
var holder = [],
storedKeys = {},
i = 0; j = 0; l1 = arr1.length, l2 = arr2.length;
var merge = function (obj, ref) {
for (var x in obj) {
if (!(x in ref || x instanceof Object)) {
ref[x] = obj[x];
} else {
merge(obj[x], ref[x]);
}
}
storedKeys[obj.key] = ref;
}
for (; i < l1; i++) {
merge(arr1[i], storedKeys[arr1[i].key] || {});
}
for (; j < l2; j++) {
merge(arr2[j], storedKeys[arr2[j].key] || {});
}
delete storedKeys[undefined];
for (var obj in storedKeys)
holder.push(storedKeys[obj]);
return holder;
}
Test Case
var arr1 = [
{
"key" : 1,
"prop1" : "x",
"test" : {
"one": 1,
"test2": {
"maybe" : false,
"test3": { "nothing" : true }
}
}
},
{
"key" : 2,
"prop1": "y",
"test" : { "one": 1 }
}],
arr2 = [
{
"key" : 1,
"prop2" : "y",
"test" : { "two" : 2 }
},
{
"key" : 2,
"prop2" : "z",
"test" : { "two": 2 }
}];
console.log(mergeByKey(arr1, arr2, "key"));
Results
/*
Object
key: 1
prop1: "x"
prop2: "y"
test: Object
one: 1
test2: Object
maybe: false
test3: Object
nothing: true
__proto__: Object
__proto__: Object
two: 2
__proto__: Object
__proto__: Object
Object
key: 2
prop1: "y"
prop2: "z"
test: Object
one: 1
two: 2
__proto__: Object
__proto__: Object
*/
This correctly merges the objects, along with all child objects. This solutions assumes that objects with matching keys have the same hierarchies. It also does not handle the merging of two arrays.
fiddle

You could do something like this:
function merge(array1, array2) {
var keyedResult = {};
function _merge(element) {
if(!keyedResult[element.key]) {
keyedResult[element.key] = {};
}
var entry = keyedResult[element.key];
for(var property in element) if(element.hasOwnProperty(property)) {
if(property !== "key") {
entry[property] = element[property];
}
}
entry["key"] = element.key;
}
array1.forEach(_merge);
array2.forEach(_merge);
var result = [];
for(var key in keyedResult) if(keyedResult.hasOwnProperty(key)) {
result.push(keyedResult[key]);
}
return result.sort(function(a, b) {
return a.key - b.key;
});
}
You could eliminate the sort if you don't care about the order. Another option is to use an array instead of the map I have used (keyedResult) if you have numeric keys and don't care about the array being sparse (i.e., if the keys are non-consecutive numbers). Here the key would also be the index of the array.
This solution also runs in O(n).
fiddle

It would be preferable to use existing infrastructure such as Underscore's _.groupBy and _.extend to handle cases like this, rather than re-inventing the wheel.
function merge(array1, array2) {
// merge the arrays
// [ {'key':1, 'property1': 'x'}, {'key':2, 'property1': 'y'}, {'key':2, 'property2': 'a'}, {'key':1, 'property2': 'b'} ]
var merged_array = array1.concat(array2);
// Use _.groupBy to create an object indexed by key of relevant array entries
// {1: [{ }, { }], 2: [{ }, { }]}
var keyed_objects = _.groupBy(merged_array, 'key');
// for each entry in keyed_objects, merge objects
return Object.keys(keyed_objects).map(function(key) {
return _.extend.apply({}, keyed_objects[key]);
});
}
The idea here is using _.extend.apply to pass the array of objects grouped under a particular key as arguments to _.extend, which will merge them all into a single object.

How can I use lodash/underscore to sort by multiple nested fields?

I want to do something like this:
var data = [
{
sortData: {a: 'a', b: 2}
},
{
sortData: {a: 'a', b: 1}
},
{
sortData: {a: 'b', b: 5}
},
{
sortData: {a: 'a', b: 3}
}
];
data = _.sortBy(data, ["sortData.a", "sortData.b"]);
_.map(data, function(element) {console.log(element.sortData.a + " " + element.sortData.b);});
And have it output this:
"a 1"
"a 2"
"a 3"
"b 5"
Unfortunately, this doesn't work and the array remains sorted in its original form. This would work if the fields weren't nested inside the sortData. How can I use lodash/underscore to sort an array of objects by more than one nested field?
I've turned this into a lodash feature request: https://github.com/lodash/lodash/issues/581

Update: See the comments below, this is not a good solution in most cases.
Someone kindly answered in the issue I created. Here's his answer, inlined:
_.sortBy(data, function(item) {
return [item.sortData.a, item.sortData.b];
});
I didn't realize that you're allowed to return an array from that function. The documentation doesn't mention that.

If you need to specify the sort direction, you can use _.orderBy with the array of functions syntax from Lodash 4.x:
_.orderBy(data, [
function (item) { return item.sortData.a; },
function (item) { return item.sortData.b; }
], ["asc", "desc"]);
This will sort first ascending by property a, and for objects that have the same value for property a, will sort them descending by property b.
It works as expected when the a and b properties have different types.
Here is a jsbin example using this syntax.

There is a _.sortByAll method in lodash version 3:
https://github.com/lodash/lodash/blob/3.10.1/doc/README.md#_sortbyallcollection-iteratees
Lodash version 4, it has been unified:
https://lodash.com/docs#sortBy
Other option would be to sort values yourself:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort
function compareValues(v1, v2) {
return (v1 > v2)
? 1
: (v1 < v2 ? -1 : 0);
};
var data = [
{ a: 2, b: 1 },
{ a: 2, b: 2 },
{ a: 1, b: 3 }
];
data.sort(function (x, y) {
var result = compareValues(x.a, y.a);
return result === 0
? compareValues(x.b, y.b)
: result;
});
// data after sort:
// [
// { a: 1, b: 3 },
// { a: 2, b: 1 },
// { a: 2, b: 2 }
// ];

The awesome, simple way is:
_.sortBy(data, [function(item) {
return item.sortData.a;
}, function(item) {
return item.sortData.b;
}]);
I found it from check the source code of lodash, it always check the function one by one.
Hope that help.

With ES6 easy syntax and lodash
sortBy(item.sortData, (item) => (-item.a), (item) => (-item.b))

I think this could work in most cases with underscore:
var properties = ["sortData.a", "sortData.b"];
data = _.sortBy(data, function (d) {
var predicate = '';
for (var i = 0; i < properties.length; i++)
{
predicate += (i == properties.length - 1
? 'd.' + properties[i]
: 'd.' + properties[i] + ' + ')
}
return eval(predicate)
});
It works and you can see it in Plunker

If the problem is an integer is converted to a string, add zeroes before the integer to make it have the same length as the longest in the collection:
var maxLength = _.reduce(data, function(result, item) {
var bString = _.toString(item.sortData.b);
return result > bString.length ? result : bString.length;
}, 0);
_.sortBy(data, function(item) {
var bString = _.toString(item.sortData.b);
if(maxLength > bString.length) {
bString = [new Array(maxLength - bString.length + 1).join('0'), bString].join('');
}
return [item.sortData.a, bString];
});

I've found a good way to sort array by multiple nested fields.
const array = [
{id: '1', name: 'test', properties: { prop1: 'prop', prop2: 'prop'}},
{id: '2', name: 'test2', properties: { prop1: 'prop second', prop2: 'prop second'}}
]
I suggest to use 'sorters' object which will describe a key and sort order. It's comfortable to use it with some data table.
const sorters = {
'id': 'asc',
'properties_prop1': 'desc',//I'm describing nested fields with '_' symbol
}
dataSorted = orderBy(array, Object.keys(sorters).map(sorter => {
return (row) => {
if (sorter.includes('_')) { //checking for nested field
const value = row["properties"][sorter.split('_')[1]];
return value || null;
};
return row[sorter] || null;// checking for empty values
};
}), Object.values(sorters));
This function will sort an array with multiple nested fields, for the first arguments it takes an array to modify, seconds one it's actually an array of functions, each function have argument that actually an object from 'array' and return a value or null for sorting. Last argument of this function is 'sorting orders', each 'order' links with functions array by index. How the function looks like simple example after mapping:
orderBy(array, [(row) => row[key] || null, (row) => row[key] || null , (row) => row[key] || null] , ['asc', 'desc', 'asc'])
P.S. This code can be improved, but I would like to keep it like this for better understanding.