Develop Reference

Array.prototype.indexOf() cannot find array inside of multi-dimensional array [duplicate]

Let's say we have the following js array
var ar = [
[2,6,89,45],
[3,566,23,79],
[434,677,9,23]
];
var val = [3,566,23,79];
Is there a js builtin function or jQuery one with which you can search the array ar for val?
Thanks
***UPDATE*************
Taking fusion's response I created this prototype
Array.prototype.containsArray = function(val) {
var hash = {};
for(var i=0; i<this.length; i++) {
hash[this[i]] = i;
}
return hash.hasOwnProperty(val);
}

you could create a hash.
var ar = [
[2,6,89,45],
[3,566,23,79],
[434,677,9,23]
];
var hash = {};
for(var i = 0 ; i < ar.length; i += 1) {
hash[ar[i]] = i;
}
var val = [434,677,9,23];
if(hash.hasOwnProperty(val)) {
document.write(hash[val]);
}

You can also use a trick with JSON serializing. It is short and simple, but kind of hacky.
It works, because "[0,1]" === "[0,1]".
Here is the working demo snippet:
Array.prototype.indexOfForArrays = function(search)
{
var searchJson = JSON.stringify(search); // "[3,566,23,79]"
var arrJson = this.map(JSON.stringify); // ["[2,6,89,45]", "[3,566,23,79]", "[434,677,9,23]"]
return arrJson.indexOf(searchJson);
};
var arr = [
[2,6,89,45],
[3,566,23,79],
[434,677,9,23]
];
document.body.innerText = arr.indexOfForArrays([3,566,23,79]);

function indexOfArray(val, array) {
var hash = {};
for (var i = 0; i < array.length; i++) {
hash[array[i]] = i;
}
return (hash.hasOwnProperty(val)) ? hash[val] : -1;
};
I consider this more useful for than containsArray(). It solves the same problem (using a hash table) but returns the index (rather than only boolean true/false).

Can you try this?
var ar = [
[2,6,89,45],
[3,566,23,79],
[434,677,9,23]
];
var val = [3,566,23,79];
var sval = val.join("");
for(var i in ar)
{
var sar = ar[i].join("");
if (sar==sval)
{
alert("found!");
break;
}
}

Why don't you use javascript array functions?
function filterArrayByValues(array, values) {
return array.filter(function (arrayItem) {
return values.some(function (value) {
return value === arrayItem;
});
});
}
Or if your array is more complicated, and you want compare only one property but as result return whole object:
function filterArrayByValues(array, values, propertyName) {
return array.filter(function (arrayItem) {
return values.some(function (value) {
return value === arrayItem[propertyName];
});
});
}
More about used functions: filter() and some()

You can use Array.prototype.some(), Array.prototype.every() to check each element of each array.
var ar = [
[2, 6, 89, 45],
[3, 566, 23, 79],
[434, 677, 9, 23]
];
var val = [3, 566, 23, 79];
var bool = ar.some(function(arr) {
return arr.every(function(prop, index) {
return val[index] === prop
})
});
console.log(bool);

I guess there is no such JS functionality available. but you can create one
function arrEquals( one, two )
{
if( one.length != two.length )
{
return false;
}
for( i = 0; i < one.length; i++ )
{
if( one[i] != two[i] )
{
return false;
}
}
return true;
}

The problem with this is that of object/array equality in Javascript. Essentially, the problem is that two arrays are not equal, even if they have the same values. You need to loop through the array and compare the members to your search key (val), but you'll need a way of accurately comparing arrays.
The easiest way round this is to use a library that allows array/object comparison. underscore.js has a very attractive method to do this:
for (var i = 0; i < ar.length; i++) {
if (_.isEqual(ar[i], val)) {
// value is present
}
}
If you don't want to use another library (though I would urge you to -- or at least borrow the message from the Underscore source), you could do this with JSON.stringify...
var valJSON = JSON.stringify(val);
for (var i = 0; i < ar.length; i++) {
if (valJSON === JSON.stringify(ar[i]) {
// value is present
}
}
This will almost certainly be significantly slower, however.

You can use toString convertion to compare elements
var ar = [
[2,6,89,45],
[3,566,23,79],
[434,677,9,23]
];
var val = [3,566,23,79];
s = !ar.every(a => (a.toString() != val.toString()));
console.log(s) // true

Use this instead
if (ar.join(".").indexOf(val) > -1) {
return true;
} else {
return false;
}

Use lodash isEqual
const isValIncludedInAr = ar.some(element => isEqual(element, val))

const arrayOne = [2,6,89,45];
const arrayTwo = [3,566,23,79];
const arrayThree = [434,677,9,23];
const data = new Set([arrayOne, arrayTwo, arrayThree]);
// Check array if exist
console.log( data.has(arrayTwo) ); // It will return true.
// If you want to make a set into array it's simple
const arrayData = [...data];
console.log(arrayData); // It will return [[2,6,89,45], [3,566,23,79], [434,677,9,23]]

How can I find which index in an array contains an object whose value for a specific key is x? [duplicate]

I would like to find index in array. Positions in array are objects, and I want to filter on their properties. I know which keys I want to filter and their values. Problem is to get index of array which meets the criteria.
For now I made code to filter data and gives me back object data, but not index of array.
var data = [
{
"text":"one","siteid":"1","chid":"default","userid":"8","time":1374156747
},
{
"text":"two","siteid":"1","chid":"default","userid":"7","time":1374156735
}
];
var filterparams = {userid:'7', chid: 'default'};
function getIndexOfArray(thelist, props){
var pnames = _.keys(props)
return _.find(thelist, function(obj){
return _.all(pnames, function(pname){return obj[pname] == props[pname]})
})};
var check = getIndexOfArray(data, filterparams ); // Want to get '2', not key => val

Using Lo-Dash in place of underscore you can do it pretty easily with _.findIndex().
var index = _.findIndex(array, { userid: '7', chid: 'default' })

here is thefiddle hope it helps you
for(var intIndex=0;intIndex < data.length; intIndex++){
eachobj = data[intIndex];
var flag = true;
for (var k in filterparams) {
if (eachobj.hasOwnProperty(k)) {
if(eachobj[k].toString() != filterparams[k].toString()){
flag = false;
}
}
}
if(flag){
alert(intIndex);
}
}

I'm not sure, but I think that this is what you need:
var data = [{
"text":"one","siteid":"1","chid":"default","userid":"8","time":1374156747
}, {
"text":"two","siteid":"1","chid":"default","userid":"7","time":1374156735
}];
var filterparams = {userid:'7', chid: 'default'};
var index = data.indexOf( _.findWhere( data, filterparams ) );

I don't think you need underscore for that just regular ole js - hope this is what you are looking for
var data = [
{
"text":"one","siteid":"1","chid":"default","userid":"8","time":1374156747
},
{
"text":"two","siteid":"1","chid":"default","userid":"7","time":1374156735
}
];
var userid = "userid"
var filterparams = {userid:'7', chid: 'default'};
var index;
for (i=0; i < data.length; i++) {
for (prop in data[i]) {
if ((prop === userid) && (data[i]['userid'] === filterparams.userid)) {
index = i
}
}
}
alert(index);

Finding my object inside JSON by its ID

[
{"ID":"5","Name":"Jay"},
{"ID":"30","Name":"Sharon"},
{"ID":"32","Name":"Paul"}
]
So I have this kind of JSON.
I need to easily supply the value for a required key.
For example:
30 would yield => "Sharon"
5 would yield => "Jay"
etc. What is the right way to do this?

Iterate the array and check if the ID matches
function getById(id) {
var O = null;
for (var i=0; i<arr.length; i++) {
if ( arr[i].ID == id ) return O = arr[i];
}
return O;
}
getById('30'); // returns {"ID":"30","Name":"Sharon"}
FIDDLE
or in newer browsers:
function getById(arr, id) {
return arr.filter(function(o) { return o.ID == id });
}
FIDDLE

Try a linear search:
var searchId = "30";
for(var i = 0; i < json.length; i++)
{
if(json[i].ID == searchId)
{
// Found it.
//
break;
}
}

If the IDs will be unique, and if you're going to need to do this frequently, then you may want to convert your collection to key/value pairs where the ID is the key.
var byId = data.reduce(function(res, obj) {
res[obj.ID] = obj;
return res
}, {});
Now you can simply use the ID to look up the object.
var target = byId["30"];

You could probably just write something to loop through it.
var data = [ {"ID":"5","Name":"Jay"},{"ID":"30","Name":"Sharon"}, {"ID":"32","Name":"Paul"} ];
for(var i in data){
if(data[i]["ID"] == 30){
return data[i]["Name"];
}
}

undersocre.js can find a object in collection by one line code
Reference: http://underscorejs.org/#find
Code:
var people = [
{"ID":"5","Name":"Jay"},
{"ID":"30","Name":"Sharon"},
{"ID":"32","Name":"Paul"}
];
_.find(people, function(person) { return person.ID === '5'; });
FIDDLE

The best way to remove array element by value

I have an array like this
arr = ["orange","red","black","white"]
I want to augment the array object defining a deleteElem() method which acts like this:
arr2 = arr.deleteElem("red"); // ["orange","black","white"] (with no hole)
What is the best way to accomplish this task using just the value parameter (no index)?

Here's how it's done:
var arr = ["orange","red","black","white"];
var index = arr.indexOf("red");
if (index >= 0) {
arr.splice( index, 1 );
}
This code will remove 1 occurency of "red" in your Array.

Back when I was new to coding I could hardly tell what splice was doing, and even today it feels less readable.
But readability counts.
I would rather use the filter method like so:
arr = ["orange","red","black","white","red"]
arr = arr.filter(val => val !== "red");
console.log(arr) // ["orange","black","white"]
Note how all occurrences of "red" are removed from the array.
From there, you can easily work with more complex data such as array of objects.
arr = arr.filter(obj => obj.prop !== "red");

There is an underscore method for this, http://underscorejs.org/#without
arr = ["orange","red","black","white"];
arr = _.without(arr, "red");

The trick is to go through the array from end to beginning, so you don't mess up the indices while removing elements.
var deleteMe = function( arr, me ){
var i = arr.length;
while( i-- ) if(arr[i] === me ) arr.splice(i,1);
}
var arr = ["orange","red","black", "orange", "white" , "orange" ];
deleteMe( arr , "orange");
arr is now ["red", "black", "white"]

Array.prototype.deleteElem = function(val) {
var index = this.indexOf(val);
if (index >= 0) this.splice(index, 1);
return this;
};
var arr = ["orange","red","black","white"];
var arr2 = arr.deleteElem("red");

My approach, let's see what others have to say. It supports an "equals" method as well.
// Remove array value
// #param {Object} val
Array.prototype.removeByValue = function (val) {
for (var i = 0; i < this.length; i++) {
var c = this[i];
if (c == val || (val.equals && val.equals(c))) {
this.splice(i, 1);
break;
}
}
};
Read https://stackoverflow.com/a/3010848/356726 for the impact on iterations when using prototype with Array.

Or simply check all items, create a new array with non equal and return it.
var arr = ['orange', 'red', 'black', 'white'];
console.info('before: ' + JSON.stringify(arr));
var deleteElem = function ( val ) {
var new_arr = [];
for ( var i = 0; i < this.length; i++ ) {
if ( this[i] !== val ) {
new_arr.push(this[i]);
}
}
return new_arr;
};
arr = deleteElem('red');
console.info('after: ' + JSON.stringify(arr));
http://jsfiddle.net/jthavn3m/

The best way is to use splice and rebuild new array, because after splice, the length of array does't change.
Check out my answer:
function remove_array_value(array, value) {
var index = array.indexOf(value);
if (index >= 0) {
array.splice(index, 1);
reindex_array(array);
}
}
function reindex_array(array) {
var result = [];
for (var key in array) {
result.push(array[key]);
}
return result;
}
example:
var example_arr = ['apple', 'banana', 'lemon']; // length = 3
remove_array_value(example_arr, 'banana');
banana is deleted and array length = 2

If order the array (changing positions) won't be a problem you can solve like:
var arr = ["orange","red","black","white"];
arr.remove = function ( item ) {
delete arr[item];
arr.sort();
arr.pop();
console.log(arr);
}
arr.remove('red');
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Here you go:
arr.deleteElem = function ( val ) {
for ( var i = 0; i < this.length; i++ ) {
if ( this[i] === val ) {
this.splice( i, 1 );
return i;
}
}
};
Live demo: http://jsfiddle.net/4vaE2/3/
The deleteElem method returns the index of the removed element.
var idx = arr.deleteElem( 'red' ); // idx is 1

Getting a union of two arrays in JavaScript [duplicate]

This question already has answers here:
How to merge two arrays in JavaScript and de-duplicate items
(89 answers)
Closed 4 years ago.
Say I have an array of [34, 35, 45, 48, 49] and another array of [48, 55]. How can I get a resulting array of [34, 35, 45, 48, 49, 55]?

With the arrival of ES6 with sets and splat operator (at the time of being works only in Firefox, check compatibility table), you can write the following cryptic one liner:
var a = [34, 35, 45, 48, 49];
var b = [48, 55];
var union = [...new Set([...a, ...b])];
console.log(union);
Little explanation about this line: [...a, ...b] concatenates two arrays, you can use a.concat(b) as well. new Set() create a set out of it and thus your union. And the last [...x] converts it back to an array.

If you don't need to keep the order, and consider 45 and "45" to be the same:
function union_arrays (x, y) {
var obj = {};
for (var i = x.length-1; i >= 0; -- i)
obj[x[i]] = x[i];
for (var i = y.length-1; i >= 0; -- i)
obj[y[i]] = y[i];
var res = []
for (var k in obj) {
if (obj.hasOwnProperty(k)) // <-- optional
res.push(obj[k]);
}
return res;
}
console.log(union_arrays([34,35,45,48,49], [44,55]));

If you use the library underscore you can write like this
var unionArr = _.union([34,35,45,48,49], [48,55]);
console.log(unionArr);
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"></script>
Ref: http://underscorejs.org/#union

I'm probably wasting time on a dead thread here. I just had to implement this and went looking to see if I was wasting my time.
I really like KennyTM's answer. That's just how I would attack the problem. Merge the keys into a hash to naturally eliminate duplicates and then extract the keys. If you actually have jQuery you can leverage its goodies to make this a 2 line problem and then roll it into an extension. The each() in jQuery will take care of not iterating over items where hasOwnProperty() is false.
jQuery.fn.extend({
union: function(array1, array2) {
var hash = {}, union = [];
$.each($.merge($.merge([], array1), array2), function (index, value) { hash[value] = value; });
$.each(hash, function (key, value) { union.push(key); } );
return union;
}
});
Note that both of the original arrays are left intact. Then you call it like this:
var union = $.union(array1, array2);

If you wants to concatenate two arrays without any duplicate value,Just try this
var a=[34, 35, 45, 48, 49];
var b=[48, 55];
var c=a.concat(b).sort();
var res=c.filter((value,pos) => {return c.indexOf(value) == pos;} );

function unique(arrayName)
{
var newArray=new Array();
label: for(var i=0; i<arrayName.length;i++ )
{
for(var j=0; j<newArray.length;j++ )
{
if(newArray[j]==arrayName[i])
continue label;
}
newArray[newArray.length] = arrayName[i];
}
return newArray;
}
var arr1 = new Array(0,2,4,4,4,4,4,5,5,6,6,6,7,7,8,9,5,1,2,3,0);
var arr2= new Array(3,5,8,1,2,32,1,2,1,2,4,7,8,9,1,2,1,2,3,4,5);
var union = unique(arr1.concat(arr2));
console.log(union);

Adapted from: https://stackoverflow.com/a/4026828/1830259
Array.prototype.union = function(a)
{
var r = this.slice(0);
a.forEach(function(i) { if (r.indexOf(i) < 0) r.push(i); });
return r;
};
Array.prototype.diff = function(a)
{
return this.filter(function(i) {return a.indexOf(i) < 0;});
};
var s1 = [1, 2, 3, 4];
var s2 = [3, 4, 5, 6];
console.log("s1: " + s1);
console.log("s2: " + s2);
console.log("s1.union(s2): " + s1.union(s2));
console.log("s2.union(s1): " + s2.union(s1));
console.log("s1.diff(s2): " + s1.diff(s2));
console.log("s2.diff(s1): " + s2.diff(s1));
// Output:
// s1: 1,2,3,4
// s2: 3,4,5,6
// s1.union(s2): 1,2,3,4,5,6
// s2.union(s1): 3,4,5,6,1,2
// s1.diff(s2): 1,2
// s2.diff(s1): 5,6

I like Peter Ajtai's concat-then-unique solution, but the code's not very clear. Here's a nicer alternative:
function unique(x) {
return x.filter(function(elem, index) { return x.indexOf(elem) === index; });
};
function union(x, y) {
return unique(x.concat(y));
};
Since indexOf returns the index of the first occurence, we check this against the current element's index (the second parameter to the filter predicate).

Shorter version of kennytm's answer:
function unionArrays(a, b) {
const cache = {};
a.forEach(item => cache[item] = item);
b.forEach(item => cache[item] = item);
return Object.keys(cache).map(key => cache[key]);
};

You can use a jQuery plugin: jQuery Array Utilities
For example the code below
$.union([1, 2, 2, 3], [2, 3, 4, 5, 5])
will return [1,2,3,4,5]

function unite(arr1, arr2, arr3) {
newArr=arr1.concat(arr2).concat(arr3);
a=newArr.filter(function(value){
return !arr1.some(function(value2){
return value == value2;
});
});
console.log(arr1.concat(a));
}//This is for Sorted union following the order :)

function unionArrays() {
var args = arguments,
l = args.length,
obj = {},
res = [],
i, j, k;
while (l--) {
k = args[l];
i = k.length;
while (i--) {
j = k[i];
if (!obj[j]) {
obj[j] = 1;
res.push(j);
}
}
}
return res;
}
var unionArr = unionArrays([34, 35, 45, 48, 49], [44, 55]);
console.log(unionArr);
Somewhat similar in approach to alejandro's method, but a little shorter and should work with any number of arrays.

function unionArray(arrayA, arrayB) {
var obj = {},
i = arrayA.length,
j = arrayB.length,
newArray = [];
while (i--) {
if (!(arrayA[i] in obj)) {
obj[arrayA[i]] = true;
newArray.push(arrayA[i]);
}
}
while (j--) {
if (!(arrayB[j] in obj)) {
obj[arrayB[j]] = true;
newArray.push(arrayB[j]);
}
}
return newArray;
}
var unionArr = unionArray([34, 35, 45, 48, 49], [44, 55]);
console.log(unionArr);
Faster
http://jsperf.com/union-array-faster

I would first concatenate the arrays, then I would return only the unique value.
You have to create your own function to return unique values. Since it is a useful function, you might as well add it in as a functionality of the Array.
In your case with arrays array1 and array2 it would look like this:
array1.concat(array2) - concatenate the two arrays
array1.concat(array2).unique() - return only the unique values. Here unique() is a method you added to the prototype for Array.
The whole thing would look like this:
Array.prototype.unique = function () {
var r = new Array();
o: for(var i = 0, n = this.length; i < n; i++)
{
for(var x = 0, y = r.length; x < y; x++)
{
if(r[x]==this[i])
{
continue o;
}
}
r[r.length] = this[i];
}
return r;
}
var array1 = [34,35,45,48,49];
var array2 = [34,35,45,48,49,55];
// concatenate the arrays then return only the unique values
console.log(array1.concat(array2).unique());

Just wrote before for the same reason (works with any amount of arrays):
/**
* Returns with the union of the given arrays.
*
* #param Any amount of arrays to be united.
* #returns {array} The union array.
*/
function uniteArrays()
{
var union = [];
for (var argumentIndex = 0; argumentIndex < arguments.length; argumentIndex++)
{
eachArgument = arguments[argumentIndex];
if (typeof eachArgument !== 'array')
{
eachArray = eachArgument;
for (var index = 0; index < eachArray.length; index++)
{
eachValue = eachArray[index];
if (arrayHasValue(union, eachValue) == false)
union.push(eachValue);
}
}
}
return union;
}
function arrayHasValue(array, value)
{ return array.indexOf(value) != -1; }

Simple way to deal with merging single array values.
var values[0] = {"id":1235,"name":"value 1"}
values[1] = {"id":4323,"name":"value 2"}
var object=null;
var first=values[0];
for (var i in values)
if(i>0)
object= $.merge(values[i],first)

You can try these:
function union(a, b) {
return a.concat(b).reduce(function(prev, cur) {
if (prev.indexOf(cur) === -1) prev.push(cur);
return prev;
}, []);
}
or
function union(a, b) {
return a.concat(b.filter(function(el) {
return a.indexOf(el) === -1;
}));
}

ES2015 version
Array.prototype.diff = function(a) {return this.filter(i => a.indexOf(i) < 0)};
Array.prototype.union = function(a) {return [...this.diff(a), ...a]}

If you want a custom equals function to match your elements, you can use this function in ES2015:
function unionEquals(left, right, equals){
return left.concat(right).reduce( (acc,element) => {
return acc.some(elt => equals(elt, element))? acc : acc.concat(element)
}, []);
}
It traverses the left+right array. Then for each element, will fill the accumulator if it does not find that element in the accumulator. At the end, there are no duplicate as specified by the equals function.
Pretty, but probably not very efficient with thousands of objects.

I think it would be simplest to create a new array, adding the unique values only as determined by indexOf.
This seems to me to be the most straightforward solution, though I don't know if it is the most efficient. Collation is not preserved.
var a = [34, 35, 45, 48, 49],
b = [48, 55];
var c = union(a, b);
function union(a, b) { // will work for n >= 2 inputs
var newArray = [];
//cycle through input arrays
for (var i = 0, l = arguments.length; i < l; i++) {
//cycle through each input arrays elements
var array = arguments[i];
for (var ii = 0, ll = array.length; ii < ll; ii++) {
var val = array[ii];
//only add elements to the new array if they are unique
if (newArray.indexOf(val) < 0) newArray.push(val);
}
}
return newArray;
}

[i for( i of new Set(array1.concat(array2)))]
Let me break this into parts for you
// This is a list by comprehension
// Store each result in an element of the array
[i
// will be placed in the variable "i", for each element of...
for( i of
// ... the Set which is made of...
new Set(
// ...the concatenation of both arrays
array1.concat(array2)
)
)
]
In other words, it first concatenates both and then it removes the duplicates (a Set, by definition cannot have duplicates)
Do note, though, that the order of the elements is not guaranteed, in this case.