Get the index of the object inside an array, matching a condition

Question

I have an array like this:

[{prop1:"abc",prop2:"qwe"},{prop1:"bnmb",prop2:"yutu"},{prop1:"zxvz",prop2:"qwrq"},...]

How can I get the index of the object that matches a condition, without iterating over the entire array?

For instance, given prop2=="yutu" , I want to get index 1 .

I saw .indexOf() but think it's used for simple arrays like ["a1","a2",...] . I also checked $.grep() but this returns objects, not the index.

Answer 1

As of 2016, you're supposed to use Array.findIndex (an ES2015/ES6 standard) for this:

 a = [ {prop1:"abc",prop2:"qwe"}, {prop1:"bnmb",prop2:"yutu"}, {prop1:"zxvz",prop2:"qwrq"}]; index = a.findIndex(x => x.prop2 ==="yutu"); console.log(index);

It's supported in Google Chrome, Firefox and Edge. For Internet Explorer, there's a polyfill on the linked page.

Performance note

Function calls are expensive, therefore with really big arrays a simple loop will perform much better than findIndex :

 let test = []; for (let i = 0; i < 1e6; i++) test.push({prop: i}); let search = test.length - 1; let count = 100; console.time('findIndex/predefined function'); let fn = obj => obj.prop === search; for (let i = 0; i < count; i++) test.findIndex(fn); console.timeEnd('findIndex/predefined function'); console.time('findIndex/dynamic function'); for (let i = 0; i < count; i++) test.findIndex(obj => obj.prop === search); console.timeEnd('findIndex/dynamic function'); console.time('loop'); for (let i = 0; i < count; i++) { for (let index = 0; index < test.length; index++) { if (test[index].prop === search) { break; } } } console.timeEnd('loop');

As with most optimizations, this should be applied with care and only when actually needed.

Answer 2

How can I get the index of the object tha match a condition (without iterate along the array)?

You cannot, something has to iterate through the array (at least once).

If the condition changes a lot, then you'll have to loop through and look at the objects therein to see if they match the condition. However, on a system with ES5 features (or if you install a shim), that iteration can be done fairly concisely:

var index;
yourArray.some(function(entry, i) {
    if (entry.prop2 == "yutu") {
        index = i;
        return true;
    }
});

That uses the new(ish) Array#some function , which loops through the entries in the array until the function you give it returns true. The function I've given it saves the index of the matching entry, then returns true to stop the iteration.

Or of course, just use a for loop. Your various iteration options are covered in this other answer .

But if you're always going to be using the same property for this lookup, and if the property values are unique, you can loop just once and create an object to map them:

var prop2map = {};
yourArray.forEach(function(entry) {
    prop2map[entry.prop2] = entry;
});

(Or, again, you could use a for loop or any of your other options .)

Then if you need to find the entry with prop2 = "yutu" , you can do this:

var entry = prop2map["yutu"];

I call this "cross-indexing" the array. Naturally, if you remove or add entries (or change their prop2 values), you need to update your mapping object as well.

Answer 3

TJ Crowder 所说的，每个地方都会有某种隐藏的迭代，用lodash这变成：

var index = _.findIndex(array, {prop2: 'yutu'})

Answer 4

var CarId = 23;

//x.VehicleId property to match in the object array
var carIndex = CarsList.map(function (x) { return x.VehicleId; }).indexOf(CarId);

And for basic array numbers you can also do this:

var numberList = [100,200,300,400,500];
var index = numberList.indexOf(200); // 1

You will get -1 if it cannot find a value in the array.

Answer 5

var index;
yourArray.some(function (elem, i) {
    return elem.prop2 === 'yutu' ? (index = i, true) : false;
});

Iterate over all elements of array. It returns either the index and true or false if the condition does not match.

Important is the explicit return value of true (or a value which boolean result is true). The single assignment is not sufficient, because of a possible index with 0 (Boolean(0) === false), which would not result an error but disables the break of the iteration.

Edit

An even shorter version of the above:

yourArray.some(function (elem, i) {
    return elem.prop2 === 'yutu' && ~(index = i);
});

Answer 6

Using Array.map() and Array.indexOf(string)

 const arr = [{ prop1: "abc", prop2: "qwe" }, { prop1: "bnmb", prop2: "yutu" }, { prop1: "zxvz", prop2: "qwrq" }] const index = arr.map(i => i.prop2).indexOf("yutu"); console.log(index);

Answer 7

I have seen many solutions in the above.

Here I am using map function to find the index of the search text in an array object.

I am going to explain my answer with using students data.

• step 1 : create array object for the students(optional you can create your own array object).
  var students = [{name:"Rambabu",htno:"1245"},{name:"Divya",htno:"1246"},{name:"poojitha",htno:"1247"},{name:"magitha",htno:"1248"}];

• step 2 : Create variable to search text
  var studentNameToSearch = "Divya";

• step 3 : Create variable to store matched index(here we use map function to iterate).
  var matchedIndex = students.map(function (obj) { return obj.name; }).indexOf(studentNameToSearch);

 var students = [{name:"Rambabu",htno:"1245"},{name:"Divya",htno:"1246"},{name:"poojitha",htno:"1247"},{name:"magitha",htno:"1248"}]; var studentNameToSearch = "Divya"; var matchedIndex = students.map(function (obj) { return obj.name; }).indexOf(studentNameToSearch); console.log(matchedIndex); alert("Your search name index in array is:"+matchedIndex)

Answer 8

You can use the Array.prototype.some() in the following way (as mentioned in the other answers):

https://jsfiddle.net/h1d69exj/2/

function findIndexInData(data, property, value) {
    var result = -1;
    data.some(function (item, i) {
        if (item[property] === value) {
            result = i;
            return true;
        }
    });
    return result;
}
var data = [{prop1:"abc",prop2:"qwe"},{prop1:"bnmb",prop2:"yutu"},{prop1:"zxvz",prop2:"qwrq"}]



alert(findIndexInData(data, 'prop2', "yutu")); // shows index of 1

Answer 9

function findIndexByKeyValue(_array, key, value) {
    for (var i = 0; i < _array.length; i++) { 
        if (_array[i][key] == value) {
            return i;
        }
    }
    return -1;
}
var a = [
    {prop1:"abc",prop2:"qwe"},
    {prop1:"bnmb",prop2:"yutu"},
    {prop1:"zxvz",prop2:"qwrq"}];
var index = findIndexByKeyValue(a, 'prop2', 'yutu');
console.log(index);

Answer 10

Try this code

var x = [{prop1:"abc",prop2:"qwe"},{prop1:"bnmb",prop2:"yutu"},{prop1:"zxvz",prop2:"qwrq"}]
let index = x.findIndex(x => x.prop1 === 'zxvz')

Answer 11

Another easy way is :

 function getIndex(items) {
        for (const [index, item] of items.entries()) {
            if (item.prop2 === 'yutu') {
                return index;
            }
        }
    }

const myIndex = getIndex(myArray);

Answer 12

The best & fastest way to do this is:

const products = [
  { prop1: 'telephone', prop2: 996 },
  { prop1: 'computadora', prop2: 1999 },
  { prop1: 'bicicleta', prop2: 995 },
];

const index = products.findIndex(el => el.prop2 > 1000);

console.log(index); // 1

Answer 13

Georg have already mentioned ES6 have Array.findIndex for this. And some other answers are workaround for ES5 using Array.some method.

One more elegant approach can be

var index;
for(index = yourArray.length; index-- > 0 && yourArray[index].prop2 !== "yutu";);

At the same time I will like to emphasize, Array.some may be implemented with binary or other efficient searching technique. So, it might perform better over for loop in some browser.

Answer 14

Why do you not want to iterate exactly ? The new Array.prototype.forEach are great for this purpose!

You can use a Binary Search Tree to find via a single method call if you want. This is a neat implementation of BTree and Red black Search tree in JS - https://github.com/vadimg/js_bintrees - but I'm not sure whether you can find the index at the same time.

Answer 15

One step using Array.reduce() - no jQuery

var items = [{id: 331}, {id: 220}, {id: 872}];

var searchIndexForId = 220;
var index = items.reduce(function(searchIndex, item, index){
  if(item.id === searchIndexForId) { 
    console.log('found!');
    searchIndex = index;
  }
  return searchIndex;
}, null);

will return null if index was not found.

Answer 16

var list =  [
                {prop1:"abc",prop2:"qwe"},
                {prop1:"bnmb",prop2:"yutu"},
                {prop1:"zxvz",prop2:"qwrq"}
            ];

var findProp = p => {
    var index = -1;
    $.each(list, (i, o) => {
        if(o.prop2 == p) {
            index = i;
            return false; // break
        }
    });
    return index; // -1 == not found, else == index
}