Getting the Index of a Particular Item in Array

How to find the array index with a value?

You can use indexOf:

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

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

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

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);

Getting the index of a particular item in array

You can use FindIndex

 var index = Array.FindIndex(myArray, row => row.Author == "xyz");

Edit: I see you have an array of string, you can use any code to match, here an example with a simple contains:

 var index = Array.FindIndex(myArray, row => row.Contains("Author='xyz'"));

Maybe you need to match using a regular expression?

How do I get the index of object in array using angular?

As mdn says:

The findIndex() method returns the index of the first element in the
array that satisfies the provided testing function. Otherwise, it
returns -1, indicating that no element passed the test.

If you have array of objects like this, then try to use findIndex:

const a = [
    { firstName: "Adam", LastName: "Howard" },
    { firstName: "Ben", LastName: "Skeet" },
    { firstName: "Joseph", LastName: "Skeet" }
];

let index = a.findIndex(x => x.LastName === "Skeet");
console.log(index);

Getting index of an array's element based on its properties

var index = myArray.map(function(el) {
  return el.id;
}).indexOf(4);

For IE below version 9, map need a patch, or just use a loop.

How can I get the index of an object by its property in JavaScript?

As the other answers suggest, looping through the array is probably the best way. But I would put it in its own function, and make it a little more abstract:

function findWithAttr(array, attr, value) {
    for(var i = 0; i < array.length; i += 1) {
        if(array[i][attr] === value) {
            return i;
        }
    }
    return -1;
}

var Data = [
    {id_list: 2, name: 'John', token: '123123'},
    {id_list: 1, name: 'Nick', token: '312312'}
];

With this, not only can you find which one contains 'John', but you can find which contains the token '312312':

findWithAttr(Data, 'name', 'John'); // returns 0
findWithAttr(Data, 'token', '312312'); // returns 1
findWithAttr(Data, 'id_list', '10'); // returns -1

The function returns -1 when not found, so it follows the same construct as Array.prototype.indexOf().

How to find the index of an element in an array in Java?

In this case, you could create e new String from your array of chars and then do an indeoxOf("e") on that String:

System.out.println(new String(list).indexOf("e")); 

But in other cases of primitive data types, you'll have to iterate over it.

Finding the index of an item in a list

>>> ["foo", "bar", "baz"].index("bar")
1

See the documentation for the built-in .index() method of the list:

list.index(x[, start[, end]])

Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Caveats

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If the list is long, and if there is no guarantee that the value will be near the beginning, this can slow down the code.

This problem can only be completely avoided by using a different data structure. However, if the element is known to be within a certain part of the list, the start and end parameters can be used to narrow the search.

For example:

>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514

The second call is orders of magnitude faster, because it only has to search through 10 elements, rather than all 1 million.

Only the index of the first match is returned

A call to index searches through the list in order until it finds a match, and stops there. If there could be more than one occurrence of the value, and all indices are needed, index cannot solve the problem:

>>> [1, 1].index(1) # the `1` index is not found.
0

Instead, use a list comprehension or generator expression to do the search, with enumerate to get indices:

>>> # A list comprehension gives a list of indices directly:
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> # A generator comprehension gives us an iterable object...
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> # which can be used in a `for` loop, or manually iterated with `next`:
>>> next(g)
0
>>> next(g)
2

The list comprehension and generator expression techniques still work if there is only one match, and are more generalizable.

Raises an exception if there is no match

As noted in the documentation above, using .index will raise an exception if the searched-for value is not in the list:

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If this is a concern, either explicitly check first using item in my_list, or handle the exception with try/except as appropriate.

The explicit check is simple and readable, but it must iterate the list a second time. See What is the EAFP principle in Python? for more guidance on this choice.

How to get value at a specific index of array In JavaScript?

You can access an element at a specific index using the bracket notation accessor.

var valueAtIndex1 = myValues[1];

On newer browsers/JavaScript engines (see browser compatibility here), you can also use the .at() method on arrays.

var valueAtIndex1 = myValues.at(1);

On positive indexes, both methods work the same (the first one being more common). Array.prototype.at() however allows you to access elements starting from the end of the array by passing a negative number. Passing -1 will give the last element of the array, passing -2 the second last, etc.

See more details at the MDN documentation.

---

Related Topics