How to get subarray from array?

I have var ar = [1, 2, 3, 4, 5] and want some function getSubarray(array, fromIndex, toIndex), that result of call getSubarray(ar, 1, 3) is new array [2, 3, 4].

Answers:

Answer

For a simple use of slice, use my extension to Array Class:

Array.prototype.subarray = function(start, end) {
    if (!end) { end = -1; } 
    return this.slice(start, this.length + 1 - (end * -1));
};

Then:

var bigArr = ["a", "b", "c", "fd", "ze"];

Test1:

bigArr.subarray(1, -1);

< ["b", "c", "fd", "ze"]

Test2:

bigArr.subarray(2, -2);

< ["c", "fd"]

Test3:

bigArr.subarray(2);

< ["c", "fd","ze"]

Might be easier for developers coming from another language (i.e. Groovy).

Answer

The question is actually asking for a New array, so I believe a better solution would be to combine Abdennour TOUMI's answer with a clone function:

function clone(obj) {
  if (null == obj || "object" != typeof obj) return obj;
  const copy = obj.constructor();
  for (const attr in obj) {
    if (obj.hasOwnProperty(attr)) copy[attr] = obj[attr];
  }
  return copy;
}

// With the `clone()` function, you can now do the following:

Array.prototype.subarray = function(start, end) {
  if (!end) {
    end = this.length;
  } 
  const newArray = clone(this);
  return newArray.slice(start, end);
};

// Without a copy you will lose your original array.

// **Example:**

const array = [1, 2, 3, 4, 5];
console.log(array.subarray(2)); // print the subarray [3, 4, 5, subarray: function]

console.log(array); // print the original array [1, 2, 3, 4, 5, subarray: function]

[http://stackoverflow.com/questions/728360/most-elegant-way-to-clone-a-javascript-object]

Answer

Take a look at Array.slice(begin, end)

const ar  = [1, 2, 3, 4, 5];

// slice from 1..3 - add 1 as the end index is not included

const ar2 = ar.slice(1, 3 + 1);

console.log(ar2);

Answer

const array_one = [11, 22, 33, 44, 55];
const start = 1;
const end = array_one.length - 1;
const array_2 = array_one.slice(start, end);
console.log(array_2);

Tags

Recent Questions

Top Questions

Home Tags Terms of Service Privacy Policy DMCA Contact Us Javascript

©2020 All rights reserved.