This question already has answers here:
How Es6 rest and spread is working
(1 answer)
Spread Syntax ES6
(8 answers)
Closed 5 years ago.
I have come accross this syntax in a tutorial. Some say its not ES6 syntax. it was in a reduce function.
I need a clear explanation. What is going on in these parentethes ?
{...curr, ...acc}
full code..
const endShape = _(raw)
.filter(({key}) =>!/garbage/.test(key))
.map(({key,value})=>({[key]:value}))
.reduce((acc,curr)=>({...curr, ...acc}));
console.log(endShape);
Spread Operator Shorthand
The spread operator, introduced in ES6, has several use cases that make JavaScript code more efficient and fun to use. It can be used to replace certain array functions. The spread operator is simply a series of three dots.
Longhand
const odd = [1, 3, 5];
const nums = [2 ,4 , 6].concat(odd);
// cloning arrays
const arr = [1, 2, 3, 4];
const arr2 = arr.slice();
Shorthand
// joining arrays
const odd = [1, 3, 5 ];
const nums = [2 ,4 , 6, ...odd];
console.log(nums); // [ 2, 4, 6, 1, 3, 5 ]
// cloning arrays
const arr = [1, 2, 3, 4];
const arr2 = [...arr];
Related
This question already has answers here:
Destructuring to get the last element of an array in es6
(17 answers)
Get first and last elements in array, ES6 way [duplicate]
(1 answer)
Closed 8 months ago.
Suppose I have an array like this: [2, 4, 6, 8, 10].
I want to access the first and last element of this array using destructuring, currently I'm doing this:
const array = [2, 4, 6, 8, 10];
const [first, , , , last] = array;
console.log(first, last);
But this is only works with arrays of length 5 and is not generic enough.
In Python I could do something like this:
array = [2, 4, 6, 8, 10]
first, *mid, last = array
print(first, last)
But in JS this is not possible since rest elements should be the last. So, is there some way to do this in JS or this is not possible?
You can use object destructuring and grab the 0 key (which is the first element) and rename it to first & then using computed property names you can grab the array.length - 1 key (which is the last element) and rename it to last.
const array = [2, 4, 6, 8, 10];
const { 0: first, [array.length - 1]: last } = array;
console.log(first, last);
You can also grab the length via destructuring and then use that to grab the last element.
const array = [2, 4, 6, 8, 10];
const { length, 0: first, [length - 1]: last } = array;
console.log(first, last);
Another simple approach to access the last element would be to use the Array.prototype.at method. This is not related to destructuring but it's worth knowing.
const array = [2, 4, 6, 8, 10];
console.log(array.at(-1));
Not necessarily using destructuring, but still concise in my opinion:
const arr = [1, 2, 3, 4, 5];
const [ first, last ] = [ arr.at(0), arr.at(-1) ];
console.log(first, last);
No, you can't use destructuring like that without convoluted workarounds.
A shorter, more readable alternative is to just pop:
const array = [2, 4, 6, 8, 10];
const [first, ...middle] = array;
const last = middle.pop();
console.log(first, middle, last);
Or, just access them by index:
const array = [2, 4, 6, 8, 10];
const first = array[0];
const last = array[array.length - 1];
console.log(first, last);
This question already has an answer here:
Destructure an array parameter [duplicate]
(1 answer)
Closed 2 years ago.
Does any version of Javascript support destructuring of arrays in arrow functions? E.g.
const items = [ [1, 2], [3, 4] ];
const sums = items.map( [a, b] => a + b );
You can, but you have to surround the parameter in parentheses:
const items = [ [1, 2], [3, 4] ];
const sums = items.map(([a, b]) => a + b );
console.log(sums);
This question already has answers here:
Limit items in a .map loop
(5 answers)
Closed 3 years ago.
I need to use map starting from the second index. I know how to use loop for with I start from 1 but I want to use map in this case.
Is there a correct way to do this with map?
I'm using JavaScript and ReactJS.
Just slice before you map:
const arr = [1, 2, 3, 4];
const res = arr.slice(1).map(e => e * 2);
console.log(res);
Alternatively, use shift after mapping the full array:
const arr = [1, 2, 3, 4];
const res = arr.slice(1).map(e => e * 2);
res.shift();
console.log(res);
I am new to Javascript and is confused why the following won't work?
var array = [1, 2, 3, 4]
var spread = ...array;
I was expecting it would become 1, 2, 3, 4. Instead, it gave an error message Unexpected token .... Can anyone explain this to me?
Thank you so much!
This is the correct way, however you're not gaining anything doing that.
var array = [1, 2, 3, 4]
var spread = [...array];
console.log(spread);
If you really want to destructure that array, you need destructuring assignment:
var array = [1, 2, 3, 4]
var [one, two, three, four] = array;
console.log(one, two, three, four);
The correct way of doing what you want is:
var array = [1, 2, 3, 4]
var spread = [...array];
The syntax for using spread is:
For function calls:
myFunction(...iterableObj);
For array literals or strings:
[...iterableObj, '4', 'five', 6];
For object literals (new in ECMAScript 2018):
let objClone = { ...obj };
So, based on the syntax, for an array by using spread you are missing the square brackets []:
var array = [1, 2, 3, 4]
var spread = [...array];
console.log(spread);
This question already has answers here:
Simplest code for array intersection in javascript
(40 answers)
Compute intersection of two arrays in JavaScript [duplicate]
(4 answers)
Finding matches between multiple JavaScript Arrays
(13 answers)
Closed 5 years ago.
Let's take two arrays for example:
a = [1, 2, 3, 4, 5]
b = [4, 5, 6, 7, 8]
Now there is duplicates as we see 4 and 5. How to make new array from them instead of getting rid of them. What is the easiest solution?
So new array should be like:
newArray = [4, 5]
Thank you guys in advance!
You can do it using Array.filter() and Array.includes()
let a = [1, 2, 3, 4, 5];
let b = [4, 5, 6, 7, 8];
let arr = a.filter(function(x){
return b.includes(x);
})
console.log(arr);