'How to generate range of numbers from 0 to n in ES2015 only?
I have always found the range function missing from JavaScript as it is available in python and others? Is there any concise way to generate range of numbers in ES2015 ?
EDIT: MY question is different from the mentioned duplicate as it is specific to ES2015 and not ECMASCRIPT-5. Also I need the range to be starting from 0 and not specific starting number (though it would be good if that is there)
Solution 1:[1]
I also found one more intuitive way using Array.from:
const range = n => Array.from({length: n}, (value, key) => key)
Now this range function will return all the numbers starting from 0 to n-1
A modified version of the range to support start and end is:
const range = (start, end) => Array.from({length: (end - start)}, (v, k) => k + start);
EDIT As suggested by @marco6, you can put this as a static method if it suits your use case
Array.range = (start, end) => Array.from({length: (end - start)}, (v, k) => k + start);
and use it as
Array.range(3, 9)
Solution 2:[2]
With Delta/Step
smallest and one-liner
[...Array(N)].map((_, i) => from + i * step);
Examples and other alternatives
[...Array(10)].map((_, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
Array.from(Array(10)).map((_, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
Array.from(Array(10).keys()).map(i => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
[...Array(10).keys()].map(i => 4 + i * -2);
//=> [4, 2, 0, -2, -4, -6, -8, -10, -12, -14]
Array(10).fill(0).map((_, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
Array(10).fill().map((_, i) => 4 + i * -2);
//=> [4, 2, 0, -2, -4, -6, -8, -10, -12, -14]
Range Function
const range = (from, to, step) =>
[...Array(Math.floor((to - from) / step) + 1)].map((_, i) => from + i * step);
range(0, 9, 2);
//=> [0, 2, 4, 6, 8]
// can also assign range function as static method in Array class (but not recommended )
Array.range = (from, to, step) =>
[...Array(Math.floor((to - from) / step) + 1)].map((_, i) => from + i * step);
Array.range(2, 10, 2);
//=> [2, 4, 6, 8, 10]
Array.range(0, 10, 1);
//=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Array.range(2, 10, -1);
//=> []
Array.range(3, 0, -1);
//=> [3, 2, 1, 0]
As Iterators
class Range {
constructor(total = 0, step = 1, from = 0) {
this[Symbol.iterator] = function* () {
for (let i = 0; i < total; yield from + i++ * step) {}
};
}
}
[...new Range(5)]; // Five Elements
//=> [0, 1, 2, 3, 4]
[...new Range(5, 2)]; // Five Elements With Step 2
//=> [0, 2, 4, 6, 8]
[...new Range(5, -2, 10)]; // Five Elements With Step -2 From 10
//=>[10, 8, 6, 4, 2]
[...new Range(5, -2, -10)]; // Five Elements With Step -2 From -10
//=> [-10, -12, -14, -16, -18]
// Also works with for..of loop
for (i of new Range(5, -2, 10)) console.log(i);
// 10 8 6 4 2
As Generators Only
const Range = function* (total = 0, step = 1, from = 0) {
for (let i = 0; i < total; yield from + i++ * step) {}
};
Array.from(Range(5, -2, -10));
//=> [-10, -12, -14, -16, -18]
[...Range(5, -2, -10)]; // Five Elements With Step -2 From -10
//=> [-10, -12, -14, -16, -18]
// Also works with for..of loop
for (i of Range(5, -2, 10)) console.log(i);
// 10 8 6 4 2
// Lazy loaded way
const number0toInf = Range(Infinity);
number0toInf.next().value;
//=> 0
number0toInf.next().value;
//=> 1
// ...
From-To with steps/delta
using iterators
class Range2 {
constructor(to = 0, step = 1, from = 0) {
this[Symbol.iterator] = function* () {
let i = 0,
length = Math.floor((to - from) / step) + 1;
while (i < length) yield from + i++ * step;
};
}
}
[...new Range2(5)]; // First 5 Whole Numbers
//=> [0, 1, 2, 3, 4, 5]
[...new Range2(5, 2)]; // From 0 to 5 with step 2
//=> [0, 2, 4]
[...new Range2(5, -2, 10)]; // From 10 to 5 with step -2
//=> [10, 8, 6]
using Generators
const Range2 = function* (to = 0, step = 1, from = 0) {
let i = 0,
length = Math.floor((to - from) / step) + 1;
while (i < length) yield from + i++ * step;
};
[...Range2(5, -2, 10)]; // From 10 to 5 with step -2
//=> [10, 8, 6]
let even4to10 = Range2(10, 2, 4);
even4to10.next().value;
//=> 4
even4to10.next().value;
//=> 6
even4to10.next().value;
//=> 8
even4to10.next().value;
//=> 10
even4to10.next().value;
//=> undefined
Solution 3:[3]
For numbers 0 to 5
[...Array(5).keys()];
=> [0, 1, 2, 3, 4]
Solution 4:[4]
A lot of these solutions build on instantiating real Array objects, which can get the job done for a lot of cases but can't support cases like range(Infinity). You could use a simple generator to avoid these problems and support infinite sequences:
function* range( start, end, step = 1 ){
if( end === undefined ) [end, start] = [start, 0];
for( let n = start; n < end; n += step ) yield n;
}
Examples:
Array.from(range(10)); // [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
Array.from(range(10, 20)); // [ 10, 11, 12, 13, 14, 15, 16, 17, 18, 19 ]
i = range(10, Infinity);
i.next(); // { value: 10, done: false }
i.next(); // { value: 11, done: false }
i.next(); // { value: 12, done: false }
i.next(); // { value: 13, done: false }
i.next(); // { value: 14, done: false }
Solution 5:[5]
So, in this case, it would be nice if Number object would behave like an Array object with the spread operator.
For instance Array object used with the spread operator:
let foo = [0,1,2,3];
console.log(...foo) // returns 0 1 2 3
It works like this because Array object has a built-in iterator.
In our case, we need a Number object to have a similar functionality:
[...3] //should return [0,1,2,3]
To do that we can simply create Number iterator for that purpose.
Number.prototype[Symbol.iterator] = function *() {
for(let i = 0; i <= this; i++)
yield i;
}
Now it is possible to create ranges from 0 to N with the spread operator.
[...N] // now returns 0 ... N array
http://jsfiddle.net/01e4xdv5/4/
Cheers.
Solution 6:[6]
You can use a generator function, which creates the range lazily only when needed:
function* range(x, y) {
while (true) {
if (x <= y)
yield x++;
else
return null;
}
}
const infiniteRange = x =>
range(x, Infinity);
console.log(
Array.from(range(1, 10)) // [1,2,3,4,5,6,7,8,9,10]
);
console.log(
infiniteRange(1000000).next()
);You can use a higher order generator function to map over the range generator:
function* range(x, y) {
while (true) {
if (x <= y)
yield x++;
else
return null;
}
}
const genMap = f => gx => function* (...args) {
for (const x of gx(...args))
yield f(x);
};
const dbl = n => n * 2;
console.log(
Array.from(
genMap(dbl) (range) (1, 10)) // [2,4,6,8,10,12,14,16,18,20]
);If you are fearless you can even generalize the generator approach to address a much wider range (pun intended):
const rangeBy = (p, f) => function* rangeBy(x) {
while (true) {
if (p(x)) {
yield x;
x = f(x);
}
else
return null;
}
};
const lte = y => x => x <= y;
const inc = n => n + 1;
const dbl = n => n * 2;
console.log(
Array.from(rangeBy(lte(10), inc) (1)) // [1,2,3,4,5,6,7,8,9,10]
);
console.log(
Array.from(rangeBy(lte(256), dbl) (2)) // [2,4,8,16,32,64,128,256]
);Keep in mind that generators/iterators are inherently stateful that is, there is an implicit state change with each invocation of next. State is a mixed blessing.
Solution 7:[7]
Range with step ES6, that works similar to python list(range(start, stop[, step])):
const range = (start, stop, step = 1) => {
return [...Array(stop - start).keys()]
.filter(i => !(i % Math.round(step)))
.map(v => start + v)
}
Examples:
range(0, 8) // [0, 1, 2, 3, 4, 5, 6, 7]
range(4, 9) // [4, 5, 6, 7, 8]
range(4, 9, 2) // [4, 6, 8]
range(4, 9, 3) // [4, 7]
Solution 8:[8]
To support delta
const range = (start, end, delta) => {
return Array.from(
{length: (end - start) / delta}, (v, k) => (k * delta) + start
)
};
Solution 9:[9]
How about just mapping ....
Array(n).map((value, index) ....) is 80% of the way there. But for some odd reason it does not work. But there is a workaround.
Array(n).map((v,i) => i) // does not work
Array(n).fill().map((v,i) => i) // does dork
For a range
Array(end-start+1).fill().map((v,i) => i + start) // gives you a range
Odd, these two iterators return the same result: Array(end-start+1).entries() and Array(end-start+1).fill().entries()
Solution 10:[10]
You can also do it with a one liner with step support like this one:
((from, to, step) => ((add, arr, v) => add(arr, v, add))((arr, v, add) => v < to ? add(arr.concat([v]), v + step, add) : arr, [], from))(0, 10, 1)
The result is [0, 1, 2, 3, 4, 5, 6 ,7 ,8 ,9].
Solution 11:[11]
This function will return an integer sequence.
const integerRange = (start, end, n = start, arr = []) =>
(n === end) ? [...arr, n]
: integerRange(start, end, start < end ? n + 1 : n - 1, [...arr, n]);
$> integerRange(1, 1)
<- Array [ 1 ]
$> integerRange(1, 3)
<- Array(3) [ 1, 2, 3 ]
$> integerRange(3, -3)
<- Array(7) [ 3, 2, 1, 0, -1, -2, -3 ]
Solution 12:[12]
const keys = Array(n).keys();
[...Array.from(keys)].forEach(callback);
in Typescript
Solution 13:[13]
Here's another variation that doesn't use Array.
let range = (n, l=[], delta=1) => {
if (n < 0) {
return l
}
else {
l.unshift(n)
return range(n - delta, l)
}
}
Solution 14:[14]
Generators now allow you to generate the number sequence lazily and using less memory for large ranges.
While the question specifically states ES2015, I expect a lot of Typescript users will end up here and the conversion to ES is straightforward...
function range(end: number): IterableIterator<number>;
// tslint:disable-next-line:unified-signatures
function range(begin: number, end: number): IterableIterator<number>;
function *range(begin: number, end: number = NaN): IterableIterator<number> {
let num = 0;
if (isNaN(end)) {
end = begin;
} else {
num = begin;
}
while (num < end) {
yield num++;
}
}
The first two function declarations are just to provide more informative completion suggestions in your IDE.
Solution 15:[15]
Few more ways to do
// Using `repeat` and `map`
const gen = n => [...'.'.repeat(n)].map((_,i) => i);
console.log('gen ', gen(5));
// Using `repeat` and `split`
const gen2 = n => ' '.repeat(n).split('').map((_,i) => i);
console.log('gen2 ', gen2(5));
// Using `concat` with recursive approach
const gen3 = n => n ? gen3(n-1).concat(n-1) : [];
console.log('gen3 ', gen3(5));
const range = (start, end, step = 1) =>
start > end ? [] : [start].concat(range(start + step, end, step));
console.log('range', range(2, 10,2));Sources
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