How do I repeat the colors array in order, given an unknown length of items?
const items = [1, 2, ...n]
const colors = ['blue', 'green', 'red']
// return ['blue', 'green', 'red', 'blue', 'green'] where items.length = 5
6 Answers
const items = [1, 2, 3,4,5,6,7,8]
const colors = ['blue', 'green', 'red']
const result = items.map((_,i) => colors[i%colors.length]);
console.log(result);
1 Comment
ndemasie
Perfect! I was stuck in a block. Many thanks! Edit: I'll accept the answer as soon as I'm allowed :)
You can map over a new array with the length you want and take the modulus of the index by the length of the colors array. You do not seem to need the items array at all.
let len = 5;
const colors = ['blue', 'green', 'red'];
const res = Array(len).fill().map((_,idx)=>colors[idx % colors.length]);
console.log(res);Array.from can also be used in this case.
let length = 5;
const colors = ['blue', 'green', 'red'];
const res = Array.from({length}, (_,idx)=>colors[idx % colors.length]);
console.log(res);2 Comments
Scott Sauyet
Or similarly,
Array.from({length: 5}, (_, i) => colors[i % colors.length])
Unmitigated
@ScottSauyet Thanks for the suggestion.
When the target size is large, you may get better performance by doubling the array over and over again:
function stretch(arr, n) {
while (arr.length < n) arr = arr.concat(arr);
return arr.slice(0, n);
}
const items = [1, 2, 3, 4, 5];
const colors = ['blue', 'green', 'red'];
console.log(stretch(colors, items.length));1 Comment
Scott Sauyet
A nice approach for a large array. For clarity, I would choose a
map / Array.from solution over this unless I expected large arrays. But when wanted, this is quite useful.Create a new array and map the values by using the remainder operator %.
const
items = [1, 2, 3, 4, 5, 6, 7, 8],
colors = ['blue', 'green', 'red'],
result = Array.from(items, (_, i) => colors[i % colors.length]);
console.log(result);2 Comments
Scott Sauyet
Do you find this cleaner than a
map call?
Nina Scholz
yes, because it take only a part structure (the length) of the original array. by using
map, i miss the part where an element is used for the result. (this may apply here as well ...)You can take mode for index so you can repeat it.
const items = [1, 2, 3,4,5,6];
const colors = ['blue', 'green', 'red'];
var result=[];
items.forEach(myFunction)
function myFunction(item, index) {
console.log(colors[(index%colors.length)])
result.push(colors[(index%colors.length)])
}
2 Comments
Scott Sauyet
That's a
map in disguise. You might as well just use it directly.
trincot
Now you have a function with side effects.
Using map()
const items = [1, 2, 3, 4, 5, 6, 7, 8],
colors = ['blue', 'green', 'red'];
output = items.map((_,i) => colors[i % colors.length])
console.log(output);
3 Comments
Scott Sauyet
What does the spread add here? See the answer from Ali for comparison.
Scott Sauyet
I'm suggesting that your same code would work if you removed the initial
[... and trailing ]. The map call already creates a new array that does what we want. The spread is entirely superfluous.
pietrodito
You can also remove True or Correct from your last comment!
itemshere? You only seem to use its length...