How can I get the following effect?
On the input of the number 5 and the following array:
const list = ['a', 'b', 'c'];
It should produce the following output:
['a', 'b', 'c', 'a', 'b'] // Total 5
I tried for and while, but I don't know how to restart the loop when the loop exceeds the length of the list.
Figure out how many times to repeat the entire list in order to have enough elements; do that repetition using Array.fill and .flat; then truncate using .slice.
function repeatToLength(source, size) {
const iterations = Math.ceil(size / source.length);
return Array(iterations).fill(source).flat().slice(0, size);
}
console.log(repeatToLength(['a', 'b', 'c'], 5));
You can do this by repeatedly adding items from your starting list (baseList below) until you reach the desired length.
You can get the correct index in the base list by using the modulo operator (%) to get the remainder of division from dividing your output list's length by the base list's length, for example:
outputLength % baseLength == indexIntoBaseList
0 % 3 == 0
1 % 3 == 1
2 % 3 == 2
3 % 3 == 0
4 % 3 == 1
5 % 3 == 2
This gives you a repeating pattern of indices into your base list that you can use to add the correct item.
function RepeatListToLength(baseList, maxLength) {
var output = [];
while(output.length < maxLength) {
output.push(baseList[output.length % baseList.length]);
}
return output;
}
console.log(RepeatListToLength([1,2,3,4,5], 13));
Here's a recursive variant. The repeatToLength function calls itself with a new array consisting of two old arrays until it reaches enough length and then slices the final array to the exact length required:
const repeatToLength = (arr, len) => arr.length < len ?
repeatToLength([...arr, ...arr], len) : arr.slice(0, len);
console.log(repeatToLength(['a', 'b', 'c'], 5));After doing some speed test between my recursive solution and Karls I though I give another go and write a version that creates a big empty array and then maps it using the original array and the modulus operator. (I stole the modulus idea from Nicks answer.)
It was marginally faster than the recursive solution in Chrome, and a bit faster in Firefox (still not beating Karls solution in speed in Firefox):
const repeatToLength = (arr, len) => [...new Array(len)]
.map((x, i) => arr[i % arr.length]);
console.log(repeatToLength(['a', 'b', 'c'], 5));Mine seems to be quite a bit faster in Chrome, but Karls is slightly faster in Firefox (MacOS - latest browser versions, May 2023).
Warning: This speed comparison test may take a few seconds to run depending on your cpu speed and browser.
const thomas = (arr, len) => arr.length < len ?
thomas([...arr, ...arr], len) : arr.slice(0, len);
function karl(source, size) {
const iterations = Math.ceil(size / source.length);
return Array(iterations).fill(source).flat().slice(0, size);
}
const thomas2 = (arr, len) => [...new Array(len)]
.map((x, i) => arr[i % arr.length]);
function time(func, args, runNumberOfTimes) {
let start = performance.now();
for (let i = 0; i < runNumberOfTimes; i++) {
func(...args);
}
return Math.round(performance.now() - start) + ' ms';
}
let args;
console.log('Small array big repeat, run 2,000 times each');
args = [[1, 2, 3, 4, 5], 10000];
console.log('Array size: 5, make an array 10,000 items long.');
console.log('Karls solution', time(karl, args, 2000));
console.log('Thomas solution', time(thomas, args, 2000));
console.log('Thomas2 solution', time(thomas2, args, 2000));
console.log('\nBig array, smaller repeat, run 100 times each');
console.log('Array size: 10,000, make an array 100,000 items long');
args = ['x'.repeat(10000).split(''), 100000];
console.log('Karls solution', time(karl, args, 100));
console.log('Thomas solution', time(thomas, args, 100));
console.log('Thomas2 solution', time(thomas2, args, 100));