Generate list of items with 0 prefix using padStart

I implemented the way to generate a list of items with iterable counts with prefix 0. What is the best way to generate such kind of list?

Current behaviour:

const generateList = (length, n, i) =>
let b = n+i

return b.toString().padStart(length.toString().length + n.toString.length, 0)


Array(10).fill(null).map((x, i) => generateList(10,2, i))


Output result:

["002", "003", "004", "005", "006", "007", "008", "009", "010", "011"]


Do u have any idea to make it another way?

n.toString.length

1

n

3 Answers
3

You could determine the number of characters needed at the start and used the predetermined value to format the output for the array.

function createList(startValue, endValue)
let
// The minimum output length, for a single digit number, is 2 chars.
outputLength = 2,
testValue = 10,
// Create an empty array which has as many items as numbers we need to
// generate for the output. Add 1 to the end value as this is to be
// inclusive of the range to create. If the +1 is not done the resulting
// array is 1 item too small.
emptyArray = Array(endValue - startValue + 1);

// As long as test value is less than the end value, keep increasing the
// output size by 1 and continue to the next multiple of 10.
while (testValue <= endValue)
outputLength++;
testValue = testValue * 10;


// Create a new array, with the same length as the empty array created
// earlier. For each position place a padded number into the output array.
return Array.from(emptyArray, (currentValue, index) =>
// Pad the current value to the determined max length.
return (startValue + index).toString().padStart(outputLength, '0');
);


function createListWithLength(length, startValue = 0)
return createList(startValue, startValue + length);


console.log(createList(2,10));
console.log(createListWithLength(30));
console.log(createListWithLength(10, 995));

(endValue + 1) - startValue

Array.from(Array(length), fillerFn)

null

for

Array.from

for..of

Have a look at generators:

function* range(from, to)
for (var i=from; i<to; i++)
yield i;

function* paddedRange(from, to)
const length = (to-1).toString(10) + 1 /* at least one pad */;
for (const i of range(from, to))
yield i.padStart(length, '0');


console.log(Array.from(paddedRange(2, 12)));


You can also inline the loop from range into paddedRange, or you can make it return an array directly:

range

paddedRange

function paddedRange(from, to)
const length = (to-1).toString(10) + 1 /* at least one pad */;
return Array.from(range(from, to), i => i.padStart(length, '0'));


console.log(paddedRange(2, 12));


The main simplification is that you should compute the padding length only once and give it a denotative name, instead of computing it for every number again. Also ranges are usually given by their lower and upper end instead of their begin and a length, but you can easily switch back if you need the latter for some reason.

var

let

var

let

let

var

let

Not sure, but maybe something like this

const generateList = length => Array(length).fill('0').map((item, index) => item + index);

console.log(generateList(20));

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