JS 均匀地减少一个数组

Question

I want to reduce this array down to 5 numbers, but as evenly as possible. So I would need the start number, the end and then some in the middle.

const range = [8000, 9000, 10000, 11000, 12000, 13000, 14000, 15000, 16000, 17000, 18000, 19000, 20000, 21000, 22000, 23000, 24000, 25000, 26000, 27000, 28000, 29000, 30000]

Basically the above array is a data set for a slider I have, I want to show the numbers below so the user can see where to slider the slider too, however there are too many numbers to display nicely.

No idea how to do this, here is my attempt but is results in strangely spaced numbers due to not being even.

const steps = range[range.length - 1] / 5;
const start = range[0];
const stop = range[range.length - 1];

let marks = Array(Math.ceil((stop - start) / steps)).fill(start).map((x, y) => x + y * steps);
marks.push(stop);
console.log(marks);

Answer 1

Since you're using the first and last values, only 4 steps will actually be taken to receive 5 values (n-1). Let steps be a fractional value representing how many indices to step over between each iteration, starting at 0 and stopping at the highest index in the array.

 const range = [8000, 9000, 10000, 11000, 12000, 13000, 14000, 15000, 16000, 17000, 18000, 19000, 20000, 21000, 22000, 23000, 24000, 25000, 26000, 27000, 28000, 29000, 30000] const values = 5; const steps = (range.length-1) / (values-1); let marks=[]; for(let i=0; i<values; i++){ console.log("Fractional index", i*steps); //just for display purposes marks.push(range[Math.round(i*steps)]); } console.log(marks);

Answer 2

Divide your range by n steps. Then step through the list and add the values at each step.

 const n=5; const range = [8000, 9000, 10000, 11000, 12000, 13000, 14000, 15000, 16000, 17000, 18000, 19000, 20000, 21000, 22000, 23000, 24000, 25000, 26000, 27000, 28000, 29000, 30000] const steps = Math.floor(range.length / n); const stop = range[range.length - 1]; let marks=[] for(let i=0;i<n-1;i++){ marks.push(range[i*steps]) } marks.push(stop) console.log(marks);