refactor: 删除多余注释

manyyuri · May 5, 2024 · e8894af · e8894af
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diff --git a/.eslintcache b/.eslintcache
@@ -1 +1 @@
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diff --git a/src/scale/time.js b/src/scale/time.js
diff --git a/src/scale/utils.js b/src/scale/utils.js
@@ -2,53 +2,33 @@ export function normalize(value, start, stop) {
  return (value - start) / (stop - start);
 }
 
-// step0 是生成指定数量的刻度的间隔
-// step1 是最后生成的刻度的间隔
-// 我们希望 step1 满足两个条件：
-// 1. step1 = 10 ^ n * b (其中 b=1,2,5)
-// 2. step0 和 step1 的误差尽量的小
 export function tickStep(min, max, count) {
- const e10 = Math.sqrt(50); // 7.07
- const e5 = Math.sqrt(10); // 3.16
- const e2 = Math.sqrt(2); // 1.41
+ const e10 = Math.sqrt(50);
+ const e5 = Math.sqrt(10);
+ const e2 = Math.sqrt(2);
 
- // 获得目标间隔 step0，设 step0 = 10 ^ m
  const step0 = Math.abs(max - min) / Math.max(0, count);
- // 获得 step1 的初始值 = 10 ^ n < step0，其中 n 为满足条件的最大整数
  let step1 = 10 ** Math.floor(Math.log(step0) / Math.LN10);
- // 计算 step1 和 step0 的误差，error = 10 ^ m / 10 ^ n = 10 ^ (m - n)
  const error = step0 / step1;
- // 根据当前的误差改变 step1 的值，从而减少误差
- // 1. 当 m - n >= 0.85 = log(e10) 的时候，step1 * 10
- // 可以减少log(10) = 1 的误差
  if (error >= e10) step1 *= 10;
- // 2. 当 0.85 > m - n >= 0.5 = log(e5) 的时候，step1 * 5
- // 可以减少 log(5) = 0.7 的误差
  else if (error >= e5) step1 *= 5;
- // 3. 当 0.5 > m - n >= 0.15 = log(e2) 的时候，step1 * 2
- // 那么可以减少 log(2) = 0.3 的误差
  else if (error >= e2) step1 *= 2;
- // 4. 当 0.15 > m - n > 0 的时候，step1 * 1
  return step1;
 }
 
 export function ticks(min, max, count) {
  const step = tickStep(min, max, count);
- // 让 start 和 stop 都是 step 的整数倍
- // 这样生成的 ticks 都是 step 的整数倍
- // 可以让可读性更强
  const start = Math.ceil(min / step);
  const stop = Math.floor(max / step);
  const n = Math.ceil(stop - start + 1);
- // n 不一定等于 count，所以生成的 ticks 的数量可能和指定的不一样
  const values = new Array(n);
  for (let i = 0; i < n; i += 1) {
  values[i] = round((start + i) * step);
  }
  return values;
 }
 
-// 简单解决 js 的精读问题：0.1 + 0.2 !== 0.3
+// 简单解决 js 的精度问题：0.1 + 0.2 !== 0.3
 export function round(n) {
  return Math.round(n * 1e12) / 1e12;
 }