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- 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); @@ -48,7 +28,7 @@ export function ticks(min, max, count) { 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; }