大佬教程收集整理的这篇文章主要介绍了根据给定的最少计数四舍五入一个数字?,大佬教程大佬觉得挺不错的,现在分享给大家,也给大家做个参考。
我正在编写一个函数,根据给定的最少计数对数字进行四舍五入。
例如,
1.2345 with a least count of 0.01 should becoume 1.23
1.23 with a least count of 0.5 should become 1
1.52 with a least count of 0.2 should become 1.4
etc...
这是我的解决方案,但感觉有点做作,有人可以就如何改进它提出任何建议吗?跟进,可以使用正则表达式吗,这会是一个更稳定的实现吗?
const round = (num,leastCount) => {
const numdecimals = `${leastCount}`.split('.')?.[1]?.length || 0;
const newnumber = Math.round(num * Math.pow(10,numdecimals) + number.EPSILON - num * Math.pow(10,numdecimals) % (leastCount * Math.pow(10,numdecimals))) / Math.pow(10,numdecimals)
return newnumber
}
console.log(round(1.2345,0.01)) // Expected: 1.23
console.log(round(5.66,1)) // Expected: 5
console.log(round(2.375,0.005)) // Expected: 2.375
console.log(round(1.33,0.1)); // Expected: 1.3
console.log(round(2,0.001)); // Expected: 2
console.log(round(1.7512323,0.0001)); // Expected: 1.7512
console.log(round(1.52,0.2)); // Expected: 1.4
@H_489_19@
@H_489_19@
一个简单的递增 while 循环可能是最简单的解决方案。
增加 n
的乘数 leastCount
,直到结果乘积大于目标 num
:
const round = (num,leastCount) => {
if (num == 0) return 0;
let n = 0;
while (n * leastCount <= Math.abs(num)) {
n++;
}
return (n - 1) * leastCount * (num < 0 ? -1 : 1);
}
console.log(round(1.2345,0.01)) // Expected: 1.23
console.log(round(5.66,1)) // Expected: 5
console.log(round(2.375,0.005)) // Expected: 2.375
console.log(round(1.33,0.1)); // Expected: 1.3
console.log(round(2,0.001)); // Expected: 2
console.log(round(1.7512323,0.0001)); // Expected: 1.7512
console.log(round(1.52,0.2)); // Expected: 1.4
@H_489_19@
@H_489_19@
回复评论
奇怪的分数是由 JavaScript handles floating point values 方式引起的,可以用不同的方式处理,具体取决于您的实现和需求。
我在下面的示例中使用了 this solution,以及一种不同甚至更简单的数学方法:
const round = (num,leastCount) => {
if (num == 0) return 0;
let intermediate = Math.floor(num / leastCount) * leastCount; // Find value
return parseFloat(intermediate.toPrecision(12)); // Round of excessive decimals
}
console.log(round(1.2345,0.2)); // Expected: 1.4
@H_489_19@
@H_489_19@
@H_489_19@
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