用国际单位制将数字格式化为三个sig图

Question

我想用国际单位制将一个整数格式化为3位有效数字。例如：

1     => '1'
999   => '999'
1234  => '1.23k'
1235  => '1.24k'

下面是我对一个函数的粗略修改，仍然不能满足我的所有测试。我遇到的问题涉及检测何时9…将四舍五入到10…，因此需要在末尾少一个小数点。

tests = [[9,'9'],[1,'1'],[10,'10'],[99,'99'],[100,'100'],[999,'999'],
        [1000,'1.00k'],[1004,'1.00k'],[1009,'1.01k'],[1472,'1.47k'],
        [1926,'1.93k'],[1999,'2.00k'],[2000,'2.00k'],[9813,'9.81k'],
        [9990,'9.99k'],[9999,'10.0k'],[10000,'10.0k'],[10010,'10.0k'],
        [60712,'60.7k'],[98712,'98.7k'],[99949,'99.9k'],[99950,'100k'],
        [99999,'100k'],[100000,'100k'],[400499,'400k'],[999499,'999k'],
        [999500,'1.00M'],[999999,'1.00M'],[1000000,'1.00M'],
        [1234567,'1.23M'],[12345678,'12.3M'],[123456789,'123M']]

tests.forEach( ([n,expected]) => {
    const actual = siRound(n)
    console.log(n, actual, actual==expected ? '=' : '≠', expected)
})

function siRound(x) {
    if (x<1e3) return x+'';
    const digits = Math.log10(x) | 0
    const tier = digits/3 | 0
    let str = (x / 10**(tier*3)).toFixed(2-(digits%3))
    // Turn "10.00" into "10.0" and "100.0" into "100"
    str = str.replace(/^(.{3})\..+|^(.{4}).+/, '$1$2')
    return str + (['','k','M','G','T'])[tier]
}

Answer 1

基于@HereticMonkey指出存在toPrecision（），这里有一个可行的答案，尽管我觉得有点恶心。如果有人发布了一个答案，我会把它换成一个更干净、完全有效的答案。

// Note: intentionally returns integers below 1,000 unchanged.
// For true precision across all numbers, remove the first line.
function siRound(n, precision=3) {
    if (x<1e3) return x+''
    let tier = Math.log10(x)/3 | 0
    let str = (x / 10**(tier*3)).toPrecision(precision)
    if (str>=1e3) str = (x / 10**(++tier*3)).toPrecision(precision)
    return str + (['','k','M','G','T','P','E','Z','Y'])[tier]
}

Answer 2

tests = [[9,'9'],[1,'1'],[10,'10'],[99,'99'],[100,'100'],[999,'999'],
        [1000,'1.00k'],[1004,'1.00k'],[1009,'1.01k'],[1472,'1.47k'],
        [1926,'1.93k'],[1999,'2.00k'],[2000,'2.00k'],[9813,'9.81k'],
        [9990,'9.99k'],[9999,'10.0k'],[10000,'10.0k'],[10010,'10.0k'],
        [60712,'60.7k'],[98712,'98.7k'],[99949,'99.9k'],[99950,'100k'],
        [99999,'100k'],[100000,'100k'],[400499,'400k'],[999499,'999k'],
        [999500,'1.00M'],[999999,'1.00M'],[1000000,'1.00M'],
        [1234567,'1.23M'],[12345678,'12.3M'],[123456789,'123M']]

const siRound = x => {
  if (x < 1e3) {
    return x;
  }
  let roudedX = x;
  let tier = 0;
  while (roudedX >= 1000) {
    roudedX /= 1000;
    tier += 1;
  }
  roudedX = Number.parseFloat(roudedX.toPrecision(3));
  if (Math.abs(roudedX - 1e3) < Number.EPSILON) {
    roudedX = 1;
    tier += 1;
  }
  return roudedX.toPrecision(3) + ['', 'k', 'M', 'G', 'T'][tier];
}

tests.forEach(([n,expected]) => {
  const actual = siRound(n)
  console.log(n, actual, actual==expected ? '=' : '≠', expected)
})

Answer 3

让你的方法像递归一样。添加此项，返回前检查：

if (str.length>4) { 
      siRound(str)
}

tests = [[9,'9'],[1,'1'],[10,'10'],[99,'99'],[100,'100'],[999,'999'],
        [1000,'1.00k'],[1004,'1.00k'],[1009,'1.01k'],[1472,'1.47k'],
        [1926,'1.93k'],[1999,'2.00k'],[2000,'2.00k'],[9813,'9.81k'],
        [9990,'9.99k'],[9999,'10.0k'],[10000,'10.0k'],[10010,'10.0k'],
        [60712,'60.7k'],[98712,'98.7k'],[99949,'99.9k'],[99950,'100k'],
        [99999,'100k'],[100000,'100k'],[400499,'400k'],[999499,'999k'],
        [999500,'1.00M'],[999999,'1.00M'],[1000000,'1.00M'],
        [1234567,'1.23M'],[12345678,'12.3M'],[123456789,'123M']]

tests.forEach( ([n,expected]) => {
    const actual = siRound(n)
    console.log(n, actual, actual==expected ? '=' : '≠', expected)
})

function siRound(x) {
    if (x<1e3) return x+'';
    const digits = Math.log10(x) | 0
    const tier = digits/3 | 0
    let str = (x / 10**(tier*3)).toFixed(2-(digits%3))
    // Turn "10.00" into "10.0" and "100.0" into "100"
    str = str.replace(/^(.{3})\..+|^(.{4}).+/, '$1$2')
    if (str.length>4) { ///Add this check
      siRound(str)
    }
    return str + (['','k','M','G','T'])[tier]
}