如何在Kotlin的十进制结束时删除额外的零？

Question

我正在创建一个函数，该功能将大量超过1,000，然后返回该圆形数字的字符串。例如，“ 2374293”将返回为“ 237万”
但是，我不希望在小数结尾处诸如“ 25.00”或“ 100.50”之类的额外零。

例如：

我想要什么：

 Input  ->  Output

"11000" -> "11k"
"11400" -> "11.4k"

我得到什么：

 Input  ->  Output

"11000" -> "11.00k"
"11400" -> "11.40k"

在需要时如何删除这些零和小数点（如果是整数）？

这是我目前的代码：

    private fun roundBigNumb(numb: Long): String {
    val newNumb = numb.toDouble()
    return when {
        numb in 1000..999994 -> {
            BigDecimal(newNumb/1000).setScale(2, RoundingMode.HALF_EVEN).toString()+"k"
        }
        numb in 999995..999999 -> {
            "999.99k"
        }
        numb in 1000000..999994999 -> {
            BigDecimal(newNumb/1000000).setScale(2, RoundingMode.HALF_EVEN).toString()+"m"
        }

        numb in 999995000..999999999 -> {
            "999.99m"
        }

        numb in 1000000000..999994999999 -> {
            BigDecimal(newNumb/1000000000).setScale(2, RoundingMode.HALF_EVEN).toString()+"b"
        }

        numb in 999995000000..999999999999 -> {                    
            "999.99b"
        }

        numb in 1000000000000..999994999999999 -> {
            BigDecimal(newNumb/1000000000000).setScale(2, RoundingMode.HALF_EVEN).toString()+"t"
        }

        numb in 999995000000000..999999999999999 -> {
            "999.99t"
        }

        numb >= 1000000000000000 -> "∞"

        else -> numb.toString()
    }
}

Answer 1

一旦将数字转换为具有多达2个小数位置的字符串（如您所做的那样），就可以使用droplast删除尾随的零和小数位。

这是一个示例，

fun prettyFormat(input: Double): String {
    if( input == 0.0 ) return "0"
    
    val prefix = if( input < 0 ) "-" else ""
    val num = abs(input)

    // figure out what group of suffixes we are in and scale the number
    val pow = floor(log10(num)/3).roundToInt()
    val base = num / 10.0.pow(pow * 3)

    // Using consistent rounding behavior, always rounding down since you want
    // 999999999 to show as 999.99M and not 1B
    val roundedDown = floor(base*100)/100.0

    // Convert the number to a string with up to 2 decimal places
    var baseStr = BigDecimal(roundedDown).setScale(2, RoundingMode.HALF_EVEN).toString()

    // Drop trailing zeros, then drop any trailing '.' if present
    baseStr = baseStr.dropLastWhile { it == '0' }.dropLastWhile { it == '.' }

    val suffixes = listOf("","k","M","B","T")

    return when {
        pow < suffixes.size -> "$prefix$baseStr${suffixes[pow]}"
        else -> "${prefix}infty"
    }
}

如果

11411.0   = 11.41k
11000.0   = 11k
9.99996E8 = 999.99M
12.4      = 12.4
0.0       = 0
-11400.0  = -11.4k

您不关心零或负数，则可以简化一点。

Answer 2

import java.math.BigDecimal
import java.math.RoundingMode.HALF_EVEN

fun roundBigNumber(number: Long): String {
  fun calc(divisor: Long) = BigDecimal(number.toDouble() / divisor)
    .setScale(2, HALF_EVEN)
    .toString()
    .dropLastWhile { it == '0' }
    .dropLastWhile { it == '.' }
  return when {
    number in 1000..999994                     -> calc(1000) + "k"
    number in 999995..999999                   -> "999.99k"
    number in 1000000..999994999               -> calc(1000000) + "m"
    number in 999995000..999999999             -> "999.99m"
    number in 1000000000..999994999999         -> calc(1000000000) + "b"
    number in 999995000000..999999999999       -> "999.99b"
    number in 1000000000000..999994999999999   -> calc(1000000000000) + "t"
    number in 999995000000000..999999999999999 -> "999.99t"
    number >= 1000000000000000                 -> "∞"
    else                                       -> number.toString()
  }
}