小数到粗俗分数，Kotlin Android

Question

我一直在尝试将我的 Swift 方法转换为 Kotlin，但在这里没有得到任何帮助： Decimal to vulgar fraction conversion method- need help converting from Swift to Kotlin

我想我会稍微改变一下我的问题，看看是否能得到一些帮助。我需要有关代码反向 engineering/explanation 的帮助，以便我可以自己进行转换。反正我会通过这种方式学到更多。

如果有兴趣，完整的 code/story 链接在上面。我几乎想要付钱给自由职业者来帮助解决这个问题。请帮忙：）此外，您可能需要知道有一个数组，其中包含此方法舍入到的小数英寸的上标和下标输出。需要解释以下几行：

val whole = number.toInt() // Of course i know what this does :)
val sign = if (whole < 0) -1 else 1 // I converted this myself and believe this is correct Kotlin
val fraction = number - whole.toDouble() // And I know what this does

// need explanation from here down

for (i in 1..fractions.count()) {
    if (abs(fraction) > (fractions[i].1 + fractions[i - 1].1) / 2) {
        if ((fractions[i - 1].1) == (1.0)) run {
            return@run Pair("${whole + sign}", (whole + sign).toDouble())
        } else {
            return ("$whole" $fractions[i - 1].0, ${whole.toDouble() + (sign.toDouble() * fractions[i - 1].1))
        }
    }
}

Answer 1

一般来说，我倾向于把这样的计算函数写成kotlin的扩展function/property，因为这样会增加可用性。

NumberExt.kt

package ***

import kotlin.math.abs

/**
 * @author aminography
 */

val Double.vulgarFraction: Pair<String, Double>
    get() {
        val whole = toInt()
        val sign = if (whole < 0) -1 else 1
        val fraction = this - whole

        for (i in 1 until fractions.size) {
            if (abs(fraction) > (fractionValues[i] + fractionValues[i - 1]) / 2) {
                return if (fractionValues[i - 1] == 1.0) {
                    "${whole + sign}" to (whole + sign).toDouble()
                } else {
                    "$whole ${fractions[i - 1]}" to whole + sign * fractionValues[i - 1]
                }
            }
        }
        return "$whole" to whole.toDouble()
    }

val Float.vulgarFraction: Pair<String, Double>
    get() = toDouble().vulgarFraction

private val fractions = arrayOf(
    "",                           // 16/16
    "\u00B9\u2075/\u2081\u2086",  // 15/16
    "\u215E",                     // 7/8
    "\u00B9\u00B3/\u2081\u2086",  // 13/16
    "\u00BE",                     // 3/4
    "\u00B9\u00B9/\u2081\u2086",  // 11/16
    "\u215D",                     // 5/8
    "\u2079/\u2081\u2086",        // 9/16
    "\u00BD",                     // 1/2
    "\u2077/\u2081\u2086",        // 7/16
    "\u215C",                     // 3/8
    "\u2075/\u2081\u2086",        // 5/16
    "\u00BC",                     // 1/4
    "\u00B3/\u2081\u2086",        // 3/16
    "\u215B",                     // 1/8
    "\u00B9/\u2081\u2086",        // 1/16
    ""                            // 0/16
)

private val fractionValues = arrayOf(
    1.0, 
    15.0 / 16, 7.0 / 8, 13.0 / 16, 3.0 / 4, 11.0 / 16,
    5.0 / 8, 9.0 / 16, 1.0 / 2, 7.0 / 16, 3.0 / 8,
    5.0 / 16, 1.0 / 4, 3.0 / 16, 1.0 / 8, 1.0 / 16,
    0.0
)

测试

val rand = java.util.Random()
repeat(10) {
    val sign = if (rand.nextBoolean()) 1 else -1
    val number = rand.nextDouble() * rand.nextInt(100) * sign
    val vulgar = number.vulgarFraction
    println("Number: $number , Vulgar: ${vulgar.first} , Rounded: ${vulgar.second}")
}

输出：

Number: 17.88674468660217 , Vulgar: 17 ⅞ , Rounded: 17.875
Number: -56.98489542592821 , Vulgar: -57 , Rounded: -57.0
Number: 39.275953137210614 , Vulgar: 39 ¼ , Rounded: 39.25
Number: 13.422939071442359 , Vulgar: 13 ⁷/₁₆ , Rounded: 13.4375
Number: -56.70735924226373 , Vulgar: -56 ¹¹/₁₆ , Rounded: -56.6875
Number: 22.657555818202972 , Vulgar: 22 ¹¹/₁₆ , Rounded: 22.6875
Number: 2.951680466645306 , Vulgar: 2 ¹⁵/₁₆ , Rounded: 2.9375
Number: -8.8311628631306 , Vulgar: -8 ¹³/₁₆ , Rounded: -8.8125
Number: 28.639946409572655 , Vulgar: 28 ⅝ , Rounded: 28.625
Number: -28.439447873884085 , Vulgar: -28 ⁷/₁₆ , Rounded: -28.4375

说明

整体逻辑解释的有点难，我会尽量讲清楚一点。请注意，在下面的代码片段中，为了简化，我将 fractionValues[i - 1] 替换为 fractionValue。

// First look at the 'fractions' array. It starts from 16/16=1 down to 0/16=0.
// So it covers all the possible 16 cases for dividing a number by 16. 
// Note that 16/16=1 and 0/16=0 are the same in terms of division residual.

for (i in 1 until fractions.size) {
    // Here, we are searching for the proper fraction that is the nearest one to the
    // actual division residual. 
    // So, '|fraction| > (fractionValues[i] + fractionValues[i - 1]) / 2' means
    // that the fraction is closer to the greater one in the 'fractionValues' array 
    // (i.e. 'fractionValues[i - 1]').
    // Consider that we always want to find the proper 'fractionValues[i - 1]' and not 
    // 'fractionValues[i]' (According to the index of the 'for' loop which starts 
    // from 1, and not 0).

    if (abs(fraction) > (fractionValues[i] + fractionValues[i - 1]) / 2) {

        val fractionValue = fractionValues[i - 1]
        // Here we've found the proper fraction value (i.e. 'fractionValue').

        return if (fractionValue == 1.0) {
            // 'fractionValue == 1.0' means that the actual division residual was greater 
            // than 15/16 but closer to 16/16=1. So the final value should be rounded to
            // the nearest integer. Consider that in this case, the nearest integer for a
            // positive number is one more and for a negative number, one less. Finally, 
            // the summation with 'sign' does it for us :)

            "${whole + sign}" to (whole + sign).toDouble()
        } else {
            // Here we have 'fractionValue < 1.0'. The only thing is to calculate the 
            // rounded value which is the sum of 'whole' and the discovered 'fractionValue'.
            // As the value could be negative, by multiplying the 'sign' to the 
            // 'fractionValue', we will be sure that the summation is always correct.

            "$whole $fractionValue" to whole + sign * fractionValue
        }
    }
}

// Finally, if we are not able to find a proper 'fractionValue' for the input number, 
// it means the number had an integer value.
return "$whole" to whole.toDouble()

小数到粗俗分数，Kotlin Android

Decimal to vulgar fraction, Kotlin Android

android

reverse-engineering

decimal

fractions

kotlin

NumberExt.kt

测试

说明