如果使用远程数据，则比其他解决方案更好 [英] Better solution than else if with ranged data

问题描述

我有一个简单的java方法，它根据从RGB转换的HSB值返回颜色。它工作（需要一些调整），但我使用一系列其他if和嵌套if语句来返回我想要的数据。我听说过HashMaps和String Factories更好，但我看不出它们如何处理远程数据。有没有更好的解决方案适用于这样的远程数据？

I have a simple java method that returns colors based on the HSB value converted from an RGB. It works (needs some tweaking), but I use a series of else if and nested if statements to return the data I want. I had heard that HashMaps and String Factories were better, but I couldn't see how these worked with ranged data. Is there a better solution that works with ranged data like this?

代码段：

public static String getColorName() {
    getHSB(rgb);
    if(hsbH >= 45 && hsbH < 75) {
        if(hsbS > 0 && hsbS < 45 && hsbB > 70){
            return "White/Off White";
        } else if(hsbS > 0 && hsbS < 45 && hsbB < 10) {
            return "Dark Yellow";
        } else {
            return "Yellow";
        }
    } else if(hsbH >= 15 && hsbH < 45) {
        if(hsbS > 0 && hsbS < 45 && hsbB > 70){
            return "White/Off White";
        } else if(hsbS > 0 && hsbS < 45 && hsbB < 10) {
            return "Dark Orange";
        } else {
            return "Orange";
        }
...

推荐答案

仔细看，代码中有很多重复和非常明显的结构！这就是我提出的，据我所知，大部分工作是在我最喜欢的IDE中使用自动重构完成的：

Look carefully, there is a lot of repetition and very obvious structure in your code! Here is what I came up with, as far as I remember most of the job was done using automatic refactorings in my favourite IDE:

public static String getColorName() {
    getHSB(rgb);
    if (hsbH < 15)
        return colorName(hsbB, hsbS, "Red");
    if (hsbH < 45)
        return colorName(hsbB, hsbS, "Orange");
    if (hsbH < 75)
        return colorName(hsbB, hsbS, "Yellow");
    //...
}

private static String colorName(int hsbB, int hsbS, String color) {
    final boolean smallSaturation = hsbS > 0 && hsbS < 45;
    if (smallSaturation) {
        if (hsbB > 70)
            return "White/Off White";
        if (hsbB < 10)
            return "Dark " + color;
    }
    return color;
}

如果您使用 Sean Patrick Floyd 的建议使用 TreeMap 这段代码会更简单（我可以帮助自己）：

If you use Sean Patrick Floyd's advice of using TreeMap this code will be even simpler (I could help myself):

public static String getColorName(final int hsbH, final int hsbB, final int hsbS) {
    NavigableMap<Integer, String> colorRanges = new TreeMap<Integer, String>();
    colorRanges.put(0, "Red");
    colorRanges.put(15, "Orange");
    colorRanges.put(75, "Yellow");
    //...
    return colorName(hsbB, hsbS, colorRanges.floorEntry(hsbH).getValue());
}

请注意 colorRanges 范围应定义一次并重复使用。

Note that colorRanges ranges should be defined once and reused.

冒险在这里投票是一种很好的方式，你可以使用Scala和简单的DSL：

Risking being downvoted here is a nice way you can write this literally using Scala and simple DSL:

implicit def toIntBetween(x: Int) = new {
    def between(left: Int) = new {
        def and(right: Int) = {
            x >= left && x < right
        }
    }
}

def getColorName = {
    if(hsbH between 45 and 75) {
        //...
    }
}

花式 if（hsbH之间） 45和75）构造实际上转换为：

Fancy if(hsbH between 45 and 75) construct actually translates to:

if(toIntBetween(hsbH).between(45).and(75))

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