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<li><em>Go语言圣经读书笔记(挖坑中...): <a href="https://github.com/chai2010/gopl-notes-zh">https://github.com/chai2010/gopl-notes-zh</a></em></li>
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<hr>
<h2 id="76-sortinterface接口"><a class="header" href="#76-sortinterface接口">7.6. sort.Interface接口</a></h2>
<p>排序操作和字符串格式化一样是很多程序经常使用的操作。尽管一个最短的快排程序只要15行就可以搞定，但是一个健壮的实现需要更多的代码，并且我们不希望每次我们需要的时候都重写或者拷贝这些代码。</p>
<p>幸运的是，sort包内置的提供了根据一些排序函数来对任何序列排序的功能。它的设计非常独到。在很多语言中，排序算法都是和序列数据类型关联，同时排序函数和具体类型元素关联。相比之下，Go语言的sort.Sort函数不会对具体的序列和它的元素做任何假设。相反，它使用了一个接口类型sort.Interface来指定通用的排序算法和可能被排序到的序列类型之间的约定。这个接口的实现由序列的具体表示和它希望排序的元素决定，序列的表示经常是一个切片。</p>
<p>一个内置的排序算法需要知道三个东西：序列的长度，表示两个元素比较的结果，一种交换两个元素的方式；这就是sort.Interface的三个方法：</p>
<pre><code class="language-go">package sort
type Interface interface {
Len() int
Less(i, j int) bool // i, j are indices of sequence elements
Swap(i, j int)
}
</code></pre>
<p>为了对序列进行排序，我们需要定义一个实现了这三个方法的类型，然后对这个类型的一个实例应用sort.Sort函数。思考对一个字符串切片进行排序，这可能是最简单的例子了。下面是这个新的类型StringSlice和它的Len，Less和Swap方法</p>
<pre><code class="language-go">type StringSlice []string
func (p StringSlice) Len() int { return len(p) }
func (p StringSlice) Less(i, j int) bool { return p[i] &lt; p[j] }
func (p StringSlice) Swap(i, j int) { p[i], p[j] = p[j], p[i] }
</code></pre>
<p>现在我们可以通过像下面这样将一个切片转换为一个StringSlice类型来进行排序：</p>
<pre><code class="language-go">sort.Sort(StringSlice(names))
</code></pre>
<p>这个转换得到一个相同长度，容量，和基于names数组的切片值；并且这个切片值的类型有三个排序需要的方法。</p>
<p>对字符串切片的排序是很常用的需要，所以sort包提供了StringSlice类型，也提供了Strings函数能让上面这些调用简化成sort.Strings(names)。</p>
<p>这里用到的技术很容易适用到其它排序序列中，例如我们可以忽略大小写或者含有的特殊字符。（本书使用Go程序对索引词和页码进行排序也用到了这个技术，对罗马数字做了额外逻辑处理。）对于更复杂的排序，我们使用相同的方法，但是会用更复杂的数据结构和更复杂地实现sort.Interface的方法。</p>
<p>我们会运行上面的例子来对一个表格中的音乐播放列表进行排序。每个track都是单独的一行，每一列都是这个track的属性像艺术家，标题，和运行时间。想象一个图形用户界面来呈现这个表格，并且点击一个属性的顶部会使这个列表按照这个属性进行排序；再一次点击相同属性的顶部会进行逆向排序。让我们看下每个点击会发生什么响应。</p>
<p>下面的变量tracks包含了一个播放列表。（One of the authors apologizes for the other author’s musical tastes.）每个元素都不是Track本身而是指向它的指针。尽管我们在下面的代码中直接存储Tracks也可以工作，sort函数会交换很多对元素，所以如果每个元素都是指针而不是Track类型会更快，指针是一个机器字码长度而Track类型可能是八个或更多。</p>
<p><u><i>gopl.io/ch7/sorting</i></u></p>
<pre><code class="language-go">type Track struct {
Title string
Artist string
Album string
Year int
Length time.Duration
}
var tracks = []*Track{
{&quot;Go&quot;, &quot;Delilah&quot;, &quot;From the Roots Up&quot;, 2012, length(&quot;3m38s&quot;)},
{&quot;Go&quot;, &quot;Moby&quot;, &quot;Moby&quot;, 1992, length(&quot;3m37s&quot;)},
{&quot;Go Ahead&quot;, &quot;Alicia Keys&quot;, &quot;As I Am&quot;, 2007, length(&quot;4m36s&quot;)},
{&quot;Ready 2 Go&quot;, &quot;Martin Solveig&quot;, &quot;Smash&quot;, 2011, length(&quot;4m24s&quot;)},
}
func length(s string) time.Duration {
d, err := time.ParseDuration(s)
if err != nil {
panic(s)
}
return d
}
</code></pre>
<p>printTracks函数将播放列表打印成一个表格。一个图形化的展示可能会更好点，但是这个小程序使用text/tabwriter包来生成一个列整齐对齐和隔开的表格，像下面展示的这样。注意到<code>*tabwriter.Writer</code>是满足io.Writer接口的。它会收集每一片写向它的数据；它的Flush方法会格式化整个表格并且将它写向os.Stdout（标准输出）。</p>
<pre><code class="language-go">func printTracks(tracks []*Track) {
const format = &quot;%v\t%v\t%v\t%v\t%v\t\n&quot;
tw := new(tabwriter.Writer).Init(os.Stdout, 0, 8, 2, ' ', 0)
fmt.Fprintf(tw, format, &quot;Title&quot;, &quot;Artist&quot;, &quot;Album&quot;, &quot;Year&quot;, &quot;Length&quot;)
fmt.Fprintf(tw, format, &quot;-----&quot;, &quot;------&quot;, &quot;-----&quot;, &quot;----&quot;, &quot;------&quot;)
for _, t := range tracks {
fmt.Fprintf(tw, format, t.Title, t.Artist, t.Album, t.Year, t.Length)
}
tw.Flush() // calculate column widths and print table
}
</code></pre>
<p>为了能按照Artist字段对播放列表进行排序，我们会像对StringSlice那样定义一个新的带有必须的Len，Less和Swap方法的切片类型。</p>
<pre><code class="language-go">type byArtist []*Track
func (x byArtist) Len() int { return len(x) }
func (x byArtist) Less(i, j int) bool { return x[i].Artist &lt; x[j].Artist }
func (x byArtist) Swap(i, j int) { x[i], x[j] = x[j], x[i] }
</code></pre>
<p>为了调用通用的排序程序，我们必须先将tracks转换为新的byArtist类型，它定义了具体的排序：</p>
<pre><code class="language-go">sort.Sort(byArtist(tracks))
</code></pre>
<p>在按照artist对这个切片进行排序后，printTrack的输出如下</p>
<pre><code>Title Artist Album Year Length
----- ------ ----- ---- ------
Go Ahead Alicia Keys As I Am 2007 4m36s
Go Delilah From the Roots Up 2012 3m38s
Ready 2 Go Martin Solveig Smash 2011 4m24s
Go Moby Moby 1992 3m37s
</code></pre>
<p>如果用户第二次请求“按照artist排序”，我们会对tracks进行逆向排序。然而我们不需要定义一个有颠倒Less方法的新类型byReverseArtist，因为sort包中提供了Reverse函数将排序顺序转换成逆序。</p>
<pre><code class="language-go">sort.Sort(sort.Reverse(byArtist(tracks)))
</code></pre>
<p>在按照artist对这个切片进行逆向排序后，printTrack的输出如下</p>
<pre><code>Title Artist Album Year Length
----- ------ ----- ---- ------
Go Moby Moby 1992 3m37s
Ready 2 Go Martin Solveig Smash 2011 4m24s
Go Delilah From the Roots Up 2012 3m38s
Go Ahead Alicia Keys As I Am 2007 4m36s
</code></pre>
<p>sort.Reverse函数值得进行更近一步的学习，因为它使用了（§6.3）章中的组合，这是一个重要的思路。sort包定义了一个不公开的struct类型reverse，它嵌入了一个sort.Interface。reverse的Less方法调用了内嵌的sort.Interface值的Less方法，但是通过交换索引的方式使排序结果变成逆序。</p>
<pre><code class="language-go">package sort
type reverse struct{ Interface } // that is, sort.Interface
func (r reverse) Less(i, j int) bool { return r.Interface.Less(j, i) }
func Reverse(data Interface) Interface { return reverse{data} }
</code></pre>
<p>reverse的另外两个方法Len和Swap隐式地由原有内嵌的sort.Interface提供。因为reverse是一个不公开的类型，所以导出函数Reverse返回一个包含原有sort.Interface值的reverse类型实例。</p>
<p>为了可以按照不同的列进行排序，我们必须定义一个新的类型例如byYear：</p>
<pre><code class="language-go">type byYear []*Track
func (x byYear) Len() int { return len(x) }
func (x byYear) Less(i, j int) bool { return x[i].Year &lt; x[j].Year }
func (x byYear) Swap(i, j int) { x[i], x[j] = x[j], x[i] }
</code></pre>
<p>在使用sort.Sort(byYear(tracks))按照年对tracks进行排序后，printTrack展示了一个按时间先后顺序的列表：</p>
<pre><code>Title Artist Album Year Length
----- ------ ----- ---- ------
Go Moby Moby 1992 3m37s
Go Ahead Alicia Keys As I Am 2007 4m36s
Ready 2 Go Martin Solveig Smash 2011 4m24s
Go Delilah From the Roots Up 2012 3m38s
</code></pre>
<p>对于我们需要的每个切片元素类型和每个排序函数，我们需要定义一个新的sort.Interface实现。如你所见，Len和Swap方法对于所有的切片类型都有相同的定义。下个例子，具体的类型customSort会将一个切片和函数结合，使我们只需要写比较函数就可以定义一个新的排序。顺便说下，实现了sort.Interface的具体类型不一定是切片类型；customSort是一个结构体类型。</p>
<pre><code class="language-go">type customSort struct {
t []*Track
less func(x, y *Track) bool
}
func (x customSort) Len() int { return len(x.t) }
func (x customSort) Less(i, j int) bool { return x.less(x.t[i], x.t[j]) }
func (x customSort) Swap(i, j int) { x.t[i], x.t[j] = x.t[j], x.t[i] }
</code></pre>
<p>让我们定义一个多层的排序函数，它主要的排序键是标题，第二个键是年，第三个键是运行时间Length。下面是该排序的调用，其中这个排序使用了匿名排序函数：</p>
<pre><code class="language-go">sort.Sort(customSort{tracks, func(x, y *Track) bool {
if x.Title != y.Title {
return x.Title &lt; y.Title
}
if x.Year != y.Year {
return x.Year &lt; y.Year
}
if x.Length != y.Length {
return x.Length &lt; y.Length
}
return false
}})
</code></pre>
<p>这下面是排序的结果。注意到两个标题是“Go”的track按照标题排序是相同的顺序，但是在按照year排序上更久的那个track优先。</p>
<pre><code>Title Artist Album Year Length
----- ------ ----- ---- ------
Go Moby Moby 1992 3m37s
Go Delilah From the Roots Up 2012 3m38s
Go Ahead Alicia Keys As I Am 2007 4m36s
Ready 2 Go Martin Solveig Smash 2011 4m24s
</code></pre>
<p>尽管对长度为n的序列排序需要 O(n log n)次比较操作，检查一个序列是否已经有序至少需要n-1次比较。sort包中的IsSorted函数帮我们做这样的检查。像sort.Sort一样，它也使用sort.Interface对这个序列和它的排序函数进行抽象，但是它从不会调用Swap方法：这段代码示范了IntsAreSorted和Ints函数在IntSlice类型上的使用：</p>
<pre><code class="language-go">values := []int{3, 1, 4, 1}
fmt.Println(sort.IntsAreSorted(values)) // &quot;false&quot;
sort.Ints(values)
fmt.Println(values) // &quot;[1 1 3 4]&quot;
fmt.Println(sort.IntsAreSorted(values)) // &quot;true&quot;
sort.Sort(sort.Reverse(sort.IntSlice(values)))
fmt.Println(values) // &quot;[4 3 1 1]&quot;
fmt.Println(sort.IntsAreSorted(values)) // &quot;false&quot;
</code></pre>
<p>为了使用方便，sort包为[]int、[]string和[]float64的正常排序提供了特定版本的函数和类型。对于其他类型，例如[]int64或者[]uint，尽管路径也很简单，还是依赖我们自己实现。</p>
<p><strong>练习 7.8：</strong> 很多图形界面提供了一个有状态的多重排序表格插件：主要的排序键是最近一次点击过列头的列，第二个排序键是第二最近点击过列头的列，等等。定义一个sort.Interface的实现用在这样的表格中。比较这个实现方式和重复使用sort.Stable来排序的方式。</p>
<p><strong>练习 7.9：</strong> 使用html/template包（§4.6）替代printTracks将tracks展示成一个HTML表格。将这个解决方案用在前一个练习中，让每次点击一个列的头部产生一个HTTP请求来排序这个表格。</p>
<p><strong>练习 7.10：</strong> sort.Interface类型也可以适用在其它地方。编写一个IsPalindrome(s sort.Interface) bool函数表明序列s是否是回文序列，换句话说反向排序不会改变这个序列。假设如果!s.Less(i, j) &amp;&amp; !s.Less(j, i)则索引i和j上的元素相等。</p>
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