blocks|key|454072|text|Tupple上没有展平。但是如果你知道它的结构，你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|454073|implicit+def+flatten1[A,+B,+C](t:+((A,+B),+C)):+(A,+B,+C)+=+(t._1._1,+t._1._2,+t._2)
implicit+def+flatten2[A,+B,+C](t:+(A,+(B,+C))):+(A,+B,+C)+=+(t._1,+t._2._1,+t._2._2)|code-block|syntax|javascript|454074|这将展平任何类型的Tupple。还可以将implicit关键字添加到定义中。这只适用于三个元素。可以展平Tupple，如下所示：|454075|(1,+("hello",+42.0))+++=>+(1,+"hello",+42.0)
(("test",+3.7f),+"hi")+=>+("test",+3.7f,+"hi")|454076|多个嵌套的Tupple不能展平到地面，因为在返回类型中只有三个元素：|454077|((1,+(2,+3)),4)++++++++=>+(1,+(2,+3),+4)|454078|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

There is no flatten on a Tupple. But if you know the structure, you can do something like this:

<pre><code>implicit def flatten1[A, B, C](t: ((A, B), C)): (A, B, C) = (t._1._1, t._1._2, t._2)
implicit def flatten2[A, B, C](t: (A, (B, C))): (A, B, C) = (t._1, t._2._1, t._2._2)
</code></pre>

This will flatten Tupple with any types. You can also add the implicit keyword to the definition. This works only for three elements. You can flatten Tupple like:

<pre><code>(1, ("hello", 42.0)) =&gt; (1, "hello", 42.0)
(("test", 3.7f), "hi") =&gt; ("test", 3.7f, "hi")
</code></pre>

Multiple nested Tupple cannot be flatten to the ground, because there are only three elements in the return type:

<pre><code>((1, (2, 3)),4) =&gt; (1, (2, 3), 4)
</code></pre>

blocks|key|4283257|text|如果你使用shapeless，我想this正是你所需要的。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4283258|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/milessabin/shapeless|1|https://github.com/milessabin/shapeless/blob/master/examples/src/main/scala/shapeless/examples/flatten.scala^0|5|9|0|H|4|1|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]|$A|R|B|S|1|T]]|C|$]]|$1|D|3|-4|5|6|7|U|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]]]

If you use <a href="https://github.com/milessabin/shapeless" rel="noreferrer">shapeless</a>, <a href="https://github.com/milessabin/shapeless/blob/master/examples/src/main/scala/shapeless/examples/flatten.scala" rel="noreferrer">this</a> is exactly what you need, I think.

blocks|key|3802784|text|不确定这样做的效率，但您可以使用tuple.productIterator.toList将Tuple转换为List，然后flatten嵌套列表：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3802785|scala>+val+tuple+=+("top",+("nested",+42.0))
tuple:+(String,+(String,+Double))+=+(top,(nested,42.0))

scala>+tuple.productIterator.map({
+++++%7C+++case+(item:+Product)+=>+item.productIterator.toList
+++++%7C+++case+(item:+Any)+=>+List(item)
+++++%7C+}).toList.flatten
res0:+List[Any]+=+List(top,+nested,+42.0)|code-block|syntax|javascript|3802786|entityMap^0|G|S|19|5|1H|4|1O|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

Not sure about the effiency of this, but you can convert <code>Tuple</code> to <code>List</code> with <code>tuple.productIterator.toList</code>, then <code>flatten</code> the nested lists:

<pre><code>scala&gt; val tuple = ("top", ("nested", 42.0))
tuple: (String, (String, Double)) = (top,(nested,42.0))

scala&gt; tuple.productIterator.map({
 | case (item: Product) =&gt; item.productIterator.toList
 | case (item: Any) =&gt; List(item)
 | }).toList.flatten
res0: List[Any] = List(top, nested, 42.0)
</code></pre>

blocks|key|3802897|text|在我看来，简单的模式匹配是可行的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3802898|scala>+val+motto+=+(("dog",+"food"),+"tastes+good")
val+motto:+((String,+String),+String)+=+((dog,food),tastes+good)

scala>+motto+match+{
+++++%7C+case+((it,+really),+does)+=>+(it,+really,+does)
+++++%7C+}
val+res0:+(String,+String,+String)+=+(dog,food,tastes+good)|code-block|syntax|javascript|3802899|或者，如果你有一个这样的元组集合：|3802900|scala>+val+motto+=+List(
+++++%7C+(("dog",+"food"),+"tastes+good"))+:%2B+(("cat",+"food"),+"tastes+bad")
val+motto:+List[((String,+String),+String)]+=+List(((dog,food),tastes+good),+((cat,food),tastes+bad))

scala>+motto.map+{
+++++%7C+case+((one,+two),+three)+=>+(one,+two,+three)
+++++%7C+}
val+res2:+List[(String,+String,+String)]+=+List((dog,food,tastes+good),+(cat,food,tastes+bad))|3802901|我想即使你有几个箱子也会很方便。|3802902|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

In my opinion simple pattern matching would work
<pre><code>scala&gt; val motto = ((&quot;dog&quot;, &quot;food&quot;), &quot;tastes good&quot;)
val motto: ((String, String), String) = ((dog,food),tastes good)

scala&gt; motto match {
 | case ((it, really), does) =&gt; (it, really, does)
 | }
val res0: (String, String, String) = (dog,food,tastes good)
</code></pre>
Or if you have a collection of such tuples:
<pre><code>scala&gt; val motto = List(
 | ((&quot;dog&quot;, &quot;food&quot;), &quot;tastes good&quot;)) :+ ((&quot;cat&quot;, &quot;food&quot;), &quot;tastes bad&quot;)
val motto: List[((String, String), String)] = List(((dog,food),tastes good), ((cat,food),tastes bad))

scala&gt; motto.map {
 | case ((one, two), three) =&gt; (one, two, three)
 | }
val res2: List[(String, String, String)] = List((dog,food,tastes good), (cat,food,tastes bad))
</code></pre>
I think it would be convenient even if you have several cases.

blocks|key|3803032|text|的补充|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3803033|粘贴此实用程序代码：|offset|length|style|BOLD|3803034|import+shapeless._
import+ops.tuple.FlatMapper
import+syntax.std.tuple._
trait+LowPriorityFlatten+extends+Poly1+{
++implicit+def+default[T]+=+at[T](Tuple1(_))
}
object+flatten+extends+LowPriorityFlatten+{
++implicit+def+caseTuple[P+<:+Product](implicit+lfm:+Lazy[FlatMapper[P,+flatten.type]])+=
++++at[P](lfm.value(_))
}|code-block|syntax|javascript|3803035|然后，您可以展平任何嵌套的元组：|3803036|scala>+val+a+=+flatten(((1,2),((3,4),(5,(6,(7,8))))))
a:+(Int,+Int,+Int,+Int,+Int,+Int,+Int,+Int)+=+(1,2,3,4,5,6,7,8)|3803037|3803038|3803039|请注意，此解决方案不适用于自定义的case+class类型，该类型将在输出中转换为String。|blockquote|CODE|3803040|3803041|scala>+val+b=flatten(猫(“c”)，狗(“d”))，猫(“c”))+b：(String，String，String)+=+(c，d，c)|3803042|entityMap^0|0|0|A|0|0|0|0|0|0|H|A|15|6|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|12|8|@$D|13|E|14|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|15|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|16|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|17|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|19|8|@]|9|@]|A|$]]|$1|S|3|T|5|U|7|1A|8|@$D|1B|E|1C|F|V]|$D|1D|E|1E|F|V]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1G|8|@]|9|@]|A|$]]|$1|Z|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|10|$]]

Complement of <a href="https://stackoverflow.com/a/13699521/8932910">answer above</a>
Paste this utility code:
<pre><code>import shapeless._
import ops.tuple.FlatMapper
import syntax.std.tuple._
trait LowPriorityFlatten extends Poly1 {
 implicit def default[T] = at[T](Tuple1(_))
}
object flatten extends LowPriorityFlatten {
 implicit def caseTuple[P &lt;: Product](implicit lfm: Lazy[FlatMapper[P, flatten.type]]) =
 at[P](lfm.value(_))
}
</code></pre>
then you are able to flatten any nested tuple:
<pre><code>scala&gt; val a = flatten(((1,2),((3,4),(5,(6,(7,8))))))
a: (Int, Int, Int, Int, Int, Int, Int, Int) = (1,2,3,4,5,6,7,8)
</code></pre>
<hr />
<blockquote>
Note that this solution does not work for self-defined <code>case class</code> type, which would be converted to <code>String</code> in the output.
<pre><code>scala&gt; val b = flatten(((Cat(&quot;c&quot;), Dog(&quot;d&quot;)), Cat(&quot;c&quot;)))
b: (String, String, String) = (c,d,c)
</code></pre>
</blockquote>

I have a nested tuple structure like <code>(String,(String,Double))</code> and I want to transform it to <code>(String,String,Double)</code>. I have various kinds of nested tuple, and I don't want to transform each manually. Is there any convenient way to do that?

How to flatten a nested tuple?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个像(String,(String,Double))这样的嵌套元组结构，我想把它转换成(String,String,Double)。我有各种各样的嵌套元组，我不想手动转换每个元组。有什么方便的方法可以做到这一点？

问如何展平嵌套的元组？
EN

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问如何展平嵌套的元组？EN