在scala中将列表转换为多个元组

我有以下元组：1,3白痴、ListAction、冒险、恐怖，我需要将它们转换为以下格式的列表：

List(
  (1,"3idiots","Action"),
  (1,"3idiots","Adventure")
)

---

假设您有多个元组，如下所示：

val tuples = List(
    (1, "3idiots", List("Action", "Adventure", "Horror")),
    (2, "foobar", List("Foo", "Bar"))
)

List(
    (1, "3idiots", "Action"),
    (1, "3idiots" , "Adventure"),
    (1, "3idiots", "Horror"),
    (2, "foobar", "Foo"),
    (2, "foobar", "Bar")
)

你想要这样的结果：

val tuples = List(
    (1, "3idiots", List("Action", "Adventure", "Horror")),
    (2, "foobar", List("Foo", "Bar"))
)

List(
    (1, "3idiots", "Action"),
    (1, "3idiots" , "Adventure"),
    (1, "3idiots", "Horror"),
    (2, "foobar", "Foo"),
    (2, "foobar", "Bar")
)

您的解决方案是使用flatMap，它可以将列表列表转换为单个列表：

tuples.flatMap(t =>
    t._3.map(s =>
        (t._1, t._2, s)
    )
)

---

或者更短：tuples.flatmap=>t._3.mapt._1，t._2，

假设您有多个这样的元组：

val tuples = List(
    (1, "3idiots", List("Action", "Adventure", "Horror")),
    (2, "foobar", List("Foo", "Bar"))
)

List(
    (1, "3idiots", "Action"),
    (1, "3idiots" , "Adventure"),
    (1, "3idiots", "Horror"),
    (2, "foobar", "Foo"),
    (2, "foobar", "Bar")
)

你想要这样的结果：

val tuples = List(
    (1, "3idiots", List("Action", "Adventure", "Horror")),
    (2, "foobar", List("Foo", "Bar"))
)

List(
    (1, "3idiots", "Action"),
    (1, "3idiots" , "Adventure"),
    (1, "3idiots", "Horror"),
    (2, "foobar", "Foo"),
    (2, "foobar", "Bar")
)

您的解决方案是使用flatMap，它可以将列表列表转换为单个列表：

tuples.flatMap(t =>
    t._3.map(s =>
        (t._1, t._2, s)
    )
)

---

或者更短：tuples.flatMapt=>t._3.mapt._1，t._2，u

这应该满足您的需要：

val input = (1,"3idiots",List("Action","Adventure","Horror"))    
val result = input._3.map(x => (input._1,input._2,x))
// gives List((1,3idiots,Action), (1,3idiots,Adventure), (1,3idiots,Horror))

---

这应该满足您的要求：

val input = (1,"3idiots",List("Action","Adventure","Horror"))    
val result = input._3.map(x => (input._1,input._2,x))
// gives List((1,3idiots,Action), (1,3idiots,Adventure), (1,3idiots,Horror))

---

若要添加到先前的答案中，您也可以在本例中使用以进行理解；这可能会让事情变得更清楚：

for(
     (a,b,l) <- ts; 
     s <- l
) yield (a,b,s)

您将获得：

List(
      (a,1,foo), 
      (a,1,bar), 
      (a,1,baz),
      (b,2,foo1), 
      (b,2,bar1), 
      (b,2,baz1)
 )

---

若要添加到先前的答案中，您也可以在本例中使用以进行理解；这可能会让事情变得更清楚：

for(
     (a,b,l) <- ts; 
     s <- l
) yield (a,b,s)

您将获得：

List(
      (a,1,foo), 
      (a,1,bar), 
      (a,1,baz),
      (b,2,foo1), 
      (b,2,bar1), 
      (b,2,baz1)
 )

你可以用这个

val question = (1,"3idiots",List("Action","Adventure","Horror"))
val result = question._3.map(x=> (question._1 , question._2 ,x))

你可以用这个

val question = (1,"3idiots",List("Action","Adventure","Horror"))
val result = question._3.map(x=> (question._1 , question._2 ,x))

---

请添加代码？那么您尝试了什么？你被困在哪里？你认为哪些功能有用？您已经阅读了文档，对吗？提示：FlatMap和恐怖如何？请添加代码？那么您尝试了什么？你被困在哪里？你认为哪些功能有用？您已经阅读了文档，对吗？提示：那恐怖呢？