Question

我有以下元组：(1,"3idiots",List("Action","Adventure","Horror")我需要将其转换为以下格式的列表：

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

Answer 1

要添加到之前的答案，您还可以在这种情况下使用 for-comprehension ;它可能会使事情更清楚恕我直言：

WITH updated as(
    UPDATE mytable SET status = 'A'
FROM
  (
    SELECT id FROM mytable
    WHERE status = 'B'
    ORDER BY mycolumn
    LIMIT 100
    FOR UPDATE
  ) sub
  LEFT JOIN jointable j USING (id)
WHERE mytable.id = sub.id
GROUP BY (mytable.id)
RETURNING mytable.id, array_agg(j.value)
)
select *
from updated
ORDER BY mycolumn

所以如果你有：

        var s:String = '364219431799224845593594391459942853555190499729152736515819';
        for (var i:int = 0; i < s.length; i += 2) {
            var sub:String = s.charAt(i) + s.charAt(i + 1);
            trace(sub);//convert 
        }

你会得到：

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

Answer 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"))
)

你想要这样的结果：

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)
    )
)

或更短：tuples.flatMap(t => t._3.map((t._1, t._2, _)))

Answer 3

这应该做你想要的：

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))

Answer 4

你可以使用它。

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

将列表转换为scala中的多个元组

4 个答案: