Question

我是scala的新手。我想为所有波纹管数据创建地图，包含PINCODE作为键，所有其他字段作为值。

<!DOCTYPE html>
<html>
<body>
<table border="1">
  <tr>
    <th>PINCODE</th>
    <th>Locality</th> 
    <th>PO_TYPE</th>
    <th>TALUK</th>
    <th>DISTRICT</th>
  </tr>
  <tr>
    <td>500001</td>
    <td>Hyderabad G.P.O.</td>
    <td>Branch Post Office</td>
    <td>Hyderabad</td>
    <td>HYDERABAD</td>
  </tr>
  <tr>
    <td>500001</td>
    <td>Gandhi Bhawan</td>
    <td>Branch Post Office</td>
    <td>Nampally</td>
    <td>HYDERABAD</td>
  </tr>
  <tr>
    <td>500001</td>
    <td>Hindi Bhawan</td>
    <td>Branch Post Office</td>
    <td>Nampally</td>
    <td>HYDERABAD</td>
  </tr>
  <tr>
    <td>500002</td>
    <td>Hyderabad Jubilee</td>
    <td>Branch Post Office</td>
    <td>HYDERABAD</td>
    <td>HYDERABAD</td>
  </tr>
  <tr>
    <td>500002</td>
    <td>Moghalpura Branch</td>
    <td>Post Office</td>
    <td>HYDERABAD</td>
    <td>HYDERABAD</td>
  </tr>
</table>

</body>
</html>

例如:(输出类似于bellow）

（500001，（Hyderabad G.P.O.，Branch Post Office，Hyderabad，HYDERABAD），（甘地巴哈，分支邮局，海南省，海得拉巴），（Hindi Bhawan，分支邮局，Nampally，海得拉巴））

（500002，（Hyderabad Jubilee，Branch Post Office，HYDERABAD，HYDERABAD），（海德拉巴，海得拉巴邮局Moghalpura分店））

提前致谢

Answer 1

实际上，Map应该uniquely map key到value。

因此，如果您想在同一个密钥上存储多个值（假设为String s），则可以将其设为Map[String, List[String]]。

val map: Map[String, List[String]] = Map(
  "1" -> List("val_1_1", "val_1_2", "val_1_3"),
  "2" -> List("val_2_1", "val_2_2")
)

但是......至于你的情况，看起来你正在尝试拥有＆＃34;价值观＆＃34;它们不是String，而是更像地址描述。

在这种情况下，为什么不为地址创建一个类？

case class Address(
  locality: String,
  poType: String,
  taluk: String,
  district: String
)

// Now you can have you map

val map: Map[String, List[Address]] = Map(
  "500001" -> List(Address("Hyderabad G.P.O.", "Branch Post Office", "Hyderabad", "HYDERABAD"))
)

// define a function that we will use to add addresses without over-writing

def updateMapByAddingAddressWithPincode(
  map: Map[String, List[String]],
  pincode: String,
  address: Address
) = {
  val existingAddressListForPincode = map.getOrElse(pincode, default = List.empty[Address])
  map + (pincode -> address :: existingAddressListForPincode)
}

// now lets say, you want to add another address with same pincode "500001"

val newAddress = Address("Gandhi Bhawan", "Branch Post Office", "Nampally", "HYDERABAD")

val updatedMap = updateMapByAddingAddressWithPincode(
  map,
  "500001",
  newAddress
)

Answer 2

您想要MultiMap，例如：

  val mm = new mutable.HashMap[Int, mutable.Set[String]] with mutable.MultiMap[Int, String]
  mm.addBinding(500001, "a")
  mm.addBinding(500003, "b")
  mm.addBinding(500001, "c")
  val l = mm.getOrElse(500001, List())
  println(l)

对于List值类型，您可以将MultiMap值类型设置为：List[String]，如：

  val mm = new mutable.HashMap[Int, mutable.Set[List[String]]] with mutable.MultiMap[Int, List[String]]
  mm.addBinding(500001, List("a", "b"))
  mm.addBinding(500003, List("b", "c"))
  mm.addBinding(500001, List("c", "D"))
  val l = mm.getOrElse(500001, Set())
  println(l)

输出：

  Set(List(e, f), List(a, b))

Answer 3

基本上，您可以将+与此处所述的特征Map[String, Set[String]]混合使用：

http://www.scala-lang.org/api/current/scala/collection/mutable/MultiMap.html

样品：

MultiMap

单个键在Scala中的Map中包含多个值

3 个答案: