Question

我有Person的列表，想要通过id

检索某个人

val person = personL.find(_.id.equals(tempId))

之后，我希望将列表的第一个和最后一个元素作为元组得到Person的属性。

val marks: Option[(Option[String], Option[String])] = person.map { p =>
              val marks = p.school.marks
              (marks.headOption.map(_.midtermMark), marks.lastOption.map(_.finalMark))
}

这项工作还不错，但现在我想将Option[(Option[String], Option[String])]转换为简单的(Option[String], Option[String])。是否有可能通过使用上一张地图在运行中做到这一点？

Answer 1

我想：

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="ahmedchtn.pockemontn">
    <!--
         The ACCESS_COARSE/FINE_LOCATION permissions are not required to use
         Google Maps Android API v2, but you must specify either coarse or fine
         location permissions for the 'MyLocation' functionality. 
    -->
    <uses-permission android:name="android.permission.INTERNET" />
    <uses-permission android:name="android.permission.ACCESS_FINE_LOCATION"/>
    <application
        android:allowBackup="true"
        android:icon="@mipmap/ic_launcher"
        android:label="@string/app_name"
        android:roundIcon="@mipmap/ic_launcher_round"
        android:supportsRtl="true"
        android:theme="@style/AppTheme">
        <!--
             The API key for Google Maps-based APIs is defined as a string resource.
             (See the file "res/values/google_maps_api.xml").
             Note that the API key is linked to the encryption key used to sign the APK.
             You need a different API key for each encryption key, including the release key that is used to
             sign the APK for publishing.
             You can define the keys for the debug and release targets in src/debug/ and src/release/. 
        -->
        <meta-data
            android:name="com.google.android.geo.API_KEY"
            android:value="@string/google_maps_key" />

        <activity
            android:name=".MapsActivity"
            android:label="@string/title_activity_maps">
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />

                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
    </application>

</manifest>

如果您的元组选项为空，

person.map{...}.getOrElse((None, None))是默认值

Answer 2

您可能正在寻找fold：

personL
 .collectFirst {
    case Person(`tempId`, _, .., school) => school.marks
 }.fold[Option[String], Option[String]](None -> None) { marks => 
    marks.headOption.map(_.midtermMark) -> marks.lastOption.map(_.finalMark) 
 }

如何在Scala中将可选的选项元组解包为元组选项？

2 个答案: