把Scala List的几种常见方法梳理汇总如下,日常开发场景基本上够用了。

创建列表

scala> val days = List("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday")
days: List[String] = List(Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday)

创建空列表

scala> val l = Nil
l: scala.collection.immutable.Nil.type = List()

scala> val l = List()
l: List[Nothing] = List()

用字符串创建列表

scala> val l = "Hello" :: "Hi" :: "Hah" :: "WOW" :: "WOOW" :: Nil
l: List[String] = List(Hello, Hi, Hah, WOW, WOOW)

用“:::”叠加创建新列表

scala> val wow = l ::: List("WOOOW", "WOOOOW")
wow: List[String] = List(Hello, Hi, Hah, WOW, WOOW, WOOOW, WOOOOW)

通过索引获取列表值

scala> l(3)
res0: String = WOW

获取值长度为3的元素数目

scala> l.count(s => s.length == 3)
res1: Int = 2

返回去掉l头两个元素的新列表

scala> l.drop(2)
res2: List[String] = List(Hah, WOW, WOOW)

scala> l
res3: List[String] = List(Hello, Hi, Hah, WOW, WOOW)

返回去掉l后两个元素的新列表

scala> l.dropRight(2)
res5: List[String] = List(Hello, Hi, Hah)

scala> l
res6: List[String] = List(Hello, Hi, Hah, WOW, WOOW)

判断l是否存在某个元素

scala> l.exists(s => s == "Hah")
res7: Boolean = true

滤出长度为3的元素

scala> l.filter(s => s.length == 3)
res8: List[String] = List(Hah, WOW)

判断所有元素是否以“H”打头

scala> l.forall(s => s.startsWith("H"))
res10: Boolean = false

判断所有元素是否以“H”结尾

scala> l.forall(s => s.endsWith("W"))
res11: Boolean = false

打印每个元素

scala> l.foreach(s => print(s + ' '))
Hello Hi Hah WOW WOOW

取出第一个元素

scala> l.head
res17: String = Hello

取出最后一个元素

scala> l.last
res20: String = WOOW

剔除最后一个元素,生成新列表

scala> l.init
res18: List[String] = List(Hello, Hi, Hah, WOW)

剔除第一个元素,生成新列表

scala> l.tail
res49: List[String] = List(Hi, Hah, WOW, WOOW)

判断列表是否为空

scala> l.isEmpty
res19: Boolean = false

获得列表长度

scala> l.length
res21: Int = 5

修改每个元素,再反转每个元素形成新列表

scala> l.map(s => {val s1 = s + " - 01"; s1.reverse})
res29: List[String] = List(10 - olleH, 10 - iH, 10 - haH, 10 - WOW, 10 - WOOW)

生成用逗号隔开的字符串

scala> l.mkString(", ")
res30: String = Hello, Hi, Hah, WOW, WOOW

反序生成新列表

scala> l.reverse
res41: List[String] = List(WOOW, WOW, Hah, Hi, Hello)

按字母递增排序

scala> l.sortWith(_.compareTo(_) < 0)
res48: List[String] = List(Hah, Hello, Hi, WOOW, WOW)
  • flatMap,打算后面专门写点

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