https://www.cnblogs.com/weilunhui/p/5658860.html
1.++[B] 在A元素后面追加B元素
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scala> val a = List(1)a: List[Int] = List(1)scala> val b = List(2)b: List[Int] = List(2)scala> val c = a ++ bc: List[Int] = List(1, 2)scala> val d = List(‘a‘)d: List[Char] = List(a)scala> val e = c ++ de: List[AnyVal] = List(1, 2, a) |
2.++: 与++不同的是使用的右边边的数据类型
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scala> val x = List(1)x: List[Int] = List(1)scala> val y = LinkedList(2)y: scala.collection.mutable.LinkedList[Int] = LinkedList(2)scala> var z = x ++: yz: scala.collection.mutable.LinkedList[Int] = LinkedList(1, 2) //y是LinkedList |
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scala> var z = y ++: x //使用x的类型z: List[Int] =List(1, 2) |
3.+: 在List的头部增加元素
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scala> val x = List(1)x: List[Int] = List(1)scala> val y = 2 +: xy: List[Int] = List(2, 1)scala> println(x)List(1) |
4./:应用二进制运算符起始值，从左到右遍历迭代所有元素。
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scala> val a = List(1,2,3,4)a: List[Int] = List(1, 2, 3, 4)scala> val b = (5 /: a)(_+_)b: Int = 15scala> val c = (5 /: a)((x,y) => x + y)c: Int = 15 |
5.:+:在后面追加元素
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scala> val a = List(1)a: List[Int] = List(1)scala> val b = a :+ 2b: List[Int] = List(1, 2)scala> println(a)List(1) |
6.:: 为列表增加单个元素，这是一个右结合的操作符
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1 :: List(2, 3) = List(2, 3).::(1) = List(1, 2, 3) |
7.::: 在列表的前面追加一个列表，这是一个右结合的操作符
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List(1, 2) ::: List(3, 4) = List(3, 4).:::(List(1, 2)) = List(1, 2, 3, 4) |
8.:\ 与元素合并后按照操作进行对列表遍历
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scala> val a = List(1,2,3,4)a: List[Int] = List(1, 2, 3, 4)scala> val b = (a :\ 5)(_+_)b: Int = 15scala> val c = (a :\ 5)((x,y) => x + y)c: Int = 15 |
9.addString(b: StringBuilder) 对列表中的所有元素迭代将拼接成字符串
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scala> val a = List(1,2,3,4)a: List[Int] = List(1, 2, 3, 4)scala> val b = new StringBuilder()b: StringBuilder =scala> val h = a.addString(b)h: StringBuilder = 1234 |
10.addString(b: StringBuilder, sep: String) 使用分隔符输出字符串
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scala> val a = List(1,2,3,4)a: List[Int] = List(1, 2, 3, 4)scala> val b = new StringBuilder()b: StringBuilder =scala> a.addString(b, ", ")res0: StringBuilder = 1, 2, 3, 4 |
11.addString(b: StringBuilder, start: String, sep: String, end: String) 定义输出以什么开始、分割、结束
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scala> val a = List(1,2,3,4)a: List[Int] = List(1, 2, 3, 4)scala> val b = new StringBuilder()b: StringBuilder =scala> a.addString(b , "List(" , ", " , ")")res5: StringBuilder = List(1, 2, 3, 4) |
12. aggregate[B](z: ? B)(seqop: (B, A) ? B, combop: (B, B) ? B): B 对列表内元素进行求和
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List(‘a‘, ‘b‘, ‘c‘).aggregate(0)({ (sum, ch) => sum + ch.toInt }, { (p1, p2) => p1 + p2 }) |
13.apply(n: Int) 通过索引选择元素
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scala> List(‘a‘,‘b‘,‘c‘).apply(1)res2: Char = b |