小池有话说

1. 构造映射 val f = Map("foo" -> 1, "bar" -> 2)
2. 可变映射 val f=scala.collection.mutable.Map("foo"->2,"bar"->3)
3. 空的映射需要指定映射实现和类型 val f= new scala.collection.mutable.HashMap[String,Int]
4. 映射是对偶的集合，对偶是二元组。 -> 是用来创建对偶的。"fooo"->3 的值是 (String, Int) = (fooo,3)
5. 获取值用小括号 val a = f("foo") 如果不存在键就抛异常
6. 检查是否存在某个键，用 contains 方法： val a=if (f.contains("baz")) f("baz") else 0 或者用糖 val a=f.getOrElse("baz",0)
7. get 方法返回 Option 对象
8. 赋值 f("bar") = 10 或者用 += 符号更新 f += ("a"->1,"ab"->2)
9. 用 -= 移除某个键 f -= "a"
10. + - 符号如你所愿
11. for ((k,v)<-映射) 遍历
12. keySet 和 values 方法可以获得键和值
13. for ((k,v)<-map) yield(v,k) 翻转，吊吊的
14. scala.collection.immutable.SortedMap 的实现是树形映射，有序，但是不可变。如果想要可变的树形映射，请用Java中的 TreeMap
15. 如果想要按插入顺序访问所有键，请使用 scala.collection.mutable.LinkedHashMap
16. import scala.collection.JavaConversions.mapAsScalaMap 转换 import scala.collection.JavaConversions.propertiesAsScalaMap 可以将 java.util.properties 转到 Map[String,String] 反之亦有相关类库
17. 元组 val t = (1,3.14,"abc") 创建元组
18. 可以使用 _1 _2 _3 方法访问元组的元素。但通常使用模式匹配来赋值 val (first, second, third) = t
19. val symbols = Array("<","-",">"); val counts = Array(2, 10, 2); val pairs = symbols.zip(counts) 然后就可以对齐进行一起操作了 for ((s,n)<-pairs) print(s*n)
20. toMap 方法可以将对偶的集合转成映射 pairs.toMap

练习

练习

1. val m = Map("shoe"->223.3, "shirt"->107.1,"coat"->320.0); val mAfter = for((name,price) <- m) yield (name,price*0.9)
2. val words = new scala.collection.mutable.HashMap[String,Int]; val in = new java.util.Scanner(new java.io.File("myfile.txt")); while(in.hasNext()) {val w=in.next();words(w)=words.getOrElse(w,0)+1}; for((w,n)<-words) printf("%s\t%d\n",w,n)
3. var words=Map[String,Int](); val in = new java.util.Scanner(new java.io.File("myfile.txt")); while(in.hasNext()) {val w=in.next();words = words + (w -> (words.getOrElse(w,0)+1)) }; for((w,n)<-words) printf("%s\t%d\n",w,n)
4. var words=scala.collection.immutable.SortedMap[String,Int](); val in = new java.util.Scanner(new java.io.File("myfile.txt")); while(in.hasNext()) {val w=in.next();words = words + (w -> (words.getOrElse(w,0)+1)) }; for((w,n)<-words) printf("%s\t%d\n",w,n)
5. import scala.collection.JavaConversions.mapAsScalaMap; var words=new java.util.TreeMap[String,Int]; val in = new java.util.Scanner(new java.io.File("myfile.txt")); while(in.hasNext()) {val w=in.next();words(w)=words.getOrElse(w,0)+1}; for((w,n)<-words) printf("%s\t%d\n",w,n)
6. val m = new scala.collection.mutable.LinkedHashMap[String,Int](); m("Monday")=java.util.Calendar.MONDAY; m("Tuesday")=java.util.Calendar.TUESDAY;m("Wednesday")=java.util.Calendar.WEDNESDAY;m("Thursday")=java.util.Calendar.THURSDAY;m("Friday")=java.util.Calendar.FRIDAY;m("Saturday")=java.util.Calendar.SATURDAY;m("Sunday")=java.util.Calendar.SUNDAY;
7. import scala.collection.JavaConversions.mapAsScalaMap; import scala.collection.JavaConversions.asScalaSet; val ps = java.lang.System.getProperties(); val len = (s:Object) => s.toString.length; val maxlen = len(ps.keySet.maxBy(len)); for ((o,v)<-ps) printf("%-"+(maxlen+1)+"s| %s\n", o, v)
8. def minmax(values: Array[Int]) = (values.min, values.max)
9. def lteqgt(values: Array[Int], v: Int) = (values.count(_<v), values.count(_==v), values.count(_>v))
10. 字符串对比、恺撒密码？