2. 编写一段程序，从文件中读取单词。用一个可变映射来清点每一个单词出现的频率。读取这些单词的操作可以使用java.util.Scanner:
val in = new java.util.Scanner(new java.io.File("myfile.txt"))
while(in.hasNext()) 处理 in.next()
或者翻到第9章看看更Scala的做法。
最后，打印出所有单词和它们出现的次数。
3. 重复前一个练习，这次用不可变的映射。
object App{
def main(args: Array[String]){
val map = countWord("myfile.txt")
println(map.mkString("[", ", ", "]"))
}
def countWord(filename:String)={
var words = Map[String, Int]()
val in = new java.util.Scanner(new java.io.File("myfile.txt"))
while(in.hasNext()){
var key = in.next()
words += (key -> (words.getOrElse(key,0) + 1))
}
words
}
}
4. 重复前一个练习，这次用已排序的映射，以便单词可以按顺序打印出来。
object App{
def main(args: Array[String]){
val map = countWord("myfile.txt")
println(map.mkString("[", ", ", "]"))
}
def countWord(filename:String)={
var words = scala.collection.immutable.SortedMap[String, Int]()
val in = new java.util.Scanner(new java.io.File("myfile.txt"))
while(in.hasNext()){
var key = in.next()
words += (key -> (words.getOrElse(key,0) + 1))
}
words
}
}
5. 重复前一个练习，这次用java.util.TreeMap并使之适用于Scala API
import scala.collection.JavaConversions.mapAsScalaMap
object App{
def main(args: Array[String]){
val map = countWord("myfile.txt")
println(map.mkString("[", ", ", "]"))
}
def countWord(filename:String)={
var words = new java.util.TreeMap[String, Int]()
val in = new java.util.Scanner(new java.io.File("myfile.txt"))
while(in.hasNext()){
var key = in.next()
words += (key -> (words.getOrElse(key,0) + 1))
}
words
}
}
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6. 定义一个链式哈希映射，将"Monday"映射到java.util.Calendar.MONDAY，以此类推加入其他日期。
展示元素是以插入的顺序被访问的。
import scala.collection.JavaConversions.mapAsScalaMap
object App{
def main(args: Array[String]){
val map = weekday("myfile.txt")
println(map.mkString("[", ", ", "]"))
}
def weekday(filename:String)={
var weekdays = new java.util.LinkedHashMap[String,Int]()
weekdays("Monday") = java.util.Calendar.MONDAY
weekdays("Tuesday") = java.util.Calendar.TUESDAY
weekdays("Wednesday") = java.util.Calendar.WEDNESDAY
weekdays("Thursday") = java.util.Calendar.THURSDAY
weekdays("Friday") = java.util.Calendar.FRIDAY
weekdays("Saturday") = java.util.Calendar.SATURDAY
weekdays("Sunday") = java.util.Calendar.SUNDAY
weekdays
}
}
7. 打印出所有Java系统属性的表格，类似这样：
java.runtime.name | Java(TM) SE Runtime Environment
...
你需要找到最长键的长度才能正确地打印出这张表格。
import scala.collection.JavaConversions.propertiesAsScalaMap
object App{
def main(args: Array[String]){
printSystemProperties()
}
def printSystemProperties()={
val props: scala.collection.Map[String, String] = System.getProperties()
var maxlen = 0
for((k,_) <- props) if(maxlen < k.length) maxlen = k.length
maxlen += 1
for((k,v) <- props){
printf("%-" + maxlen.toString + "s | %s\n", k, v)
}
}
}
8. 编写一个函数minmax(values: Array[Int])，返回数组中最小值和最大值的对偶。
object App{
def main(args: Array[String]){
val a = Array(1,2,3,4,5,6,7)
val b = minmax(a)
println(a.mkString("[",", ","]"))
println(b)
}
def minmax(values: Array[Int])={
(values.min,values.max)
}
}
9. 编写一个函数lteqgt(values: Array[Int], v: Int)，返回数组中小于v、等于v和大于v的数量，要求三个值一起返回。
object App{
def main(args: Array[String]){
val a = Array(1,2,3,4,5,6,7)
val b = lteqgt(a, 4)
println(a.mkString("[",", ","]"))
println(b)
}
def lteqgt(values: Array[Int], v: Int)={
val lt = values.count(_ < v)
val eq = values.count(_ == v)
val gt = values.count(_ > v)
(lt, eq, gt)
}
}
10. 当你将两个字符串拉链在一起，比如"Hello".zip("World")，会是什么结果？想出一个讲得通的用例。
object App{
def main(args: Array[String]){
val a = zip("Hello", "World")
println(a)
}
def zip(s1:String, s2:String)={
val a = s1.zip(s2)
a
}
}
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