使用＆＃34;进行收益＆＃34;从Scala中的hashmap替换字符串

Question

我有一个字符串：

Template string:
I want to eat food, drink beverage at restaurant myrestaurant. 

Hashmap:
food -> pizza
beverage -> miler light beer
myrestaurant -> Papa Johns

现在，我想编写一个for-yield loop来替换模板字符串，其中包含hashmap中的所有键值。

解决它的最佳Scala方法是什么？

Answer 1

您需要替换模板中的字词，因为您继续在hashmap上进行迭代，

例如，

scala> val template = "I want to eat food, drink beverage at restaurant myrestaurant."
template: String = I want to eat food, drink beverage at restaurant myrestaurant.

scala> val hashmap = Map("food" -> "pizza", "beverage" -> "miler light beer", "myrestaurant" -> "Papa Johns")
hashmap: scala.collection.immutable.Map[String,String] = Map(food -> pizza, beverage -> miler light beer, myrestaurant -> Papa Johns)

scala> hashmap.foldLeft(template)((a: String, b: (String, String)) => a.replaceAll(b._1, b._2))
res40: String = I want to eat pizza, drink miler light beer at restaurant Papa Johns.

只有两个键（food和beverage）

的hashmap

scala> val hashmap = Map("food" -> "pizza", "beverage" -> "miler light beer")
hashmap: scala.collection.immutable.Map[String,String] = Map(food -> pizza, beverage -> miler light beer)

scala> hashmap.foldLeft(template)((a, b) => a.replaceAll(b._1, b._2))
res41: String = I want to eat pizza, drink miler light beer at restaurant myrestaurant.

Answer 2

这样的事情：

(template /: hashMap) {case (t, (k, v)) => t.replace(k,v)}

（/:是foldLeft的符号名称）

不确定为什么要为此获得收益率。