Scala 转换集到地图
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Scala Convert Set to Map
提问by Torsten Schmidt
How do I convert a Set("a","b","c") to a Map("a"->1,"b"->2,"c"->3)? I think it should work with toMap.
如何将 Set("a","b","c") 转换为 Map("a"->1,"b"->2,"c"->3)？我认为它应该与 toMap 一起使用。
回答by Adam Rabung
zipWithIndexis probably what you are looking for. It will take your collection of letters and make a new collection of Tuples, matching value with position in the collection. You have an extra requirement though - it looks like your positions start with 1, rather than 0, so you'll need to transform those Tuples:
zipWithIndex可能是您正在寻找的。它将获取您的字母集合并创建一个新的元组集合，将值与集合中的位置相匹配。不过你有一个额外的要求——看起来你的位置从 1 开始，而不是 0，所以你需要转换这些元组：
Set("a","b","c")
.zipWithIndex //(a,0), (b,1), (c,2)
.map{case(v,i) => (v, i+1)} //increment each of those indexes
.toMap //toMap does work for a collection of Tuples
One extra consideration - Sets don't preserve position. Consider using a structure like List if you want the above position to consistently work.
一个额外的考虑 - 集合不保留位置。如果您希望上述职位始终有效，请考虑使用类似 List 的结构。
回答by Frank S. Thomas
回答by GClaramunt
toMaponly works if the Setentries are key/value pairs (e.g. Set(("a",1),("b",2),("c",3))).
toMap仅当Set条目是键/值对时才有效（例如Set(("a",1),("b",2),("c",3))）。
To get what you want, use zipWithIndex:
要获得您想要的东西，请使用zipWithIndex：
Set("a","b","c") zipWithIndex
// Set[(String, Int)] = Set((a,0), (b,1), (c,2))
or (as in you original question):
或（如你原来的问题）：
Set("a","b","c") zip (1 to 3) toMap
回答by Viktor Nordling
This would also work:
这也可以：
(('a' to 'c') zip (1 to 3)).toMap
