![](/img/trans.png)
[英]Sort elements in list based on length first and alphabetical order second
[英]Scala sort list based on second attribute and then first
我希望首先根据长度排序包含(word,word.length)的列表,然后按字母顺序排序。 所以给出: "I am a girl"
输出应该是a:1, I:1, am:2, girl:4
我有下面的代码,但不是所有的例子
val lengths = words.map(x => x.length)
val wordPairs = words.zip(lengths).toList
val mapwords = wordPairs.sort (_._2 < _._2).sortBy(_._1)
你可以按元组排序:
scala> val words = "I am a girl".split(" ")
words: Array[java.lang.String] = Array(I, am, a, girl)
scala> words.sortBy(w => w.length -> w)
res0: Array[java.lang.String] = Array(I, a, am, girl)
scala> words.sortBy(w => w.length -> w.toLowerCase)
res1: Array[java.lang.String] = Array(a, I, am, girl)
你可以在一行中做到这一点:
"I am a girl".toLowerCase.split(" ").map(x => (x,x.length)).sortWith { (x: (String,Int), y: (String,Int)) => x._1 < y._1 }
或者分为两行:
val wordPairs = "I am a girl".split(" ").map(x => (x,x.length))
val result = wordPairs.toLowerCase.sortWith { (x: (String,Int), y: (String,Int)) => x._1 < y._1 }
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.