如何在 Scala 中用两个字段对列表进行排序？

如何在 Scala 中按两个字段对列表进行排序，在这个例子中，我将按 lastName 和 firstName 进行排序？

case class Row(var firstName: String, var lastName: String, var city: String)


var rows = List(new Row("Oscar", "Wilde", "London"),
new Row("Otto",  "Swift", "Berlin"),
new Row("Carl",  "Swift", "Paris"),
new Row("Hans",  "Swift", "Dublin"),
new Row("Hugo",  "Swift", "Sligo"))


rows.sortBy(_.lastName)

我试过这种方法

rows.sortBy(_.lastName + _.firstName)

但是没有用，所以我对一个简单的解决方案很好奇。

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rows.sortBy (row => row.lastName + row.firstName)

If you want to sort by the merged names, as in your question, or 

rows.sortBy (row => (row.lastName, row.firstName))

if you first want to sort by lastName, then firstName; relevant for longer names (Wild, Wilder, Wilderman).

If you write

rows.sortBy(_.lastName + _.firstName)

with 2 underlines, the method expects two parameters:

<console>:14: error: wrong number of parameters; expected = 1
rows.sortBy (_.lastName + _.firstName)
^
小开

In general, if you use a stable sorting algorithm, you can just sort by one key, then the next.

rows.sortBy(_.firstName).sortBy(_.lastName)

The final result will be sorted by lastname, then where that is equal, by firstname.

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最佳答案 
rows.sortBy(r => (r.lastName, r.firstName))
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Perhaps this works only for a List of Tuples, but

scala> var zz = List((1, 0.1), (2, 0.5), (3, 0.6), (4, 0.3), (5, 0.1))
zz: List[(Int, Double)] = List((1,0.1), (2,0.5), (3,0.6), (4,0.3), (5,0.1))


scala> zz.sortBy( x => (-x._2, x._1))
res54: List[(Int, Double)] = List((3,0.6), (2,0.5), (4,0.3), (1,0.1), (5,0.1))

appears to work and be a simple way to express it.


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