Groovy list.sort由第一，第二和第三个元素组成 [英] Groovy list.sort by first, second then third elements

问题描述

我有一个常用的列表清单，即

  list = [[2,0,1]，[1，5 ，2]，[1,0,3]] 
  

我想按照第一个元素，然后是第二个，然后是第三个元素。

期望

  assert list == [[1,0,3]，[1,5,2]，[2,0,1]] 
  

我以 list = list.sort {a，b - > a [0] => b [0]} 但只排序第一个元素。如何完成？

谢谢

解决方案

您应该可以

  list = [[2,0,1]，[1，5， 2]，[1,0,3]] 
 
 list = list.sort {a，b  - > a [2] => b [2]} 
 list = list.sort {a，b  - > a [1]< => b [1]} 
 list = list.sort {a，b  - > a [0] => b [0]} 
 
断言列表== [[1,0,3]，[1,5,2]，[2,0,1]] 
  
 
每个都应该覆盖前面的内容，以保持组合排序完整。 
 
 
您也可以按照 Elvis算子，？： ，当前一个相等时（并且< => / code>返回 0 ）： 
 
 
  list.sort {a，b  - > a [0] => b [0]？：a [1] => b [1]？：a [2] => b [2]} 
  
 
I have a groovy list of lists i.e.
list = [[2, 0, 1], [1, 5, 2], [1, 0, 3]]
I would like sort it by order of the first element, then second, then third.


Expected
assert list == [[1, 0, 3], [1, 5, 2], [2, 0, 1]]
I started with list = list.sort{ a,b -> a[0] <=> b[0] } but that only sorts the first element. How do you finish?

Thanks
解决方案
You should be able to iterate through the desired sorting in reverse order:
list = [[2, 0, 1], [1, 5, 2], [1, 0, 3]]

list = list.sort{ a,b -> a[2] <=> b[2] }
list = list.sort{ a,b -> a[1] <=> b[1] }
list = list.sort{ a,b -> a[0] <=> b[0] }

assert list == [[1, 0, 3], [1, 5, 2], [2, 0, 1]]
Each should override the previous just enough to keep the combined sorting intact.



You can also chain them in order with the Elvis operator, ?:, which will defer to the next comparison when the previous are equal (and <=> returns 0):
list.sort { a,b -> a[0] <=> b[0] ?: a[1] <=> b[1] ?: a[2] <=> b[2] }


                        
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