CODEWITHSUNDEEP
groovy
SaDman_SUT
SaDman_SUT

Reputation: 177

How to sort list in groovy

i'm try to sort array list

eg.

def list = [1, 1, 4, 4, 3, 4, 1]

hope to sort :

[1, 1, 1, 4, 4, 4, 3]

Thank you very much.

i'm used to my code

eg.

def plnProcessGoalInstance = ......someting    
def order = plnProcessGoalInstance.plnGoal.plnTargetPlan.id.unique() //[1, 4, 3,] ,plnProcessGoalInstance.plnGoal.plnTargetPlan.id = [1, 1, 4, 4, 3, 4, 1]
def plnProcessGoalInstance = plnProcessGoalInstance.sort{ a, b -> 
           order.indexOf(a.plnGoal.plnTargetPlan.id ) <=> order.indexOf(b.plnGoal.plnTargetPlan.id )}

Thank you very much for help.

Upvotes: 3

Views: 4461

Answers (1)

tim_yates
tim_yates

Reputation: 171074

How about:

def order = [ 1, 4, 3 ]
def list = [ 1, 1, 4, 4, 3, 4, 1 ]

list.sort { a, b -> order.indexOf( a ) <=> order.indexOf( b ) }

assert list == [1, 1, 1, 4, 4, 4, 3]

Or, assuming the comment by Deruijter is correct and you want to sort by descending frequency and then by number for those with the same freq:

def list = [ 1, 1, 4, 4, 3, 4, 1 ]
def order = list.countBy { it }
                .sort { a, b -> 
                  b.value <=> a.value ?: a.key <=> b.key
                }.keySet().toList()
list.sort { a, b -> order.indexOf( a ) <=> order.indexOf( b ) }

countBy requires Groovy 1.8

Upvotes: 3

Related Questions