How to sort list in groovy

groovy

Solution

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

Problem

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.

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