Finding duplicates in a list/file. [Groovy/Java] - list

Question

I have an input file where each line is a special record.
I would gladly work on the file level but might be a more convenient way to transfer the file into a list. (each object in the list = each row in the file)
In the input file, there can be several duplicate rows.
The goal: Split the given file/list into unique records and duplicate records, i.e., Records which are present multiple times, keep one occurrence and other duplicate parts store in a new list
I found an easy way how to remove duplicates but never found a way how to store them
File inputFile = new File("....")
inputFile.eachLine { inputList.add(it) } //fill the list
List inputList = [1,1,3,3,1,2,2,3,4,1,5,6,7,7,8,9,8,10]
inputList = inputList.unique() // remove duplicates
println inputList
// inputList = [1, 3, 2, 4, 5, 6, 7, 8, 9, 10]
The output should look like: Two lists/files with removed duplicates and duplicates itself
inputList = [1,3,2,4,5,6,7,8,9,10] //only one ocurance of each line
listOfDuplicates = [1,1,1,3,3,2,7,8] //duplicates removed from original list
The output does not need to correspond with the initial order of items.
Thank you for help, Matt

Answer 1

You could simply iterate over the list yourself:
def inputList = [1,1,3,3,1,2,2,3,4,1,5,6,7,7,8,9,8,10]
def uniques = []
def duplicates = []
inputList.each { uniques.contains(it) ? duplicates << it : uniques << it }
assert inputList.size() == uniques.size() + duplicates.size()
assert uniques == [1,3,2,4,5,6,7,8,9,10] //only one ocurance of each line
assert duplicates == [1,3,1,2,3,1,7,8] //duplicates removed from original list
inputList = uniques // if desired

Answer 2

There are many ways to do this,following is the simplest way
def list = [1,1,3,3,1,2,2,3,4,1,5,6,7,7,8,9,8,10]
def unique=[]
def duplicates=[]
list.each {
if(unique.contains(it))
duplicates.add(it)
else
unique.add(it)
}
println list //[1, 1, 3, 3, 1, 2, 2, 3, 4, 1, 5, 6, 7, 7, 8, 9, 8, 10]
println unique //[1, 3, 2, 4, 5, 6, 7, 8, 9, 10]
println duplicates //[1, 3, 1, 2, 3, 1, 7, 8]
Hope this will helps you

Answer 3

Something very straight-forward:
List inputList = [1,1,3,3,1,2,2,3,4,1,5,6,7,7,8,9,8,10]
def uniques = [], duplicates = []
Iterator iter = inputList.iterator()
iter.each{
iter.remove()
inputList.contains( it ) ? ( duplicates << it ) : ( uniques << it )
}
assert [2, 3, 4, 1, 5, 6, 7, 9, 8, 10] == uniques
assert [1,1,3,3,1,2,7,8] == duplicates

Answer 4

If order of duplicates isn't important:
def list = [1,1,3,3,1,2,2,3,4,1,5,6,7,7,8,9,8,10]
def (unique, dups) = list.groupBy().values()*.with{ [it[0..0], tail()] }.transpose()*.sum()
assert unique == [1,3,2,4,5,6,7,8,9,10]
assert dups == [1,1,1,3,3,2,7,8]

Answer 5

This code should solve the problem
List listOfDuplicates = inputList.clone()
listOfDuplicates.removeAll{
listOfDuplicates.count(it) == 1
}

Answer 6

The more the merrier:
groovy:000> list.groupBy().values()*.tail().flatten()
===> [1, 1, 1, 3, 3, 2, 7, 8]
Group by identity (this is basically a "frequencies" function).
Take just the values
Clip the first element
Combine the lists