Sorting a list in Groovy in an unusual way

Question

I have a list of, let's say [Cat, Dog, Cow, Horse], that I want to be sorted in the following way

• if Cat is on the list it should come first
• if Cow is on the list it should come second
• The rest of the elements should come after in alphabetic order.

Any suggestions how this could be done in Groovy?

Answer 1

Tim's answer is pretty clever. I'm personally more a fan of just using list operations as the code it generates is slightly easier to read.

def highPriority = [ 'Cat', 'Cow' ]

def list = [ 'Armadillo', 'Dog', 'Cow', 'Zebra', 'Horse', 'Cow', 'Cat' ]

def remainder = ( list - highPriority ).sort()

list.retainAll( highPriority )

list.sort{ highPriority.indexOf( it ) } + remainder

That will give you Cow twice. If you don't want duplicates, using intersect is fairly simple.

def highPriority = [ 'Cat', 'Cow' ]

def list = [ 'Armadillo', 'Dog', 'Cow', 'Zebra', 'Horse', 'Cow', 'Cat' ]

list.intersect( highPriority ).sort{ highPriority.indexOf( it ) } + ( list - highPriority ).sort()

Answer 2

This should do it:

// Define our input list
def list = [ 'Armadillo', 'Cat', 'Dog', 'Cow', 'Zebra', 'Horse', 'Cow' ]

// Define a closure that will do the sorting
def sorter = { String a, String b, List prefixes=[ 'Cat', 'Cow' ] ->
  // Get the index into order for a and b
  // if not found, set to being Integer.MAX_VALUE
  def (aidx,bidx) = [a,b].collect { prefixes.indexOf it }.collect {
    it == -1 ? Integer.MAX_VALUE : it
  }
  // Compare the two indexes.
  // If they are the same, compare alphabetically
  aidx <=> bidx ?: a <=> b
}

// Create a new list by sorting using our closure
def sorted = list.sort false, sorter

// Print it out
println sorted

That prints:

[Cat, Cow, Cow, Armadillo, Dog, Horse, Zebra]

I've commented it to try and explain each step it takes. By adding the default prefix items as an optional parameter on the sorter closure, it means we can do stuff like this to change the default:

// Use Dog, Zebra, Cow as our prefix items
def dzc = list.sort false, sorter.rcurry( [ 'Dog', 'Zebra', 'Cow' ] )
println dzc

Which then prints the list sorted as:

[Dog, Zebra, Cow, Cow, Armadillo, Cat, Horse]

Answer 3

Here's another alternative that feels simpler to me:

// smaller values get sorted first
def priority(animal) {
    animal in ['Cat', 'Cow'] ? 0 : 1
}

def list = [ 'Armadillo', 'Cat', 'Dog', 'Cow', 'Zebra', 'Horse', 'Cow' ]

def sorted = list.sort{ a, b -> priority(a) <=> priority(b) ?: a <=> b }

assert sorted == ['Cat', 'Cow', 'Cow', 'Armadillo', 'Dog', 'Horse', 'Zebra']

Answer 4

If you don't have duplicate elements, you can try this:

def highPriority = [ 'Cat', 'Cow' ]
def list = [ 'Armadillo', 'Dog', 'Cow', 'Zebra', 'Horse', 'Cat' ]
highPriority + list.minus(highPriority).sort()