Groovy list.sort by first, second then third elements

If you want to sort arrays of arbitrary (though homogenous) length, you can use this and it will do it in a single pass:

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

list.sort { a, b -> 
    for (int i : (0..<a.size())) {
        def comparison = (a[i] <=> b[i])
        if (comparison) return comparison
    } 
    return 0
}

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

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] }