Asymmetric nearest-neighbour in Java

Question

From a sorted map, I want to retrieve a subset of n entries, starting m entries before a specified value v. For example, for the key set k = {0.2, 0.3, 0.4, 0.6, 0.8, 0.9, 1.0}, a query with n=5, m=2, v=0.5, would return {0.3, 0.4, 0.6, 0.8, 0.9}. Is there an implementation of a data structure in Java supporting such a query, without having to iterate over the whole (big) set?

What do I need this for? Interpolation. I want to interpolate at v based on the values in the map. However, I have many v. They are sorted, and have a spacing between them much smaller than the ones in k. So, I take a range of entries from the map, do some expensive preparatory calculations with them (for example calculating coefficients of a polynom), and can then quickly interpolate another value in that range (by evaluating the polynom with that value).

But why do I need the m entries before v? The values in k are usually equally spaced, and in order to avoid the Runge phenomenon of high oscillations at the ends of the interpolation interval, I simply cut them off, which means I need some nodes before the actual valid interval of the interpolation.

Does that make sense? What’s your suggestion?

(It would be fun if a method like java.util.TreeMap.ceilingEntry() would return an iterator, with which I could step back twice.)

Answer 1

Using headMap() and tailMap() is probably the most straightforward solution. If one fears the overhead of making the same search twice, using a list instead of a map is probably the solution. I extended Petar’s suggestion. It can now handle key-value pairs, represented by the small subclass Pair:

public class DataSet {

  // Usage example
  public static void main (String [] args) {

    DataSet nn = new DataSet ();
    nn.add(0.2,1);
    nn.add(0.3,2);
    nn.add(0.4,3);
    nn.add(0.6,4);
    nn.add(0.8,5);
    nn.add(0.9,6);
    nn.add(1.0,7);

    int n = 5;
    int m = 2;
    double v = 0.5;

    ListIterator <Pair> it = nn.iterator(v);
    for (int i=0; i<m; ++i)
      it.previous();      
    for (int i=0; i<n; ++i)
      System.out.append(it.next()+"\n");
  }

  // Implementation
  TreeSet <Pair> set = new TreeSet <Pair> (new PairComparator());
  ArrayList <Pair> list = new ArrayList <Pair> ();
  boolean listUpToDate = false;

  // Add data
  boolean add (double x, double y) {
    listUpToDate = false;
    return set.add(new Pair(x,y));
  }

  // Get iterator at specified position
  ListIterator <Pair> iterator (double v) {
    if (!listUpToDate) {
      list = new ArrayList (set);
      listUpToDate = true;
    }
    int pos = Collections.binarySearch(list,v);
    if (pos < 0)
      pos = -pos - 1;
    return list.listIterator(pos);
  }

  // Helper class
  static class Pair implements Comparable <Double> {
    double x, y;
    Pair (double x, double y) {
      this.x = x; this.y = y;
    }
    public int compareTo (Double d) {
      return Double.valueOf(x).compareTo(d);
    }
    public String toString () {
        return String.format("[%.1f,%.1f]",x,y);
    }
  }

  // Used for sorting
  class PairComparator implements Comparator <Pair> {
    public int compare (Pair n0, Pair n1) {
      return Double.valueOf(n0.x).compareTo(Double.valueOf(n1.x));
    }
  }

}

Of course, one could also just use the list, and make sure it is sorted before calling binarySearch(). But the TreeSet has also the advantage, besides the ordering, that it prevents duplicate keys.

Answer 2

This is much simpler than that:

1. Use binary search to get the position at which v would be inserted in the list so that it will stay sorted.
2. Move m positions to the left

3. Take the first n elements to the right.

   double[] k = new double[] {0.2, 0.3, 0.4, 0.6, 0.8, 0.9, 1.0};
   int n=5;
   int m=2;
   double v=0.5;
   
   int pos = Arrays.binarySearch(k, v);
   if (pos < 0)
       pos = -pos - 1;
   
   while(pos > 0 && k[pos-1]==v)
       --pos;
   
   pos = Math.max(pos-m, 0);
   
   double[] result = Arrays.copyOfRange(k, pos, Math.min(pos+n, k.length));
   

Answer 3

You can place the entries in a sorted array and use a Arrays.binarySearch().

However if you must have a NavigableMap like TreeMap, you need to do two lookups to get the headMap() and tailMap() and iterate over those.