/*
 *  RapidMiner
 *
 *  Copyright (C) 2001-2011 by Rapid-I and the contributors
 *
 *  Complete list of developers available at our web site:
 *
 *       http://rapid-i.com
 *
 *  This program is free software: you can redistribute it and/or modify
 *  it under the terms of the GNU Affero General Public License as published by
 *  the Free Software Foundation, either version 3 of the License, or
 *  (at your option) any later version.
 *
 *  This program is distributed in the hope that it will be useful,
 *  but WITHOUT ANY WARRANTY; without even the implied warranty of
 *  MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 *  GNU Affero General Public License for more details.
 *
 *  You should have received a copy of the GNU Affero General Public License
 *  along with this program.  If not, see http://www.gnu.org/licenses/.
 */
package com.rapidminer.tools;

import java.util.regex.Matcher;
import java.util.regex.Pattern;

/**
 * A heuristic number parser. That tries to analyze a given String in a more
 * flexible way than the traditional Java number parse:
 *  
 *  1. All "," are replaced by "."
 *  2. ¼,½,¾ are replaced by 0.25, 0.5, 0.75 respectively
 *  3. Ranges ([number]-[number])are matched and the arithmetic average is used as result
 *  4. All Prefixes and suffices are ignored
 *  
 *  Example: "Weight: 8,5 - 10,5 g" would result in 9.5
 *    
 * @author Michael Wurst
 *
 */
public class NumberParser {

	/**
	 * Parse a number possibly surrounded by other information.
	 * @param s the string
	 * @return a double representation or NaN if it cannot be parsed
	 */
	public static double parse(String s) {

		try {
			return Double.parseDouble(s);
		} catch (NumberFormatException e) {
			// Try other heuristics in this case
		}

		// First, replace all ',' by '.'

		s = s.replaceAll(",",".");
		s = s.replaceAll("\u00BC",".25");
		s = s.replaceAll("\u00BD",".5");
		s = s.replaceAll("\u00BE",".75");

		try {
			return Double.parseDouble(s);
		} catch (NumberFormatException e) {
			// Try other heuristics in this case
		}

		// Try to resolve ranges

		Pattern p2 = Pattern.compile("[^0-9]*([[0-9.]\\.]+)\\s?-\\s?([[0-9]\\.]+)(.*)");
		Matcher m2 = p2.matcher(s);
		if(m2.matches()) {

			try {
				return (Double.parseDouble(m2.group(1))+Double.parseDouble(m2.group(2)))/2;

			} catch (NumberFormatException e) {
				// Try other heuristics in this case
			}

		}
		// Try to ignore all suffixes

		Pattern p1 = Pattern.compile("[^0-9]*([[0-9]\\.]+)(.*)");
		Matcher m1 = p1.matcher(s);
		if(m1.matches()) {

			try {
				return Double.parseDouble(m1.group(1));
			} catch (NumberFormatException e) {
				// Try other heuristics in this case
			}
		}
		return Double.NaN;
	}

	/**
	 * This method parses the given string as double value.
	 * It first tries the normal parse method, then tests if it is the ?
	 * and would return NaN in this case.
	 * Otherwise a NumberFormatExceptin is thrown.
	 */
	public static double parseDouble(String s) throws NumberFormatException {
		try {
			return Double.parseDouble(s);
		} catch (NumberFormatException e) {}
		// try if NaN
		if (s.equals("?"))
			return Double.NaN;
		throw new NumberFormatException();
	}
}