/**
 * Copyright (C) 2001-2017 by RapidMiner and the contributors
 * 
 * Complete list of developers available at our web site:
 * 
 * http://rapidminer.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();
	}
}