mathx.java

来自「EXCEL read and write」· Java 代码 · 共 487 行

/*
* Licensed to the Apache Software Foundation (ASF) under one or more
* contributor license agreements.  See the NOTICE file distributed with
* this work for additional information regarding copyright ownership.
* The ASF licenses this file to You under the Apache License, Version 2.0
* (the "License"); you may not use this file except in compliance with
* the License.  You may obtain a copy of the License at
*
*     http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
/*
 * Created on May 19, 2005
 *
 */
package org.apache.poi.hssf.record.formula.functions;



/**
 * @author Amol S. Deshmukh < amolweb at ya hoo dot com >
 * This class is an extension to the standard math library
 * provided by java.lang.Math class. It follows the Math class
 * in that it has a private constructor and all static methods.
 */
public final class MathX {

    
    private MathX() {}
    
    
    /**
     * Returns a value rounded to p digits after decimal.
     * If p is negative, then the number is rounded to
     * places to the left of the decimal point. eg. 
     * 10.23 rounded to -1 will give: 10. If p is zero,
     * the returned value is rounded to the nearest integral
     * value.
     * <p>If n is negative, the resulting value is obtained
     * as the round value of absolute value of n multiplied
     * by the sign value of n (@see MathX.sign(double d)). 
     * Thus, -0.6666666 rounded to p=0 will give -1 not 0.
     * <p>If n is NaN, returned value is NaN.
     * @param n
     * @param p
     */
    public static double round(double n, int p) {
        double retval;
        
        if (Double.isNaN(n) || Double.isInfinite(n)) {
            retval = Double.NaN;
        }
        else {
            if (p != 0) {
                double temp = Math.pow(10, p);
                retval = Math.round(n*temp)/temp;
            }
            else {
                retval = Math.round(n);
            }
        }
        
        return retval;
    }

    /**
     * Returns a value rounded-up to p digits after decimal.
     * If p is negative, then the number is rounded to
     * places to the left of the decimal point. eg. 
     * 10.23 rounded to -1 will give: 20. If p is zero,
     * the returned value is rounded to the nearest integral
     * value.
     * <p>If n is negative, the resulting value is obtained
     * as the round-up value of absolute value of n multiplied
     * by the sign value of n (@see MathX.sign(double d)). 
     * Thus, -0.2 rounded-up to p=0 will give -1 not 0.
     * <p>If n is NaN, returned value is NaN.
     * @param n
     * @param p
     */
    public static double roundUp(double n, int p) {
        double retval;
        
        if (Double.isNaN(n) || Double.isInfinite(n)) {
            retval = Double.NaN;
        }
        else {
            if (p != 0) {
                double temp = Math.pow(10, p);
                double nat = Math.abs(n*temp);
                
                retval = sign(n) * 
                    ((nat == (long) nat)
                            ? nat / temp
                            : Math.round(nat + 0.5) / temp);
            }
            else {
                double na = Math.abs(n);
                retval = sign(n) * 
                    ((na == (long) na)
                        ? na
                        : (long) na + 1);
            }
        }
        
        return retval;
    }

    /**
     * Returns a value rounded to p digits after decimal.
     * If p is negative, then the number is rounded to
     * places to the left of the decimal point. eg. 
     * 10.23 rounded to -1 will give: 10. If p is zero,
     * the returned value is rounded to the nearest integral
     * value.
     * <p>If n is negative, the resulting value is obtained
     * as the round-up value of absolute value of n multiplied
     * by the sign value of n (@see MathX.sign(double d)). 
     * Thus, -0.8 rounded-down to p=0 will give 0 not -1.
     * <p>If n is NaN, returned value is NaN.
     * @param n
     * @param p
     */
    public static double roundDown(double n, int p) {
        double retval;
        
        if (Double.isNaN(n) || Double.isInfinite(n)) {
            retval = Double.NaN;
        }
        else {
            if (p != 0) {
                double temp = Math.pow(10, p);
                retval = sign(n) * Math.round((Math.abs(n)*temp) - 0.5)/temp;
            }
            else {
                retval = (long) n;
            }
        }
        
        return retval;
    }
    
    
    /**
     * If d < 0, returns short -1
     * <br/>
     * If d > 0, returns short 1
     * <br/>
     * If d == 0, returns short 0 
     * <p> If d is NaN, then 1 will be returned. It is the responsibility
     * of caller to check for d isNaN if some other value is desired.
     * @param d
     */
    public static short sign(double d) {
        return (short) ((d == 0)
                ? 0
                : (d < 0)
                        ? -1
                        : 1);
    }
   
    /**
     * average of all values
     * @param values
     */
    public static double average(double[] values) {
        double ave = 0;
        double sum = 0;
        for (int i=0, iSize=values.length; i<iSize; i++) {
            sum += values[i];
        }
        ave = sum / values.length;
        return ave;
    }
    
    
    /**
     * sum of all values
     * @param values
     */
    public static double sum(double[] values) {
        double sum = 0;
        for (int i=0, iSize=values.length; i<iSize; i++) {
            sum += values[i];
        }
        return sum;
    }
    
    /**
     * sum of squares of all values
     * @param values
     */
    public static double sumsq(double[] values) {
        double sumsq = 0;
        for (int i=0, iSize=values.length; i<iSize; i++) {
            sumsq += values[i]*values[i];
        }
        return sumsq;
    }
    
    
    /**
     * product of all values
     * @param values
     */
    public static double product(double[] values) {
        double product = 0;
        if (values!=null && values.length > 0) {
            product = 1;
            for (int i=0, iSize=values.length; i<iSize; i++) {
                product *= values[i];
            }
        }
        return product;
    }
    
    /**
     * min of all values. If supplied array is zero length,
     * Double.POSITIVE_INFINITY is returned.
     * @param values
     */
    public static double min(double[] values) {
        double min = Double.POSITIVE_INFINITY;
        for (int i=0, iSize=values.length; i<iSize; i++) {
            min = Math.min(min, values[i]);
        }
        return min;
    }

    /**
     * min of all values. If supplied array is zero length,
     * Double.NEGATIVE_INFINITY is returned.
     * @param values
     */
    public static double max(double[] values) {
        double max = Double.NEGATIVE_INFINITY;
        for (int i=0, iSize=values.length; i<iSize; i++) {
            max = Math.max(max, values[i]);
        }
        return max;
    }

    /**
     * Note: this function is different from java.lang.Math.floor(..).
     * <p>
     * When n and s are "valid" arguments, the returned value is: Math.floor(n/s) * s;
     * <br/>
     * n and s are invalid if any of following conditions are true:
     * <ul>
     * <li>s is zero</li>
     * <li>n is negative and s is positive</li>
     * <li>n is positive and s is negative</li>
     * </ul>
     * In all such cases, Double.NaN is returned.
     * @param n
     * @param s
     */
    public static double floor(double n, double s) {
        double f;
        
        if ((n<0 && s>0) || (n>0 && s<0) || (s==0 && n!=0)) {
            f = Double.NaN;
        }
        else {
            f = (n==0 || s==0) ? 0 : Math.floor(n/s) * s;
        }
        
        return f;
    }
    
    /**
     * Note: this function is different from java.lang.Math.ceil(..).
     * <p>
     * When n and s are "valid" arguments, the returned value is: Math.ceiling(n/s) * s;
     * <br/>
     * n and s are invalid if any of following conditions are true:
     * <ul>
     * <li>s is zero</li>
     * <li>n is negative and s is positive</li>
     * <li>n is positive and s is negative</li>
     * </ul>
     * In all such cases, Double.NaN is returned.
     * @param n
     * @param s
     */
    public static double ceiling(double n, double s) {
        double c;
        
        if ((n<0 && s>0) || (n>0 && s<0)) {
            c = Double.NaN;
        }
        else {
            c = (n == 0 || s == 0) ? 0 : Math.ceil(n/s) * s;
        }
        
        return c;
    }
    
    /**
     * <br/> for all n >= 1; factorial n = n * (n-1) * (n-2) * ... * 1 
     * <br/> else if n == 0; factorial n = 1
     * <br/> else if n < 0; factorial n = Double.NaN
     * <br/> Loss of precision can occur if n is large enough.
     * If n is large so that the resulting value would be greater 
     * than Double.MAX_VALUE; Double.POSITIVE_INFINITY is returned.
     * If n < 0, Double.NaN is returned. 
     * @param n
     */
    public static double factorial(int n) {
        double d = 1;
        
        if (n >= 0) {
            if (n <= 170) {
                for (int i=1; i<=n; i++) {
                    d *= i;
                }
            }
            else {
                d = Double.POSITIVE_INFINITY;
            }
        }
        else {
            d = Double.NaN;
        }
        return d;
    }
    

    /**
     * returns the remainder resulting from operation:
     * n / d. 
     * <br/> The result has the sign of the divisor.
     * <br/> Examples:
     * <ul>
     * <li>mod(3.4, 2) = 1.4</li>
     * <li>mod(-3.4, 2) = 0.6</li>
     * <li>mod(-3.4, -2) = -1.4</li>
     * <li>mod(3.4, -2) = -0.6</li>
     * </ul>
     * If d == 0, result is NaN
     * @param n
     * @param d
     */
    public static double mod(double n, double d) {
        double result = 0;
        
        if (d == 0) {
            result = Double.NaN;
        }
        else if (sign(n) == sign(d)) {
            double t = Math.abs(n / d);
            t = t - (long) t;
            result = sign(d) * Math.abs(t * d);
        }
        else {
            double t = Math.abs(n / d);
            t = t - (long) t;
            t = Math.ceil(t) - t;
            result = sign(d) * Math.abs(t * d);
        }
        
        return result;
    }
    
    
    /**
     * inverse hyperbolic cosine
     * @param d
     */
    public static double acosh(double d) {
        return Math.log(Math.sqrt(Math.pow(d, 2) - 1) + d);
    }
    
    /**
     * inverse hyperbolic sine
     * @param d
     */
    public static double asinh(double d) {
        double d2 = d*d;
        return Math.log(Math.sqrt(d*d + 1) + d);
    }
    
    /**
     * inverse hyperbolic tangent
     * @param d
     */
    public static double atanh(double d) {
        return Math.log((1 + d)/(1 - d)) / 2;
    }
    
    /**
     * hyperbolic cosine
     * @param d
     */
    public static double cosh(double d) {
        double ePowX = Math.pow(Math.E, d);
        double ePowNegX = Math.pow(Math.E, -d);
        d = (ePowX + ePowNegX) / 2;
        return d;
    }
    
    /**
     * hyperbolic sine
     * @param d
     */
    public static double sinh(double d) {
        double ePowX = Math.pow(Math.E, d);
        double ePowNegX = Math.pow(Math.E, -d);
        d = (ePowX - ePowNegX) / 2;
        return d;
    }
    
    /**
     * hyperbolic tangent
     * @param d
     */
    public static double tanh(double d) {
        double ePowX = Math.pow(Math.E, d);
        double ePowNegX = Math.pow(Math.E, -d);
        d = (ePowX - ePowNegX) / (ePowX + ePowNegX);
        return d;
    }
    
    /**
     * returns the sum of product of corresponding double value in each
     * subarray. It is the responsibility of the caller to ensure that
     * all the subarrays are of equal length. If the subarrays are
     * not of equal length, the return value can be unpredictable.
     * @param arrays
     */
    public static double sumproduct(double[][] arrays) {
        double d = 0;
        
        try {
            int narr = arrays.length;
            int arrlen = arrays[0].length; 
            
            for (int j=0; j<arrlen; j++) {
                double t = 1;
                for (int i=0; i<narr; i++) {
                    t *= arrays[i][j];
                }
                d += t;
            }            
        }
        catch (ArrayIndexOutOfBoundsException ae) {
            d = Double.NaN;
        }
        
        return d;
    }
    
    
    /**
     * returns the total number of combinations possible when
     * k items are chosen out of total of n items. If the number
     * is too large, loss of precision may occur (since returned
     * value is double). If the returned value is larger than
     * Double.MAX_VALUE, Double.POSITIVE_INFINITY is returned.
     * If either of the parameters is negative, Double.NaN is returned.
     * @param n
     * @param k
     */
    public static double nChooseK(int n, int k) {
        double d = 1;
        if (n<0 || k<0 || n<k) {
            d= Double.NaN;
        }
        else {
            int minnk = Math.min(n-k, k);
            int maxnk = Math.max(n-k, k);
            for (int i=maxnk; i<n; i++) {
                d *= i+1;
            }
            d /= factorial(minnk);
        }
        
        return d;
    }
    
}