/*
* 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
* @return
*/
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
* @return
*/
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
* @return
*/
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
* @return
*/
public static short sign(double d) {
return (short) ((d == 0)
? 0
: (d < 0)
? -1
: 1);
}
/**
* average of all values
* @param values
* @return
*/
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
* @return
*/
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
* @return
*/
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
* @return
*/
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
* @return
*/
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
* @return
*/
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
* @return
*/
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
* @return
*/
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
* @return
*/
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
* @return
*/
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
* @return
*/
public static double acosh(double d) {
return Math.log(Math.sqrt(Math.pow(d, 2) - 1) + d);
}
/**
* inverse hyperbolic sine
* @param d
* @return
*/
public static double asinh(double d) {
double d2 = d*d;
return Math.log(Math.sqrt(d*d + 1) + d);
}
/**
* inverse hyperbolic tangent
* @param d
* @return
*/
public static double atanh(double d) {
return Math.log((1 + d)/(1 - d)) / 2;
}
/**
* hyperbolic cosine
* @param d
* @return
*/
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
* @return
*/
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
* @return
*/
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
* @return
*/
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 sum of difference of squares of corresponding double
* value in each subarray: ie. sigma (xarr[i]^2-yarr[i]^2)
* <br/>
* It is the responsibility of the caller
* to ensure that the two subarrays are of equal length. If the
* subarrays are not of equal length, the return value can be
* unpredictable.
* @param arrays
* @return
*/
public static double sumx2my2(double[] xarr, double[] yarr) {
double d = 0;
try {
for (int i=0, iSize=xarr.length; i<iSize; i++) {
d += (xarr[i] + yarr[i]) * (xarr[i] - yarr[i]);
}
}
catch (ArrayIndexOutOfBoundsException ae) {
d = Double.NaN;
}
return d;
}
/**
* returns the sum of sum of squares of corresponding double
* value in each subarray: ie. sigma (xarr[i]^2 + yarr[i]^2)
* <br/>
* It is the responsibility of the caller
* to ensure that the two subarrays are of equal length. If the
* subarrays are not of equal length, the return value can be
* unpredictable.
* @param arrays
* @return
*/
public static double sumx2py2(double[] xarr, double[] yarr) {
double d = 0;
try {
for (int i=0, iSize=xarr.length; i<iSize; i++) {
d += (xarr[i] * xarr[i]) + (yarr[i] * yarr[i]);
}
}
catch (ArrayIndexOutOfBoundsException ae) {
d = Double.NaN;
}
return d;
}
/**
* returns the sum of squares of difference of corresponding double
* value in each subarray: ie. sigma ( (xarr[i]-yarr[i])^2 )
* <br/>
* It is the responsibility of the caller
* to ensure that the two subarrays are of equal length. If the
* subarrays are not of equal length, the return value can be
* unpredictable.
* @param arrays
* @return
*/
public static double sumxmy2(double[] xarr, double[] yarr) {
double d = 0;
try {
for (int i=0, iSize=xarr.length; i<iSize; i++) {
double t = (xarr[i] - yarr[i]);
d += t * 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
* @return
*/
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;
}
}
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