Related
I am able to calculate the distance between two points(Latitude and Longtitude) by using the below function
public static double GetDistance(double lat1, double lng1, double lat2, double lng2)
{
double sLatitude = 48.672309;
double sLongitude = 15.695585;
double eLatitude = 48.237867;
double eLongitude = 16.389477;
var coordinate1 = new GeoCoordinate(lat1, lng1);
var coordinate2 = new GeoCoordinate(lat2, lng2);
var resultInMeter = coordinate1.GetDistanceTo(coordinate2); //in meters
//convert to KM : 1 meter = 0.001 KM
var resultInKM = resultInMeter * 0.001; //in KM's
return resultInKM ;
}
this works as under
double lat1 = 48.672309;
double lng1 = 15.695585;
double lat2 = 48.237867;
double lng2 = 16.389477;
var distance = GetDistance(lat1,lng1,lat2,lng2);
Now say I have a collection of Latitudes and Longtitudes as under
double[,] arrLatLong = new double[,]
{
{22.57891304, 88.34285622},
{22.54398196, 88.38221001},
{22.58277011, 88.38303798},
{22.59950095, 88.39352995},
{22.59887647, 88.32905243},
};
How to calculate the total distance? Is there any recursive way?
Trying to use recursion will only make this problem more difficult. Plain old iteration is more than sufficient.
double dist = 0.0;
for (int i = 0; i < arrLatLong.GetLength(0) - 1; i++)
{
dist += GetDistance(
arrLatLong[i, 0],
arrLatLong[i, 1],
arrLatLong[i+1, 0],
arrLatLong[i+1, 1]);
}
If you absolutely must use a recursive method, this should do the trick:
double GetDistanceRecursive(double[,] coords, int idx)
{
if (idx + 1 >= coords.Length)
return 0.0;
double dist = GetDistance(
coords[idx, 0],
coords[idx, 1],
coords[idx+1, 0],
coords[idx+1, 1]);
dist += GetDistanceRecursive(coords, idx + 1);
return dist;
}
I am trying to convert a C++ class to C# and in the process learn something of C++. I had never run into a vector<> before and my understanding is this is like a List<> function in C#. During the conversion of the class I re-wrote the code using List futures_price = New List(Convert.ToInt32(no_steps) + 1);. As soon as I run the code, I get a "Index was out of range" error.
Having looked around on SOF, I believe the issue is regarding the parameter being out of index range relating to this, but I do not see a simple solution to solve this with the below code.
In particular, this is the line that is triggering the error: futures_prices[0] = spot_price * Math.Pow(d, no_steps);
Below is the full code:
public double futures_option_price_call_american_binomial(double spot_price, double option_strike, double r, double sigma, double time, double no_steps)
{
//double spot_price, // price futures contract
//double option_strike, // exercise price
//double r, // interest rate
//double sigma, // volatility
//double time, // time to maturity
//int no_steps
List<double> futures_prices = new List<double>(Convert.ToInt32(no_steps) + 1);
//(no_steps+1);
//double call_values = (no_steps+1);
List<double> call_values = new List<double>(Convert.ToInt32(no_steps) + 1);
double t_delta = time/no_steps;
double Rinv = Math.Exp(-r*(t_delta));
double u = Math.Exp(sigma * Math.Sqrt(t_delta));
double d = 1.0/u;
double uu= u*u;
double pUp = (1-d)/(u-d); // note how probability is calculated
double pDown = 1.0 - pUp;
futures_prices[0] = spot_price * Math.Pow(d, no_steps);
int i;
for (i=1; i<=no_steps; ++i) futures_prices[i] = uu*futures_prices[i-1]; // terminal tree nodes
for (i=0; i<=no_steps; ++i) call_values[i] = Math.Max(0.0, (futures_prices[i]-option_strike));
for (int step = Convert.ToInt32(no_steps) - 1; step >= 0; --step)
{
for (i = 0; i <= step; ++i)
{
futures_prices[i] = d * futures_prices[i + 1];
call_values[i] = (pDown * call_values[i] + pUp * call_values[i + 1]) * Rinv;
call_values[i] = Math.Max(call_values[i], futures_prices[i] - option_strike); // check for exercise
};
};
return call_values[0];
}
Here is the original source in C++:
double futures_option_price_call_american_binomial(const double& F, // price futures contract
const double& K, // exercise price
const double& r, // interest rate
const double& sigma, // volatility
const double& time, // time to maturity
const int& no_steps) { // number of steps
vector<double> futures_prices(no_steps+1);
vector<double> call_values (no_steps+1);
double t_delta= time/no_steps;
double Rinv = exp(-r*(t_delta));
double u = exp(sigma*sqrt(t_delta));
double d = 1.0/u;
double uu= u*u;
double pUp = (1-d)/(u-d); // note how probability is calculated
double pDown = 1.0 - pUp;
futures_prices[0] = F*pow(d, no_steps);
int i;
for (i=1; i<=no_steps; ++i) futures_prices[i] = uu*futures_prices[i-1]; // terminal tree nodes
for (i=0; i<=no_steps; ++i) call_values[i] = max(0.0, (futures_prices[i]-K));
for (int step=no_steps-1; step>=0; --step) {
for (i=0; i<=step; ++i) {
futures_prices[i] = d*futures_prices[i+1];
call_values[i] = (pDown*call_values[i]+pUp*call_values[i+1])*Rinv;
call_values[i] = max(call_values[i], futures_prices[i]-K); // check for exercise
};
};
return call_values[0];
};
A List<double> starts out empty until you add items to it. (passing the constructor argument just sets the capacity, preventing costly resizes)
You can't access [0] until you Add() it.
To use it the way you are, use an array instead.
As SLaks says, it's better to use an Array in this situation. C# lists are filled with Add method and values are removed through Remove method... this would be more complicated and memory/performance expensive as you are also replacing values.
public Double FuturesOptionPriceCallAmericanBinomial(Double spotPrice, Double optionStrike, Double r, Double sigma, Double time, Double steps)
{
// Avoid calling Convert multiple times as it can be quite performance expensive.
Int32 stepsInteger = Convert.ToInt32(steps);
Double[] futurePrices = new Double[(stepsInteger + 1)];
Double[] callValues = new Double[(stepsInteger + 1)];
Double tDelta = time / steps;
Double rInv = Math.Exp(-r * (tDelta));
Double u = Math.Exp(sigma * Math.Sqrt(tDelta));
Double d = 1.0 / u;
Double uu = u * u;
Double pUp = (1 - d) / (u - d);
Double pDown = 1.0 - pUp;
futurePrices[0] = spotPrice * Math.Pow(d, steps);
for (Int32 i = 1; i <= steps; ++i)
futurePrices[i] = uu * futurePrices[(i - 1)];
for (Int32 i = 0; i <= steps; ++i)
callValues[i] = Math.Max(0.0, (futurePrices[i] - optionStrike));
for (Int32 step = stepsInteger - 1; step >= 0; --step)
{
for (Int32 i = 0; i <= step; ++i)
{
futurePrices[i] = d * futurePrices[(i + 1)];
callValues[i] = ((pDown * callValues[i]) + (pUp * callValues[i + 1])) * rInv;
callValues[i] = Math.Max(callValues[i], (futurePrices[i] - option_strike));
}
}
return callValues[0];
}
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Do you know of a .net library to perform a LOESS/LOWESS regression? (preferably free/open source)
Port from java to c#
public class LoessInterpolator
{
public static double DEFAULT_BANDWIDTH = 0.3;
public static int DEFAULT_ROBUSTNESS_ITERS = 2;
/**
* The bandwidth parameter: when computing the loess fit at
* a particular point, this fraction of source points closest
* to the current point is taken into account for computing
* a least-squares regression.
*
* A sensible value is usually 0.25 to 0.5.
*/
private double bandwidth;
/**
* The number of robustness iterations parameter: this many
* robustness iterations are done.
*
* A sensible value is usually 0 (just the initial fit without any
* robustness iterations) to 4.
*/
private int robustnessIters;
public LoessInterpolator()
{
this.bandwidth = DEFAULT_BANDWIDTH;
this.robustnessIters = DEFAULT_ROBUSTNESS_ITERS;
}
public LoessInterpolator(double bandwidth, int robustnessIters)
{
if (bandwidth < 0 || bandwidth > 1)
{
throw new ApplicationException(string.Format("bandwidth must be in the interval [0,1], but got {0}", bandwidth));
}
this.bandwidth = bandwidth;
if (robustnessIters < 0)
{
throw new ApplicationException(string.Format("the number of robustness iterations must be non-negative, but got {0}", robustnessIters));
}
this.robustnessIters = robustnessIters;
}
/**
* Compute a loess fit on the data at the original abscissae.
*
* #param xval the arguments for the interpolation points
* #param yval the values for the interpolation points
* #return values of the loess fit at corresponding original abscissae
* #throws MathException if some of the following conditions are false:
* <ul>
* <li> Arguments and values are of the same size that is greater than zero</li>
* <li> The arguments are in a strictly increasing order</li>
* <li> All arguments and values are finite real numbers</li>
* </ul>
*/
public double[] smooth(double[] xval, double[] yval)
{
if (xval.Length != yval.Length)
{
throw new ApplicationException(string.Format("Loess expects the abscissa and ordinate arrays to be of the same size, but got {0} abscisssae and {1} ordinatae", xval.Length, yval.Length));
}
int n = xval.Length;
if (n == 0)
{
throw new ApplicationException("Loess expects at least 1 point");
}
checkAllFiniteReal(xval, true);
checkAllFiniteReal(yval, false);
checkStrictlyIncreasing(xval);
if (n == 1)
{
return new double[] { yval[0] };
}
if (n == 2)
{
return new double[] { yval[0], yval[1] };
}
int bandwidthInPoints = (int)(bandwidth * n);
if (bandwidthInPoints < 2)
{
throw new ApplicationException(string.Format("the bandwidth must be large enough to accomodate at least 2 points. There are {0} " +
" data points, and bandwidth must be at least {1} but it is only {2}",
n, 2.0 / n, bandwidth
));
}
double[] res = new double[n];
double[] residuals = new double[n];
double[] sortedResiduals = new double[n];
double[] robustnessWeights = new double[n];
// Do an initial fit and 'robustnessIters' robustness iterations.
// This is equivalent to doing 'robustnessIters+1' robustness iterations
// starting with all robustness weights set to 1.
for (int i = 0; i < robustnessWeights.Length; i++) robustnessWeights[i] = 1;
for (int iter = 0; iter <= robustnessIters; ++iter)
{
int[] bandwidthInterval = { 0, bandwidthInPoints - 1 };
// At each x, compute a local weighted linear regression
for (int i = 0; i < n; ++i)
{
double x = xval[i];
// Find out the interval of source points on which
// a regression is to be made.
if (i > 0)
{
updateBandwidthInterval(xval, i, bandwidthInterval);
}
int ileft = bandwidthInterval[0];
int iright = bandwidthInterval[1];
// Compute the point of the bandwidth interval that is
// farthest from x
int edge;
if (xval[i] - xval[ileft] > xval[iright] - xval[i])
{
edge = ileft;
}
else
{
edge = iright;
}
// Compute a least-squares linear fit weighted by
// the product of robustness weights and the tricube
// weight function.
// See http://en.wikipedia.org/wiki/Linear_regression
// (section "Univariate linear case")
// and http://en.wikipedia.org/wiki/Weighted_least_squares
// (section "Weighted least squares")
double sumWeights = 0;
double sumX = 0, sumXSquared = 0, sumY = 0, sumXY = 0;
double denom = Math.Abs(1.0 / (xval[edge] - x));
for (int k = ileft; k <= iright; ++k)
{
double xk = xval[k];
double yk = yval[k];
double dist;
if (k < i)
{
dist = (x - xk);
}
else
{
dist = (xk - x);
}
double w = tricube(dist * denom) * robustnessWeights[k];
double xkw = xk * w;
sumWeights += w;
sumX += xkw;
sumXSquared += xk * xkw;
sumY += yk * w;
sumXY += yk * xkw;
}
double meanX = sumX / sumWeights;
double meanY = sumY / sumWeights;
double meanXY = sumXY / sumWeights;
double meanXSquared = sumXSquared / sumWeights;
double beta;
if (meanXSquared == meanX * meanX)
{
beta = 0;
}
else
{
beta = (meanXY - meanX * meanY) / (meanXSquared - meanX * meanX);
}
double alpha = meanY - beta * meanX;
res[i] = beta * x + alpha;
residuals[i] = Math.Abs(yval[i] - res[i]);
}
// No need to recompute the robustness weights at the last
// iteration, they won't be needed anymore
if (iter == robustnessIters)
{
break;
}
// Recompute the robustness weights.
// Find the median residual.
// An arraycopy and a sort are completely tractable here,
// because the preceding loop is a lot more expensive
System.Array.Copy(residuals, sortedResiduals, n);
//System.arraycopy(residuals, 0, sortedResiduals, 0, n);
Array.Sort<double>(sortedResiduals);
double medianResidual = sortedResiduals[n / 2];
if (medianResidual == 0)
{
break;
}
for (int i = 0; i < n; ++i)
{
double arg = residuals[i] / (6 * medianResidual);
robustnessWeights[i] = (arg >= 1) ? 0 : Math.Pow(1 - arg * arg, 2);
}
}
return res;
}
/**
* Given an index interval into xval that embraces a certain number of
* points closest to xval[i-1], update the interval so that it embraces
* the same number of points closest to xval[i]
*
* #param xval arguments array
* #param i the index around which the new interval should be computed
* #param bandwidthInterval a two-element array {left, right} such that: <p/>
* <tt>(left==0 or xval[i] - xval[left-1] > xval[right] - xval[i])</tt>
* <p/> and also <p/>
* <tt>(right==xval.length-1 or xval[right+1] - xval[i] > xval[i] - xval[left])</tt>.
* The array will be updated.
*/
private static void updateBandwidthInterval(double[] xval, int i, int[] bandwidthInterval)
{
int left = bandwidthInterval[0];
int right = bandwidthInterval[1];
// The right edge should be adjusted if the next point to the right
// is closer to xval[i] than the leftmost point of the current interval
int nextRight = nextNonzero(weights, right);
if (nextRight < xval.Length && xval[nextRight] - xval[i] < xval[i] - xval[left])
{
int nextLeft = nextNonzero(weights, bandwidthInterval[0]);
bandwidthInterval[0] = nextLeft;
bandwidthInterval[1] = nextRight;
}
}
/**
* Compute the
* tricube
* weight function
*
* #param x the argument
* #return (1-|x|^3)^3
*/
private static double tricube(double x)
{
double tmp = Math.abs(x);
tmp = 1 - tmp * tmp * tmp;
return tmp * tmp * tmp;
}
/**
* Check that all elements of an array are finite real numbers.
*
* #param values the values array
* #param isAbscissae if true, elements are abscissae otherwise they are ordinatae
* #throws MathException if one of the values is not
* a finite real number
*/
private static void checkAllFiniteReal(double[] values, bool isAbscissae)
{
for (int i = 0; i < values.Length; i++)
{
double x = values[i];
if (Double.IsInfinity(x) || Double.IsNaN(x))
{
string pattern = isAbscissae ?
"all abscissae must be finite real numbers, but {0}-th is {1}" :
"all ordinatae must be finite real numbers, but {0}-th is {1}";
throw new ApplicationException(string.Format(pattern, i, x));
}
}
}
/**
* Check that elements of the abscissae array are in a strictly
* increasing order.
*
* #param xval the abscissae array
* #throws MathException if the abscissae array
* is not in a strictly increasing order
*/
private static void checkStrictlyIncreasing(double[] xval)
{
for (int i = 0; i < xval.Length; ++i)
{
if (i >= 1 && xval[i - 1] >= xval[i])
{
throw new ApplicationException(string.Format(
"the abscissae array must be sorted in a strictly " +
"increasing order, but the {0}-th element is {1} " +
"whereas {2}-th is {3}",
i - 1, xval[i - 1], i, xval[i]));
}
}
}
}
Since I'm unable to comment on other people's posts (new user), and people seem to think I should do that with this instead of editing the above answer, I'm simply going to write it as an answer even though I know this is better as a comment.
The updateBandwidthInterval method in the above answer forgets to check the left side as written in the method comment. This can give NaN issues for sumWeights. The below should fix that. I encountered this when doing a c++ implementation based on the above.
/**
* Given an index interval into xval that embraces a certain number of
* points closest to xval[i-1], update the interval so that it embraces
* the same number of points closest to xval[i]
*
* #param xval arguments array
* #param i the index around which the new interval should be computed
* #param bandwidthInterval a two-element array {left, right} such that: <p/>
* <tt>(left==0 or xval[i] - xval[left-1] > xval[right] - xval[i])</tt>
* <p/> and also <p/>
* <tt>(right==xval.length-1 or xval[right+1] - xval[i] > xval[i] - xval[left])</tt>.
* The array will be updated.
*/
private static void updateBandwidthInterval(double[] xval, int i, int[] bandwidthInterval)
{
int left = bandwidthInterval[0];
int right = bandwidthInterval[1];
// The edges should be adjusted if the previous point to the
// left is closer to x than the current point to the right or
// if the next point to the right is closer
// to x than the leftmost point of the current interval
if (left != 0 &&
xval[i] - xval[left - 1] < xval[right] - xval[i])
{
bandwidthInterval[0]++;
bandwidthInterval[1]++;
}
else if (right < xval.Length - 1 &&
xval[right + 1] - xval[i] < xval[i] - xval[left])
{
bandwidthInterval[0]++;
bandwidthInterval[1]++;
}
}
Hope someone after 5 years find this useful. This is the original code posted by Tutcugil but with the missing methods and updated.
using System;
using System.Linq;
namespace StockCorrelation
{
public class LoessInterpolator
{
public static double DEFAULT_BANDWIDTH = 0.3;
public static int DEFAULT_ROBUSTNESS_ITERS = 2;
/**
* The bandwidth parameter: when computing the loess fit at
* a particular point, this fraction of source points closest
* to the current point is taken into account for computing
* a least-squares regression.
*
* A sensible value is usually 0.25 to 0.5.
*/
private double bandwidth;
/**
* The number of robustness iterations parameter: this many
* robustness iterations are done.
*
* A sensible value is usually 0 (just the initial fit without any
* robustness iterations) to 4.
*/
private int robustnessIters;
public LoessInterpolator()
{
this.bandwidth = DEFAULT_BANDWIDTH;
this.robustnessIters = DEFAULT_ROBUSTNESS_ITERS;
}
public LoessInterpolator(double bandwidth, int robustnessIters)
{
if (bandwidth < 0 || bandwidth > 1)
{
throw new ApplicationException(string.Format("bandwidth must be in the interval [0,1], but got {0}", bandwidth));
}
this.bandwidth = bandwidth;
if (robustnessIters < 0)
{
throw new ApplicationException(string.Format("the number of robustness iterations must be non-negative, but got {0}", robustnessIters));
}
this.robustnessIters = robustnessIters;
}
/**
* Compute a loess fit on the data at the original abscissae.
*
* #param xval the arguments for the interpolation points
* #param yval the values for the interpolation points
* #return values of the loess fit at corresponding original abscissae
* #throws MathException if some of the following conditions are false:
* <ul>
* <li> Arguments and values are of the same size that is greater than zero</li>
* <li> The arguments are in a strictly increasing order</li>
* <li> All arguments and values are finite real numbers</li>
* </ul>
*/
public double[] smooth(double[] xval, double[] yval, double[] weights)
{
if (xval.Length != yval.Length)
{
throw new ApplicationException(string.Format("Loess expects the abscissa and ordinate arrays to be of the same size, but got {0} abscisssae and {1} ordinatae", xval.Length, yval.Length));
}
int n = xval.Length;
if (n == 0)
{
throw new ApplicationException("Loess expects at least 1 point");
}
checkAllFiniteReal(xval, true);
checkAllFiniteReal(yval, false);
checkStrictlyIncreasing(xval);
if (n == 1)
{
return new double[] { yval[0] };
}
if (n == 2)
{
return new double[] { yval[0], yval[1] };
}
int bandwidthInPoints = (int)(bandwidth * n);
if (bandwidthInPoints < 2)
{
throw new ApplicationException(string.Format("the bandwidth must be large enough to accomodate at least 2 points. There are {0} " +
" data points, and bandwidth must be at least {1} but it is only {2}",
n, 2.0 / n, bandwidth
));
}
double[] res = new double[n];
double[] residuals = new double[n];
double[] sortedResiduals = new double[n];
double[] robustnessWeights = new double[n];
// Do an initial fit and 'robustnessIters' robustness iterations.
// This is equivalent to doing 'robustnessIters+1' robustness iterations
// starting with all robustness weights set to 1.
for (int i = 0; i < robustnessWeights.Length; i++) robustnessWeights[i] = 1;
for (int iter = 0; iter <= robustnessIters; ++iter)
{
int[] bandwidthInterval = { 0, bandwidthInPoints - 1 };
// At each x, compute a local weighted linear regression
for (int i = 0; i < n; ++i)
{
double x = xval[i];
// Find out the interval of source points on which
// a regression is to be made.
if (i > 0)
{
updateBandwidthInterval(xval, weights, i, bandwidthInterval);
}
int ileft = bandwidthInterval[0];
int iright = bandwidthInterval[1];
// Compute the point of the bandwidth interval that is
// farthest from x
int edge;
if (xval[i] - xval[ileft] > xval[iright] - xval[i])
{
edge = ileft;
}
else
{
edge = iright;
}
// Compute a least-squares linear fit weighted by
// the product of robustness weights and the tricube
// weight function.
// See http://en.wikipedia.org/wiki/Linear_regression
// (section "Univariate linear case")
// and http://en.wikipedia.org/wiki/Weighted_least_squares
// (section "Weighted least squares")
double sumWeights = 0;
double sumX = 0, sumXSquared = 0, sumY = 0, sumXY = 0;
double denom = Math.Abs(1.0 / (xval[edge] - x));
for (int k = ileft; k <= iright; ++k)
{
double xk = xval[k];
double yk = yval[k];
double dist;
if (k < i)
{
dist = (x - xk);
}
else
{
dist = (xk - x);
}
double w = tricube(dist * denom) * robustnessWeights[k];
double xkw = xk * w;
sumWeights += w;
sumX += xkw;
sumXSquared += xk * xkw;
sumY += yk * w;
sumXY += yk * xkw;
}
double meanX = sumX / sumWeights;
double meanY = sumY / sumWeights;
double meanXY = sumXY / sumWeights;
double meanXSquared = sumXSquared / sumWeights;
double beta;
if (meanXSquared == meanX * meanX)
{
beta = 0;
}
else
{
beta = (meanXY - meanX * meanY) / (meanXSquared - meanX * meanX);
}
double alpha = meanY - beta * meanX;
res[i] = beta * x + alpha;
residuals[i] = Math.Abs(yval[i] - res[i]);
}
// No need to recompute the robustness weights at the last
// iteration, they won't be needed anymore
if (iter == robustnessIters)
{
break;
}
// Recompute the robustness weights.
// Find the median residual.
// An arraycopy and a sort are completely tractable here,
// because the preceding loop is a lot more expensive
System.Array.Copy(residuals, sortedResiduals, n);
//System.arraycopy(residuals, 0, sortedResiduals, 0, n);
Array.Sort<double>(sortedResiduals);
double medianResidual = sortedResiduals[n / 2];
if (medianResidual == 0)
{
break;
}
for (int i = 0; i < n; ++i)
{
double arg = residuals[i] / (6 * medianResidual);
robustnessWeights[i] = (arg >= 1) ? 0 : Math.Pow(1 - arg * arg, 2);
}
}
return res;
}
public double[] smooth(double[] xval, double[] yval)
{
if (xval.Length != yval.Length)
{
throw new Exception($"xval and yval len are different");
}
double[] unitWeights = Enumerable.Repeat(1.0, xval.Length).ToArray();
return smooth(xval, yval, unitWeights);
}
/**
* Given an index interval into xval that embraces a certain number of
* points closest to xval[i-1], update the interval so that it embraces
* the same number of points closest to xval[i]
*
* #param xval arguments array
* #param i the index around which the new interval should be computed
* #param bandwidthInterval a two-element array {left, right} such that: <p/>
* <tt>(left==0 or xval[i] - xval[left-1] > xval[right] - xval[i])</tt>
* <p/> and also <p/>
* <tt>(right==xval.length-1 or xval[right+1] - xval[i] > xval[i] - xval[left])</tt>.
* The array will be updated.
*/
private static void updateBandwidthInterval(double[] xval, double[] weights,
int i,
int[] bandwidthInterval)
{
int left = bandwidthInterval[0];
int right = bandwidthInterval[1];
// The right edge should be adjusted if the next point to the right
// is closer to xval[i] than the leftmost point of the current interval
int nextRight = nextNonzero(weights, right);
if (nextRight < xval.Length && xval[nextRight] - xval[i] < xval[i] - xval[left])
{
int nextLeft = nextNonzero(weights, bandwidthInterval[0]);
bandwidthInterval[0] = nextLeft;
bandwidthInterval[1] = nextRight;
}
}
private static int nextNonzero(double[] weights, int i)
{
int j = i + 1;
while (j < weights.Length && weights[j] == 0)
{
++j;
}
return j;
}
/**
* Compute the
* tricube
* weight function
*
* #param x the argument
* #return (1-|x|^3)^3
*/
private static double tricube(double x)
{
double tmp = Math.Abs(x);
tmp = 1 - tmp * tmp * tmp;
return tmp * tmp * tmp;
}
/**
* Check that all elements of an array are finite real numbers.
*
* #param values the values array
* #param isAbscissae if true, elements are abscissae otherwise they are ordinatae
* #throws MathException if one of the values is not
* a finite real number
*/
private static void checkAllFiniteReal(double[] values, bool isAbscissae)
{
for (int i = 0; i < values.Length; i++)
{
double x = values[i];
if (Double.IsInfinity(x) || Double.IsNaN(x))
{
string pattern = isAbscissae ?
"all abscissae must be finite real numbers, but {0}-th is {1}" :
"all ordinatae must be finite real numbers, but {0}-th is {1}";
throw new ApplicationException(string.Format(pattern, i, x));
}
}
}
/**
* Check that elements of the abscissae array are in a strictly
* increasing order.
*
* #param xval the abscissae array
* #throws MathException if the abscissae array
* is not in a strictly increasing order
*/
private static void checkStrictlyIncreasing(double[] xval)
{
for (int i = 0; i < xval.Length; ++i)
{
if (i >= 1 && xval[i - 1] >= xval[i])
{
throw new ApplicationException(string.Format(
"the abscissae array must be sorted in a strictly " +
"increasing order, but the {0}-th element is {1} " +
"whereas {2}-th is {3}",
i - 1, xval[i - 1], i, xval[i]));
}
}
}
}
}
Trying to find functions that will assist us to draw a 3D line through a series of points.
For each point we know: Date&Time, Latitude, Longitude, Altitude, Speed and Heading.
Data might be recorded every 10 seconds and we would like to be able to guestimate the points in between and increase granularity to 1 second. Thus creating a virtual flight path in 3D space.
I have found a number of curve fitting algorithms that will approximate a line through a series of points but they do not guarantee that the points are intersected. They also do not take into account speed and heading to determine the most likely path taken by the object to reach the next point.
From a physics viewpoint:
You have to assume something about the acceleration in your intermediate points to get the interpolation.
If your physical system is relatively well-behaved (as a car or a plane), as opposed to for example a bouncing ball, you may go ahead supposing an acceleration varying linearly with time between your points.
The vector equation for a constant varying accelerated movement is:
x''[t] = a t + b
where all magnitudes except t are vectors.
For each segment you already know v(t=t0) x(t=t0) tfinal and x(tfinal) v(tfinal)
By solving the differential equation you get:
Eq 1:
x[t_] := (3 b t^2 Tf + a t^3 Tf - 3 b t Tf^2 - a t Tf^3 - 6 t X0 + 6 Tf X0 + 6 t Xf)/(6 Tf)
And imposing the initial and final contraints for position and velocity you get:
Eqs 2:
a -> (6 (Tf^2 V0 - 2 T0 Tf Vf + Tf^2 Vf - 2 T0 X0 + 2 Tf X0 +
2 T0 Xf - 2 Tf Xf))/(Tf^2 (3 T0^2 - 4 T0 Tf + Tf^2))
b -> (2 (-2 Tf^3 V0 + 3 T0^2 Tf Vf - Tf^3 Vf + 3 T0^2 X0 -
3 Tf^2 X0 - 3 T0^2 Xf + 3 Tf^2 Xf))/(Tf^2 (3 T0^2 - 4 T0 Tf + Tf^2))}}
So inserting the values for eqs 2 into eq 1 you get the temporal interpolation for your points, based on the initial and final position and velocities.
HTH!
Edit
A few examples with abrupt velocity change in two dimensions (in 3D is exactly the same). If the initial and final speeds are similar, you'll get "straighter" paths.
Suppose:
X0 = {0, 0}; Xf = {1, 1};
T0 = 0; Tf = 1;
If
V0 = {0, 1}; Vf = {-1, 3};
V0 = {0, 1}; Vf = {-1, 5};
V0 = {0, 1}; Vf = {1, 3};
Here is an animation where you may see the speed changing from V0 = {0, 1} to Vf = {1, 5}:
Here you may see an accelerating body in 3D with positions taken at equal intervals:
Edit
A full problem:
For convenience, I'll work in Cartesian coordinates. If you want to convert from lat/log/alt to Cartesian just do:
x = rho sin(theta) cos(phi)
y = rho sin(theta) sin(phi)
z = rho cos(theta)
Where phi is the longitude, theta is the latitude, and rho is your altitude plus the radius of the Earth.
So suppose we start our segment at:
t=0 with coordinates (0,0,0) and velocity (1,0,0)
and end at
t=10 with coordinates (10,10,10) and velocity (0,0,1)
I clearly made a change in the origin of coordinates to set the origin at my start point. That is just for getting nice round numbers ...
So we replace those numbers in the formulas for a and b and get:
a = {-(3/50), -(3/25), -(3/50)} b = {1/5, 3/5, 2/5}
With those we go to eq 1, and the position of the object is given by:
p[t] = {1/60 (60 t + 6 t^2 - (3 t^3)/5),
1/60 (18 t^2 - (6 t^3)/5),
1/60 (12 t^2 - (3 t^3)/5)}
And that is it. You get the position from 1 to 10 secs replacing t by its valus in the equation above.
The animation runs:
Edit 2
If you don't want to mess with the vertical acceleration (perhaps because your "speedometer" doesn't read it), you could just assign a constant speed to the z axis (there is a very minor error for considering it parallel to the Rho axis), equal to (Zfinal - Zinit)/(Tf-T0), and then solve the problem in the plane forgetting the altitude.
What you're asking is a general interpolation problem. My guess is your actual problem isn't due to the curve-fitting algorithm being used, but rather your application of it to all discrete values recorded by the system instead of the relevant set of values.
Let's decompose your problem. You're currently drawing a point in spherically-mapped 3D space, adjusting for linear and curved paths. If we discretize the operations performed by an object with six degrees of freedom (roll, pitch, and yaw), the only operations you're particularly interested in are linear paths and curved paths accounting for pitch and yaw in any direction. Accounting for acceleration and deceleration also possible given understanding of basic physics.
Dealing with the spherical mapping is easy. Simply unwrap your points relative to their position on a plane, adjusting for latitude, longitude, and altitude. This should allow you to flatten data that would otherwise exist along a curved path, though this may not strictly be necessary for the solutions to your problem (see below).
Linear interpolation is easy. Given an arbitrary number of points backwards in time that fit a line within n error as determined by your system,* construct the line and compute the distance in time between each point. From here, attempt to fit the time points to one of two cases: constant velocity or constant acceleration.
Curve interpolation is a little more difficult, but still plausible. For cases of pitch, yaw, or combined pitch+yaw, construct a plane containing an arbitrary number of points backwards in time, within m error for curved readouts from your system.* From these data, construct a planar curve and once again account for constant velocity or acceleration along the curve.
You can do better than this by attempting to predict the expected operations of a plane in flight as part of a decision tree or neural network relative to the flight path. I'll leave that as an exercise for the reader.
Best of luck designing your system.
--
* Both error readouts are expected to be from GPS data, given the description of the problem. Accounting and adjusting for errors in these data is a separate interesting problem.
What you need (instead of modeling the physics) is to fit a spline through the data. I used a numerical recipies book (http://www.nrbook.com/a has free C and FORTRAN algorithms. Look into F77 section 3.3 to get the math needed). If you want to be simple then just fit lines through the points, but that will not result in a smooth flight path at all. Time will be your x value, and each parameter loged will have it's own cublic spline parameters.
Since we like long postings for this question here is the full code:
//driver program
static void Main(string[] args)
{
double[][] flight_data = new double[][] {
new double[] { 0, 10, 20, 30 }, // time in seconds
new double[] { 14500, 14750, 15000, 15125 }, //altitude in ft
new double[] { 440, 425, 415, 410 }, // speed in knots
};
CubicSpline altitude = new CubicSpline(flight_data[0], flight_data[1]);
CubicSpline speed = new CubicSpline(flight_data[0], flight_data[2]);
double t = 22; //Find values at t
double h = altitude.GetY(t);
double v = speed.GetY(t);
double ascent = altitude.GetYp(t); // ascent rate in ft/s
}
// Cubic spline definition
using System.Linq;
/// <summary>
/// Cubic spline interpolation for tabular data
/// </summary>
/// <remarks>
/// Adapted from numerical recipies for FORTRAN 77
/// (ISBN 0-521-43064-X), page 110, section 3.3.
/// Function spline(x,y,yp1,ypn,y2) converted to
/// C# by jalexiou, 27 November 2007.
/// Spline integration added also Nov 2007.
/// </remarks>
public class CubicSpline
{
double[] xi;
double[] yi;
double[] yp;
double[] ypp;
double[] yppp;
double[] iy;
#region Constructors
public CubicSpline(double x_min, double x_max, double[] y)
: this(Sequence(x_min, x_max, y.Length), y)
{ }
public CubicSpline(double x_min, double x_max, double[] y, double yp1, double ypn)
: this(Sequence(x_min, x_max, y.Length), y, yp1, ypn)
{ }
public CubicSpline(double[] x, double[] y)
: this(x, y, double.NaN, double.NaN)
{ }
public CubicSpline(double[] x, double[] y, double yp1, double ypn)
{
if( x.Length == y.Length )
{
int N = x.Length;
xi = new double[N];
yi = new double[N];
x.CopyTo(xi, 0);
y.CopyTo(yi, 0);
if( N > 0 )
{
double p, qn, sig, un;
ypp = new double[N];
double[] u = new double[N];
if( double.IsNaN(yp1) )
{
ypp[0] = 0;
u[0] = 0;
}
else
{
ypp[0] = -0.5;
u[0] = (3 / (xi[1] - xi[0])) *
((yi[1] - yi[0]) / (x[1] - x[0]) - yp1);
}
for (int i = 1; i < N-1; i++)
{
double hp = x[i] - x[i - 1];
double hn = x[i + 1] - x[i];
sig = hp / hn;
p = sig * ypp[i - 1] + 2.0;
ypp[i] = (sig - 1.0) / p;
u[i] = (6 * ((y[i + 1] - y[i]) / hn) - (y[i] - y[i - 1]) / hp)
/ (hp + hn) - sig * u[i - 1] / p;
}
if( double.IsNaN(ypn) )
{
qn = 0;
un = 0;
}
else
{
qn = 0.5;
un = (3 / (x[N - 1] - x[N - 2])) *
(ypn - (y[N - 1] - y[N - 2]) / (x[N - 1] - x[N - 2]));
}
ypp[N - 1] = (un - qn * u[N - 2]) / (qn * ypp[N - 2] + 1.0);
for (int k = N-2; k > 0; k--)
{
ypp[k] = ypp[k] * ypp[k + 1] + u[k];
}
// Calculate 1st derivatives
yp = new double[N];
double h;
for( int i = 0; i < N - 1; i++ )
{
h = xi[i + 1] - xi[i];
yp[i] = (yi[i + 1] - yi[i]) / h
- h / 6 * (ypp[i + 1] + 2 * ypp[i]);
}
h = xi[N - 1] - xi[N - 2];
yp[N - 1] = (yi[N - 1] - yi[N - 2]) / h
+ h / 6 * (2 * ypp[N - 1] + ypp[N - 2]);
// Calculate 3rd derivatives as average of dYpp/dx
yppp = new double[N];
double[] jerk_ij = new double[N - 1];
for( int i = 0; i < N - 1; i++ )
{
h = xi[i + 1] - xi[i];
jerk_ij[i] = (ypp[i + 1] - ypp[i]) / h;
}
Yppp = new double[N];
yppp[0] = jerk_ij[0];
for( int i = 1; i < N - 1; i++ )
{
yppp[i] = 0.5 * (jerk_ij[i - 1] + jerk_ij[i]);
}
yppp[N - 1] = jerk_ij[N - 2];
// Calculate Integral over areas
iy = new double[N];
yi[0] = 0; //Integration constant
for( int i = 0; i < N - 1; i++ )
{
h = xi[i + 1] - xi[i];
iy[i + 1] = h * (yi[i + 1] + yi[i]) / 2
- h * h * h / 24 * (ypp[i + 1] + ypp[i]);
}
}
else
{
yp = new double[0];
ypp = new double[0];
yppp = new double[0];
iy = new double[0];
}
}
else
throw new IndexOutOfRangeException();
}
#endregion
#region Actions/Functions
public int IndexOf(double x)
{
//Use bisection to find index
int i1 = -1;
int i2 = Xi.Length;
int im;
double x1 = Xi[0];
double xn = Xi[Xi.Length - 1];
bool ascending = (xn >= x1);
while( i2 - i1 > 1 )
{
im = (i1 + i2) / 2;
double xm = Xi[im];
if( ascending & (x >= Xi[im]) ) { i1 = im; } else { i2 = im; }
}
if( (ascending && (x <= x1)) || (!ascending & (x >= x1)) )
{
return 0;
}
else if( (ascending && (x >= xn)) || (!ascending && (x <= xn)) )
{
return Xi.Length - 1;
}
else
{
return i1;
}
}
public double GetIntY(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1 = yi[i];
double y2 = yi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double h2 = h * h;
double a = (x-x1)/h;
double a2 = a*a;
return h / 6 * (3 * a * (2 - a) * y1
+ 3 * a2 * y2 - a2 * h2 * (a2 - 4 * a + 4) / 4 * y1pp
+ a2 * h2 * (a2 - 2) / 4 * y2pp);
}
public double GetY(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1 = yi[i];
double y2 = yi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double h2 = h * h;
double A = 1 - (x - x1) / (x2 - x1);
double B = 1 - A;
return A * y1 + B * y2 + h2 / 6 * (A * (A * A - 1) * y1pp
+ B * (B * B - 1) * y2pp);
}
public double GetYp(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1 = yi[i];
double y2 = yi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double A = 1 - (x - x1) / (x2 - x1);
double B = 1 - A;
return (y2 - y1) / h + h / 6 * (y2pp * (3 * B * B - 1)
- y1pp * (3 * A * A - 1));
}
public double GetYpp(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double A = 1 - (x - x1) / (x2 - x1);
double B = 1 - A;
return A * y1pp + B * y2pp;
}
public double GetYppp(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
return (y2pp - y1pp) / h;
}
public double Integrate(double from_x, double to_x)
{
if( to_x < from_x ) { return -Integrate(to_x, from_x); }
int i = IndexOf(from_x);
int j = IndexOf(to_x);
double x1 = xi[i];
double xn = xi[j];
double z = GetIntY(to_x) - GetIntY(from_x); // go to nearest nodes (k) (j)
for( int k = i + 1; k <= j; k++ )
{
z += iy[k]; // fill-in areas in-between
}
return z;
}
#endregion
#region Properties
public bool IsEmpty { get { return xi.Length == 0; } }
public double[] Xi { get { return xi; } set { xi = value; } }
public double[] Yi { get { return yi; } set { yi = value; } }
public double[] Yp { get { return yp; } set { yp = value; } }
public double[] Ypp { get { return ypp; } set { ypp = value; } }
public double[] Yppp { get { return yppp; } set { yppp = value; } }
public double[] IntY { get { return yp; } set { iy = value; } }
public int Count { get { return xi.Length; } }
public double X_min { get { return xi.Min(); } }
public double X_max { get { return xi.Max(); } }
public double Y_min { get { return yi.Min(); } }
public double Y_max { get { return yi.Max(); } }
#endregion
#region Helpers
static double[] Sequence(double x_min, double x_max, int double_of_points)
{
double[] res = new double[double_of_points];
for (int i = 0; i < double_of_points; i++)
{
res[i] = x_min + (double)i / (double)(double_of_points - 1) * (x_max - x_min);
}
return res;
}
#endregion
}
You can find an approximation of a line that intersects points in 3d and 2d space using a Hough Transformation algorithm. I am only familiar with it's uses in 2d however but it will still work for 3d spaces given that you know what kind of line you are looking for. There is a basic implementation description linked. You can Google for pre-mades and here is a link to a 2d C implementation CImg.
The algorithm process (roughly)... First you find equation of a line that you think will best approximate the shape of the line (in 2d parabolic, logarithmic, exponential, etc). You take that formula and solve for one of the parameters.
y = ax + b
becomes
b = y - ax
Next, for each point you are attempting to match, you plugin the points to the y and x values. With 3 points, you would have 3 separate functions of b with respect to a.
(2, 3) : b = 3 - 2a
(4, 1) : b = 1 - 4a
(10, -5): b = -5 - 10a
Next, the theory is that you find all possible lines which pass through each of the points, which is infinitely many for each individual point however when combined in an accumulator space only a few possible parameters best fit. In practice this is done by choosing a range space for the parameters (I chose -2 <= a <= 1, 1 <= b <= 6) and begin plugging in values for the variant parameter(s) and solving for the other. You tally up the number of intersections from each function in an accumulator. The points with the highest values give you your parameters.
Accumulator after processing b = 3 - 2a
a: -2 -1 0 1
b: 1
2
3 1
4
5 1
6
Accumulator after processing b = 1 - 4a
a: -2 -1 0 1
b: 1 1
2
3 1
4
4
5 2
6
Accumulator after processing b = -5 - 10a
a: -2 -1 0 1
b: 1 1
2
3 1
4
5 3
6
The parameter set with the highest accumulated value is (b a) = (5 -1) and the function best fit to the points given is y = 5 - x.
Best of luck.
My guess is that a serious application of this would use a http://en.wikipedia.org/wiki/Kalman_filter. By the way, that probably wouldn't guarantee that the reported points were intersected either, unless you fiddled with the parameters a bit. It would expect some degree of error in each data point given to it, so where it thinks the object is at time T would not necessarily be where it was at time T. Of course, you could set the error distribution to say that you were absolutely sure you knew where it was at time T.
Short of using a Kalman filter, I would try and turn it into an optimisation problem. Work at the 1s granularity and write down equations like
x_t' = x_t + (Vx_t + Vx_t')/2 + e,
Vx_t_reported = Vx_t + f,
Vx_t' = Vx_t + g
where e, f, and g represent the noise. Then create a penalty function such as e^2 + f^2 + g^2 +...
or some weighted version such as 1.5e^2 + 3f^2 + 2.6g^2 +... according to your idea of what the errors really are and how smooth you wnat the answer to be, and find the values that make the penalty function as small as possible - with least squares if the equations turn out nicely.
I'm running some Monte Carlo simulations and making extensive use of the Excel function NORM.INV using Office Interrop. This functions takes three arguments (probability, average, standard deviation) and returns the inverse of the cumulative distribution.
I'd like to move my code into a web app, but that will require installing Excel on the server. Does anybody know of a C# statistics library that has an equivalent function to NORM.INV?
Meta.Numerics has exactly what you are looking for. Here is the code to do it using that library:
Distribution n = new NormalDistribution(mean, standardDeviation);
double x = n.InverseLeftProbability(probability);
If you are doing this in order to generate normal deviates, the GetRandomValue function is even faster.
I was also needing a C# implementation of NORMINV, the closest thing I found was a C++ implementation http://www.wilmott.com/messageview.cfm?catid=10&threadid=38771, so I made a quick and dirty translation to C#, the details here http://weblogs.asp.net/esanchez/archive/2010/07/29/a-quick-and-dirty-implementation-of-excel-norminv-function-in-c.aspx. I have made only a few basic tests so be careful if you decide to use it, anyway, hope it helps!
https://numerics.mathdotnet.com/ has a pretty neat looking library that deals with stats (so I assume the CDF), I've not used it so I can't say for definite that it's what you want, but it seems like it should be.
The inverse normal CDF, including coefficients, is described here. And the absolute value of the relative error is less than 1.15 × 10−9
public static class NormalDistributionConfidenceCalculator
{
/// <summary>
///
/// </summary>
public static double InverseNormalDistribution(double probability, double min, double max)
{
double x = 0;
double a = 0;
double b = 1;
double precision = Math.Pow(10, -3);
while ((b - a) > precision)
{
x = (a + b) / 2;
if (NormInv(x) > probability)
{
b = x;
}
else
{
a = x;
}
}
if ((max > 0) && (min > 0))
{
x = x * (max - min) + min;
}
return x;
}
/// <summary>
/// Returns the cumulative density function evaluated at A given value.
/// </summary>
/// <param name="x">A position on the x-axis.</param>
/// <param name="mean"></param>
/// <param name="sigma"></param>
/// <returns>The cumulative density function evaluated at <C>x</C>.</returns>
/// <remarks>The value of the cumulative density function at A point <C>x</C> is
/// probability that the value of A random variable having this normal density is
/// less than or equal to <C>x</C>.
/// </remarks>
public static double NormalDistribution(double x, double mean, double sigma)
{
// This algorithm is ported from dcdflib:
// Cody, W.D. (1993). "ALGORITHM 715: SPECFUN - A Portabel FORTRAN
// Package of Special Function Routines and Test Drivers"
// acm Transactions on Mathematical Software. 19, 22-32.
int i;
double del, xden, xnum, xsq;
double result, ccum;
double arg = (x - mean) / sigma;
const double sixten = 1.60e0;
const double sqrpi = 3.9894228040143267794e-1;
const double thrsh = 0.66291e0;
const double root32 = 5.656854248e0;
const double zero = 0.0e0;
const double min = Double.Epsilon;
double z = arg;
double y = Math.Abs(z);
const double half = 0.5e0;
const double one = 1.0e0;
double[] a =
{
2.2352520354606839287e00, 1.6102823106855587881e02, 1.0676894854603709582e03,
1.8154981253343561249e04, 6.5682337918207449113e-2
};
double[] b =
{
4.7202581904688241870e01, 9.7609855173777669322e02, 1.0260932208618978205e04,
4.5507789335026729956e04
};
double[] c =
{
3.9894151208813466764e-1, 8.8831497943883759412e00, 9.3506656132177855979e01,
5.9727027639480026226e02, 2.4945375852903726711e03, 6.8481904505362823326e03,
1.1602651437647350124e04, 9.8427148383839780218e03, 1.0765576773720192317e-8
};
double[] d =
{
2.2266688044328115691e01, 2.3538790178262499861e02, 1.5193775994075548050e03,
6.4855582982667607550e03, 1.8615571640885098091e04, 3.4900952721145977266e04,
3.8912003286093271411e04, 1.9685429676859990727e04
};
double[] p =
{
2.1589853405795699e-1, 1.274011611602473639e-1, 2.2235277870649807e-2,
1.421619193227893466e-3, 2.9112874951168792e-5, 2.307344176494017303e-2
};
double[] q =
{
1.28426009614491121e00, 4.68238212480865118e-1, 6.59881378689285515e-2,
3.78239633202758244e-3, 7.29751555083966205e-5
};
if (y <= thrsh)
{
//
// Evaluate anorm for |X| <= 0.66291
//
xsq = zero;
if (y > double.Epsilon) xsq = z * z;
xnum = a[4] * xsq;
xden = xsq;
for (i = 0; i < 3; i++)
{
xnum = (xnum + a[i]) * xsq;
xden = (xden + b[i]) * xsq;
}
result = z * (xnum + a[3]) / (xden + b[3]);
double temp = result;
result = half + temp;
}
//
// Evaluate anorm for 0.66291 <= |X| <= sqrt(32)
//
else if (y <= root32)
{
xnum = c[8] * y;
xden = y;
for (i = 0; i < 7; i++)
{
xnum = (xnum + c[i]) * y;
xden = (xden + d[i]) * y;
}
result = (xnum + c[7]) / (xden + d[7]);
xsq = Math.Floor(y * sixten) / sixten;
del = (y - xsq) * (y + xsq);
result = Math.Exp(-(xsq * xsq * half)) * Math.Exp(-(del * half)) * result;
ccum = one - result;
if (z > zero)
{
result = ccum;
}
}
//
// Evaluate anorm for |X| > sqrt(32)
//
else
{
xsq = one / (z * z);
xnum = p[5] * xsq;
xden = xsq;
for (i = 0; i < 4; i++)
{
xnum = (xnum + p[i]) * xsq;
xden = (xden + q[i]) * xsq;
}
result = xsq * (xnum + p[4]) / (xden + q[4]);
result = (sqrpi - result) / y;
xsq = Math.Floor(z * sixten) / sixten;
del = (z - xsq) * (z + xsq);
result = Math.Exp(-(xsq * xsq * half)) * Math.Exp(-(del * half)) * result;
ccum = one - result;
if (z > zero)
{
result = ccum;
}
}
if (result < min)
result = 0.0e0;
return result;
}
/// <summary>
/// Given a probability, a mean, and a standard deviation, an x value can be calculated.
/// </summary>
/// <returns></returns>
public static double NormInv(double probability)
{
const double a1 = -39.6968302866538;
const double a2 = 220.946098424521;
const double a3 = -275.928510446969;
const double a4 = 138.357751867269;
const double a5 = -30.6647980661472;
const double a6 = 2.50662827745924;
const double b1 = -54.4760987982241;
const double b2 = 161.585836858041;
const double b3 = -155.698979859887;
const double b4 = 66.8013118877197;
const double b5 = -13.2806815528857;
const double c1 = -7.78489400243029E-03;
const double c2 = -0.322396458041136;
const double c3 = -2.40075827716184;
const double c4 = -2.54973253934373;
const double c5 = 4.37466414146497;
const double c6 = 2.93816398269878;
const double d1 = 7.78469570904146E-03;
const double d2 = 0.32246712907004;
const double d3 = 2.445134137143;
const double d4 = 3.75440866190742;
//Define break-points
// using Epsilon is wrong; see link above for reference to 0.02425 value
//const double pLow = double.Epsilon;
const double pLow = 0.02425;
const double pHigh = 1 - pLow;
//Define work variables
double q;
double result = 0;
// if argument out of bounds.
// set it to a value within desired precision.
if (probability <= 0)
probability = pLow;
if (probability >= 1)
probability = pHigh;
if (probability < pLow)
{
//Rational approximation for lower region
q = Math.Sqrt(-2 * Math.Log(probability));
result = (((((c1 * q + c2) * q + c3) * q + c4) * q + c5) * q + c6) / ((((d1 * q + d2) * q + d3) * q + d4) * q + 1);
}
else if (probability <= pHigh)
{
//Rational approximation for lower region
q = probability - 0.5;
double r = q * q;
result = (((((a1 * r + a2) * r + a3) * r + a4) * r + a5) * r + a6) * q /
(((((b1 * r + b2) * r + b3) * r + b4) * r + b5) * r + 1);
}
else if (probability < 1)
{
//Rational approximation for upper region
q = Math.Sqrt(-2 * Math.Log(1 - probability));
result = -(((((c1 * q + c2) * q + c3) * q + c4) * q + c5) * q + c6) / ((((d1 * q + d2) * q + d3) * q + d4) * q + 1);
}
return result;
}
/// <summary>
///
/// </summary>
/// <param name="probability"></param>
/// <param name="mean"></param>
/// <param name="sigma"></param>
/// <returns></returns>
public static double NormInv(double probability, double mean, double sigma)
{
double x = NormInv(probability);
return sigma * x + mean;
}
}
I do not know of a library, but found this link - http://home.online.no/~pjacklam/notes/invnorm/ - describing an algoritm. It has implementations in a number of languages, but not C#. You could use the VB.NET version, or port it yourself.
Maybe you can try this component,http://www.smartxls.com,it has an Excel compatible runtime calculation engine and it does not need Excel installed.
add a chart control
double result = Chart1.DataManipulator.Statistics.InverseNormalDistribution(probability);
probability eg : 0.9, 0.4