你知道一个.net库来执行LOESS / LOWESS回归吗？ （最好是免费/开源）
答案 0 :(得分：16)
从java到c＃
的端口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
* <a href="http://en.wikipedia.org/wiki/Local_regression#Weight_function">tricube</a>
* 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]));
}
}
}
}
答案 1 :(得分：2)
由于我无法评论其他人的帖子（新用户），人们似乎认为我应该这样做而不是编辑上面的答案，我只是去把它写成答案即使我知道这作为评论更好。
上述答案中的updateBandwidthInterval方法忘记检查方法注释中写入的左侧。这可以为sumWeights提供NaN问题。以下应该解决这个问题。我在做基于上面的c ++实现时遇到了这个问题。
/**
* 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]++;
}
}