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Calculate the ticks of an axis for a chart with a stepsize

I've calculated a stepsize for an axis on a chart.
Also I have the Min and Max -Values. Now I need to calculate all ticks so that all values between my Min and Max can be displayed.
For Example:
Stepsize: 1000
Min: 213
Max: 4405
Expected ticks: 0,1000,2000,3000,4000,5000
Stepsize: 500
Min: -1213
Max: 1405
Expected ticks: -1500,-1000,-500,0,500,1000,1500
Until now I'm trying to calculate the first value with "try and error" like:
bool firstStepSet = false;
double firstStep = stepSize;
do
{
if (xValue >= (firstStep - (stepSize / 2)) && xValue <=
(firstStep + (stepSize / 2)))
{
firstStepSet = true;
this.myBarXValues.Add(firstStep, 0);
}
else if (xValue > stepSize)
{
firstStep += stepSize;
}
else
{
firstStep -= stepSize;
}
}
while (!firstStepSet);
And after that I'm adding steps to this list until all values fit.
This seems pretty dirty to me and I want to know if there is another solution.
So what I need is a solution which calculate the first tick that I need.

This function calculates first and last step values:
static void CalcSteps(int min, int max, int stepSize, out int firstStep, out int lastStep)
{
if (min >= 0)
{
firstStep = (min / stepSize) * stepSize;
}
else
{
firstStep = ((min - stepSize + 1) / stepSize) * stepSize;
}
if (max >= 0)
{
lastStep = ((max + stepSize - 1) / stepSize) * stepSize;
}
else
{
lastStep = (max / stepSize) * stepSize;
}
}

You can calculate axis limits using integer rounding to lower and higher values
low = stepsize * (min / stepsize) //integer division needed
high = stepsize * ((max + stepsize - 1) / stepsize)
Example Python code returns limits and number of ticks (one more than interval count)
def getminmax(minn, maxx, step):
low = (minn // step)
high = (maxx + step - 1) // step
ticks = high - low + 1
return low * step, high * step, ticks
print(getminmax(213, 4405, 1000))
print(getminmax(-1213,1405, 500))
(0, 5000, 6)
(-1500, 1500, 7)

Angle Normalization C#

I have an Angle class that has this constructor
public Angle(int deg, // Degrees, minutes, seconds
int min, // (Signs should agree
int sec) // for conventional notation.)
{
/* //Bug degree normalization
while (deg <= -180) deg += 360;
while (deg > Math.PI) deg -= 360;
//correction end */
double seconds = sec + 60 * (min + 60 * deg);
value = seconds * Math.PI / 648000.0;
normalize();
}
and I have these values for testing that constructor
int[] degrees = { 0, 180, -180, Int32.MinValue / 60, 120+180*200000};
int[] minutes = { 0, 0, 0, 0,56};
int[] seconds = { 0, 0, 0, 0,10};
Console.WriteLine("Testing constructor Angle(int deg, int min)");
for (int i = 0; i < degrees.Length; i++)
{
p = new Angle(degrees[i], minutes[i], seconds[i]);
Console.WriteLine("p = " + p);
}
/*Testing constructor Angle(int deg, int min)
p = 0°0'0"
p = 180°0'0"
p = 180°0'0"
p = 0°8'0" incorrect output
p = -73°11'50" incorrect output expected 120 56 10
*/
I do not understand why there is a bug here ? and why did they use divide Int32.MinValue by 60 and 120+180*200000 as this format ?
the comments in the constructor is a correction for the code
UPDATE: Added the code of normalize()
// For compatibility with the math libraries and other software
// range is (-pi,pi] not [0,2pi), enforced by the following function:
void normalize()
{
double twoPi = Math.PI + Math.PI;
while (value <= -Math.PI) value += twoPi;
while (value > Math.PI) value -= twoPi;
}

The problem is in this piece of code:
double seconds = sec + 60 * (min + 60 * deg);
Although you are storing seconds as a double, the conversion from int to double is taking place after sec + 60 * (min + 60 * deg) is computed as an int.
The compiler will not choose double arithmetics for you based on the type you decide to store the result in. The compiler will choose the best operator overload based on the types of the operands which in this case are all int and look for a valid implicit conversion (in this case int to double) afterwards; therefore it is choosing int arithmetics and the operation will overflow in the last two test cases:
Int32.MinValue / 60 * 60 * 60 = Int32.MinValue * 60 < Int32.MinValue which will overflow.
120 + 180 * 200000 * 60 * 60 > Int32.MaxValue which will also overflow.
Your expected results for these two cases are probably not considering this behavior.
In order to solve this issue, change your code to:
double seconds = sec + 60 * (min + 60f * deg);
Explicitly setting 60 to a double typed literal constant (60f) will force the compiler to resolve all operations to double arithmetics.
Also, it is worth pointing out that your constructor logic has some other issues:
You should be validating the input data; should it be valid to specify negative minutes or seconds? IMO that doesn't seem reasonable. Only deg should be allowed to have a negative value. You should check for this condition and act accordingly: throw an exception (preferable) or normalize sign of min and sec based on the sign of deg (ugly and potentially confusing).
Your seconds calculation doesn't seem to be correct for negative angles (again, this is tied to the previous issue and whatever sign convention you have decided to implement). Unless the convention is that negative angles must have negative deg, min and sec, the way you are computing seconds is wrong because you are always adding the minutes and seconds terms no matter the sign of deg.
UPDATE There is one more issue in your code that I missed until I had the chance to test it. Some of your test cases are failing because double doesn't have enough resolution. I think your code needs some major refactoring; normalize() should be called first. This way you will always be managing tightly bounded values that can not cause overflows or precision loss.
This is the way I would do it:
public Angle(int deg, int min, int sec)
{
//Omitting input values check.
double seconds = sec + 60 * (min + 60 * normalize(deg));
value = seconds * Math.PI / 648000f;
}
private int normalize(int deg)
{
int normalizedDeg = deg % 360;
if (normalizedDeg <= -180)
normalizedDeg += 360;
else if (normalizedDeg > 180)
normalizedDeg -= 360;
return normalizedDeg;
}

// For compatibility with the math libraries and other software
// range is (-pi,pi] not [0,2pi), enforced by the following function:
void normalize()
{double twoPi = Math.PI + Math.PI;
while (value <= -Math.PI) value += twoPi;
while (value > Math.PI) value -= twoPi;
}
This is the normalize function that I have

while loops is generally a bad idea. If you deal with small values it's okay, but imagine you have some angle like 1e+25, that'd be 1.59e+24 iterations or about 100 million years to compute if you have a decent CPU.
How it should be done instead:
static double NormalizeDegree360(double value)
{
var result = value % 360.0;
return result > 0 ? result : result + 360;
}
static double NormalizeDegree180(double value)
{
return (((value + 180) % 360) + 360) % 360 - 180;
}
static double TwoPI = 2*System.Math.PI;
static double NormalizeRadians2Pi(double value)
{
var result = value % TwoPI;
return result > 0 ? result : result + TwoPI;
}
static double NormalizeRadiansPi(double value)
{
return (((value + System.Math.PI) % TwoPI) + TwoPI) % TwoPI - System.Math.PI;
}

They're using the very large negative and positive numbers to make sure that the normalization caps angles to the range [-180, 180] degrees properly.

Average of 3 long integers

I have 3 very large signed integers.
long x = long.MaxValue;
long y = long.MaxValue - 1;
long z = long.MaxValue - 2;
I want to calculate their truncated average. Expected average value is long.MaxValue - 1, which is 9223372036854775806.
It is impossible to calculate it as:
long avg = (x + y + z) / 3; // 3074457345618258600
Note: I read all those questions about average of 2 numbers, but I don't see how that technique can be applied to average of 3 numbers.
It would be very easy with the usage of BigInteger, but let's assume I cannot use it.
BigInteger bx = new BigInteger(x);
BigInteger by = new BigInteger(y);
BigInteger bz = new BigInteger(z);
BigInteger bavg = (bx + by + bz) / 3; // 9223372036854775806
If I convert to double, then, of course, I lose precision:
double dx = x;
double dy = y;
double dz = z;
double davg = (dx + dy + dz) / 3; // 9223372036854780000
If I convert to decimal, it works, but also let's assume that I cannot use it.
decimal mx = x;
decimal my = y;
decimal mz = z;
decimal mavg = (mx + my + mz) / 3; // 9223372036854775806
Question: Is there a way to calculate the truncated average of 3 very large integers only with the usage of long type? Don't consider that question as C#-specific, just it is easier for me to provide samples in C#.

This code will work, but isn't that pretty.
It first divides all three values (it floors the values, so you 'lose' the remainder), and then divides the remainder:
long n = x / 3
+ y / 3
+ z / 3
+ ( x % 3
+ y % 3
+ z % 3
) / 3
Note that the above sample does not always work properly when having one or more negative values.
As discussed with Ulugbek, since the number of comments are exploding below, here is the current BEST solution for both positive and negative values.
Thanks to answers and comments of Ulugbek Umirov, James S, KevinZ, Marc van Leeuwen, gnasher729 this is the current solution:
static long CalculateAverage(long x, long y, long z)
{
return (x % 3 + y % 3 + z % 3 + 6) / 3 - 2
+ x / 3 + y / 3 + z / 3;
}
static long CalculateAverage(params long[] arr)
{
int count = arr.Length;
return (arr.Sum(n => n % count) + count * (count - 1)) / count - (count - 1)
+ arr.Sum(n => n / count);
}

NB - Patrick has already given a great answer. Expanding on this you could do a generic version for any number of integers like so:
long x = long.MaxValue;
long y = long.MaxValue - 1;
long z = long.MaxValue - 2;
long[] arr = { x, y, z };
var avg = arr.Select(i => i / arr.Length).Sum()
+ arr.Select(i => i % arr.Length).Sum() / arr.Length;

Patrick Hofman has posted a great solution. But if needed it can still be implemented in several other ways. Using the algorithm here I have another solution. If implemented carefully it may be faster than the multiple divisions in systems with slow hardware divisors. It can be further optimized by using divide by constants technique from hacker's delight
public class int128_t {
private int H;
private long L;
public int128_t(int h, long l)
{
H = h;
L = l;
}
public int128_t add(int128_t a)
{
int128_t s;
s.L = L + a.L;
s.H = H + a.H + (s.L < a.L);
return b;
}
private int128_t rshift2() // right shift 2
{
int128_t r;
r.H = H >> 2;
r.L = (L >> 2) | ((H & 0x03) << 62);
return r;
}
public int128_t divideby3()
{
int128_t sum = {0, 0}, num = new int128_t(H, L);
while (num.H || num.L > 3)
{
int128_t n_sar2 = num.rshift2();
sum = add(n_sar2, sum);
num = add(n_sar2, new int128_t(0, num.L & 3));
}
if (num.H == 0 && num.L == 3)
{
// sum = add(sum, 1);
sum.L++;
if (sum.L == 0) sum.H++;
}
return sum;
}
};
int128_t t = new int128_t(0, x);
t = t.add(new int128_t(0, y));
t = t.add(new int128_t(0, z));
t = t.divideby3();
long average = t.L;
In C/C++ on 64-bit platforms it's much easier with __int128
int64_t average = ((__int128)x + y + z)/3;

You can calculate the mean of numbers based on the differences between the numbers rather than using the sum.
Let's say x is the max, y is the median, z is the min (as you have). We will call them max, median and min.
Conditional checker added as per #UlugbekUmirov's comment:
long tmp = median + ((min - median) / 2); //Average of min 2 values
if (median > 0) tmp = median + ((max - median) / 2); //Average of max 2 values
long mean;
if (min > 0) {
mean = min + ((tmp - min) * (2.0 / 3)); //Average of all 3 values
} else if (median > 0) {
mean = min;
while (mean != tmp) {
mean += 2;
tmp--;
}
} else if (max > 0) {
mean = max;
while (mean != tmp) {
mean--;
tmp += 2;
}
} else {
mean = max + ((tmp - max) * (2.0 / 3));
}

Patching Patrick Hofman's solution with supercat's correction, I give you the following:
static Int64 Avg3 ( Int64 x, Int64 y, Int64 z )
{
UInt64 flag = 1ul << 63;
UInt64 x_ = flag ^ (UInt64) x;
UInt64 y_ = flag ^ (UInt64) y;
UInt64 z_ = flag ^ (UInt64) z;
UInt64 quotient = x_ / 3ul + y_ / 3ul + z_ / 3ul
+ ( x_ % 3ul + y_ % 3ul + z_ % 3ul ) / 3ul;
return (Int64) (quotient ^ flag);
}
And the N element case:
static Int64 AvgN ( params Int64 [ ] args )
{
UInt64 length = (UInt64) args.Length;
UInt64 flag = 1ul << 63;
UInt64 quotient_sum = 0;
UInt64 remainder_sum = 0;
foreach ( Int64 item in args )
{
UInt64 uitem = flag ^ (UInt64) item;
quotient_sum += uitem / length;
remainder_sum += uitem % length;
}
return (Int64) ( flag ^ ( quotient_sum + remainder_sum / length ) );
}
This always gives the floor() of the mean, and eliminates every possible edge case.

Because C uses floored division rather than Euclidian division, it may easier to compute a properly-rounded average of three unsigned values than three signed ones. Simply add 0x8000000000000000UL to each number before taking the unsigned average, subtract it after taking the result, and use an unchecked cast back to Int64 to get a signed average.
To compute the unsigned average, compute the sum of the top 32 bits of the three values. Then compute the sum of the bottom 32 bits of the three values, plus the sum from above, plus one [the plus one is to yield a rounded result]. The average will be 0x55555555 times the first sum, plus one third of the second.
Performance on 32-bit processors might be enhanced by producing three "sum" values each of which is 32 bits long, so that the final result is ((0x55555555UL * sumX)<<32) + 0x55555555UL * sumH + sumL/3; it might possibly be further enhanced by replacing sumL/3 with ((sumL * 0x55555556UL) >> 32), though the latter would depend upon the JIT optimizer [it might know how to replace a division by 3 with a multiply, and its code might actually be more efficient than an explicit multiply operation].

If you know you have N values, can you just divide each value by N and sum them together?
long GetAverage(long* arrayVals, int n)
{
long avg = 0;
long rem = 0;
for(int i=0; i<n; ++i)
{
avg += arrayVals[i] / n;
rem += arrayVals[i] % n;
}
return avg + (rem / n);
}

You could use the fact that you can write each of the numbers as y = ax + b, where x is a constant. Each a would be y / x (the integer part of that division). Each b would be y % x (the rest/modulo of that division). If you choose this constant in an intelligent way, for example by choosing the square root of the maximum number as a constant, you can get the average of x numbers without having problems with overflow.
The average of an arbitrary list of numbers can be found by finding:
( ( sum( all A's ) / length ) * constant ) +
( ( sum( all A's ) % length ) * constant / length) +
( ( sum( all B's ) / length )
where % denotes modulo and / denotes the 'whole' part of division.
The program would look something like:
class Program
{
static void Main()
{
List<long> list = new List<long>();
list.Add( long.MaxValue );
list.Add( long.MaxValue - 1 );
list.Add( long.MaxValue - 2 );
long sumA = 0, sumB = 0;
long res1, res2, res3;
//You should calculate the following dynamically
long constant = 1753413056;
foreach (long num in list)
{
sumA += num / constant;
sumB += num % constant;
}
res1 = (sumA / list.Count) * constant;
res2 = ((sumA % list.Count) * constant) / list.Count;
res3 = sumB / list.Count;
Console.WriteLine( res1 + res2 + res3 );
}
}

I also tried it and come up with a faster solution (although only by a factor about 3/4). It uses a single division
public static long avg(long a, long b, long c) {
final long quarterSum = (a>>2) + (b>>2) + (c>>2);
final long lowSum = (a&3) + (b&3) + (c&3);
final long twelfth = quarterSum / 3;
final long quarterRemainder = quarterSum - 3*twelfth;
final long adjustment = smallDiv3(lowSum + 4*quarterRemainder);
return 4*twelfth + adjustment;
}
where smallDiv3 is division by 3 using multipliation and working only for small arguments
private static long smallDiv3(long n) {
assert -30 <= n && n <= 30;
// Constants found rather experimentally.
return (64/3*n + 10) >> 6;
}
Here is the whole code including a test and a benchmark, the results are not that impressive.

This function computes the result in two divisions. It should generalize nicely to other divisors and word sizes.
It works by computing the double-word addition result, then working out the division.
Int64 average(Int64 a, Int64 b, Int64 c) {
// constants: 0x10000000000000000 div/mod 3
const Int64 hdiv3 = UInt64(-3) / 3 + 1;
const Int64 hmod3 = UInt64(-3) % 3;
// compute the signed double-word addition result in hi:lo
UInt64 lo = a; Int64 hi = a>=0 ? 0 : -1;
lo += b; hi += b>=0 ? lo<b : -(lo>=UInt64(b));
lo += c; hi += c>=0 ? lo<c : -(lo>=UInt64(c));
// divide, do a correction when high/low modulos add up
return hi>=0 ? lo/3 + hi*hdiv3 + (lo%3 + hi*hmod3)/3
: lo/3+1 + hi*hdiv3 + Int64(lo%3-3 + hi*hmod3)/3;
}

Math
(x + y + z) / 3 = x/3 + y/3 + z/3
(a[1] + a[2] + .. + a[k]) / k = a[1]/k + a[2]/k + .. + a[k]/k
Code
long calculateAverage (long a [])
{
double average = 0;
foreach (long x in a)
average += (Convert.ToDouble(x)/Convert.ToDouble(a.Length));
return Convert.ToInt64(Math.Round(average));
}
long calculateAverage_Safe (long a [])
{
double average = 0;
double b = 0;
foreach (long x in a)
{
b = (Convert.ToDouble(x)/Convert.ToDouble(a.Length));
if (b >= (Convert.ToDouble(long.MaxValue)-average))
throw new OverflowException ();
average += b;
}
return Convert.ToInt64(Math.Round(average));
}

Try this:
long n = Array.ConvertAll(new[]{x,y,z},v=>v/3).Sum()
+ (Array.ConvertAll(new[]{x,y,z},v=>v%3).Sum() / 3);

c# LOESS/LOWESS regression [closed]

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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]));
}
}
}
}
}

Round any n-digit number to (n-1) zero-digits

Sorry hard to formulate.
I need to round like this:
12 -> 10
152 -> 200
1538 -> 2000
25000 -> 30000
etc.
Twisting my head, but can't see how to make this. Must work for any n number of digits. Anyone got an elegant method for it?
c# or vb.net

How about this:
double num = 152;
int pow = (int)Math.Log10(num);
int factor = (int)Math.Pow(10, pow);
double temp = num / factor;
double result = Math.Round(temp) * factor;

I think you should try with something like this:
public int Round( int number)
{
int power = number.ToString().Length - 1;
int sz = Math.Pow(10, power);
int rounded = (int)Math.Round( number / sz );
return rounded * sz;
}
The idea is to get the size of the nearest 10 power, available by the length of the number expressed as a string. Then divide the number by that power, leaving it like 1,2 and then round it using the Math.Round method and restore the size by remultiplying it to the power.
Much like the previous answer...

I would do it this way:
double d = 25000;
int power = d.ToString().Length - 1;
double multipler = Math.Pow(10,power);
d = Math.Round(d / multipler) * multipler;
Console.WriteLine(d);

One of the way could be
Convert the number to Decimal
Divide it by 10^(n-1) (where n is number of digits)
Now use round function (Decimal.Round)
Multiply again by 10^(n-1)

Divide the number by 10n and round the result, then multiply the result back with 10n;
int MyRound(int num)
{
double divisor = Math.Pow(10, num.ToString().Length - 1);
return (int)(Math.Round(num / divisor, MidpointRounding.AwayFromZero) * divisor);
}
Note that we should use MidpointRounding.AwayFromZero when rounding because of the default banker's rounding.

int MakeOneSigFig(int value)
{
int neg = 1;
if(value <= 10 && value >= -10) { return value; }
if(value == int.MinValue) { value = int.MaxValue; neg = -1; }
if(value < 0) { value = -value; neg = -1; }
int mult = 10; // start at 10 because we've got 'firstDigit = value / 10' below
while(value > 99) { value /= 10; mult *= 10; }
int firstDigit = value / 10;
if(value % 10 >= 5) firstDigit++;
return neg * firstDigit * mult;
}
This is equivalent to MidpointRounding.AwayFromZero. This method doesn't do any double math or string conversions. If you didn't want to loop, you could replace that with the if block below. That would be more efficient, but more code and not quite as easy to read.
if(value < 100) { mult = 10; }
else if(value < 1000) { mult = 100; value /= 10; }
else if(value < 10000) { mult = 1000; value /= 100; }
else if(value < 100000) { mult = 10000; value /= 1000; }
// etc.