The code below somewhat works but not rounding correctly, my goal is if a value = 1.5 round down, if 1.51 round up.
Thanks
if (!String.IsNullOrEmpty(tbSnp_Uld.Text) && !string.IsNullOrEmpty(cbSnp_Uld.Text))
{
double d_tbSnp_Uld = Convert.ToDouble(tbSnp_Uld.Text);
double d_cbSnp_Uld = Convert.ToDouble(cbSnp_Uld.Text);
double result1 = Math.Ceiling(d_tbSnp_Uld / d_cbSnp_Uld);
double d = 0;
int floored = (int)Math.Floor(d);
int ceiled = (int)Math.Ceiling(d);
double epsilon = 0;
int lessThan = floored - Convert.ToInt32(Math.Abs(d - floored) < epsilon);
int moreThan = ceiled + Convert.ToInt32(Math.Abs(d - ceiled) < epsilon);
tbTrailer_Needed.Text = result1.ToString();
}
Here raw sample of what you trying to accomplish. Wrote this by using TDD apporach: add tests -> make it pass.
So feel free to refactor it
public static int CustomRound(double value)
{
int sign = Math.Sign(value);
double absValue = Math.Abs(value);
int absResult = (int)Math.Round(absValue - 0.01, 0, MidpointRounding.AwayFromZero);
return absResult * sign;
}
Related
I'm working on a Midpoint Riemann Sum program, and it finds the integral of a randomly generated function called f.
Here's what wrote:
public static double FindIntegral (double start, double end, function f)
{
double sum = 0;
double stepsize = 1E-2;
int numSteps = (int)((end - start) / stepsize);
for (int i = 0; i < numSteps; i++)
{
sum += f(start + (stepsize * (i + 0.5)));
}
return sum * stepsize;
}
The function returns numbers that are too low (I have a reliable checking mechanism).
I put in x^3 for f, and I got the right answer. I tried a couple of more integrable functions and got a good answer. But somehow once I put in f it doesn't work.
I got the math formula for "Riemann Midpoint Sum" from here.
My implementation below seems to get the right answer (using the example function on the page). I used a class because 1) I could make the algorithm work specifying either the step size or the number of rectangles (I preferred the latter) and 2) I didn't see any reason to hard-code either into the algorithm.
As it turns out your code seemed to work just fine (see below); Make sure the code you have here in your question is what you're executing and make sure your expected result is accurate and that you're supplying good inputs (i.e. you don't have start and end backwards or the wrong function f or something). In other words what you provided in your question looks fine. Note double is approximate in C# (floating point arithmetic, in general) so to compare equality you can't use == unless you want exact if you're using unit tests or something.
public class Program
{
public static void Main()
{
function f = x => 50 / (10 + x * x);
// 9.41404285216233
Console.Out.WriteLine(new RiemannMidpointSum(6).FindIntegral(1, 4, f));
// 9.41654853716462
Console.Out.WriteLine(new RiemannMidpointSum(1E-2).FindIntegral(1, 4, f));
// 9.41654853716462
Console.Out.WriteLine(Program.FindIntegral(1, 4, f));
}
// This is your function.
public static double FindIntegral (double start, double end, function f)
{
double sum = 0;
double stepsize = 1E-2;
int numSteps = (int)((end - start) / stepsize);
for (int i = 0; i < numSteps; i++)
{
sum += f(start + (stepsize * (i + 0.5)));
}
return sum * stepsize;
}
}
public delegate double function(double d);
public class RiemannMidpointSum
{
private int? _numberOfRectangles;
private double? _widthPerRectangle;
public RiemannMidpointSum(int numberOfRectangles)
{
// TODO: Handle non-positive input.
this._numberOfRectangles = numberOfRectangles;
}
public RiemannMidpointSum(double widthPerRectangle)
{
// TODO: Handle non-positive input.
this._widthPerRectangle = widthPerRectangle;
}
public double FindIntegral(double a, double b, function f)
{
var totalWidth = b - a;
var widthPerRectangle = this._widthPerRectangle ?? (totalWidth / this._numberOfRectangles.Value);
var numberOfRectangles = this._numberOfRectangles ?? ((int)Math.Round(totalWidth / this._widthPerRectangle.Value, 0));
double sum = 0;
foreach (var i in Enumerable.Range(0, numberOfRectangles))
{
var rectangleMidpointX = a + widthPerRectangle * i + widthPerRectangle / 2;
var rectangleHeightY = f(rectangleMidpointX);
var rectangleArea = widthPerRectangle * rectangleHeightY;
sum += rectangleArea;
}
return sum;
}
}
How to shift all the whole numbers in a double to the right of the point ?
Example i have 5342, i want the function to return 0.5342. I do not know the number of digits in the double, it's randomly generated. Should be fairly easy but i can't find any answers.
private static void Main(string[] args)
{
Console.WriteLine(MyFunction(5127));
Console.WriteLine(MyFunction(1));
Console.WriteLine(MyFunction(51283271));
Console.WriteLine(MyFunction(-512));
Console.WriteLine(MyFunction(0));
}
public static double MyFunction(double myNumber)
{
return Math.Floor(myNumber) / Math.Pow(10, Math.Abs(myNumber).ToString().Length);
}
This sounds like a pretty bizarre task, to be honest, but you could use:
while (Math.Abs(value) >= 1)
{
value = value / 10;
}
That will go into an infinite loop if the input is infinity though - and you may well lose information as you keep dividing. The latter point is important - if what you're really interested in is the decimal representation, you should consider using decimal instead of double.
You could potentially use a mixture of Math.Log and Math.Pow to do it, but the above is probably what I'd start with.
This will get you most of the way there
public static string test()
{
double dub = 5432;
string dubTxt = dub.ToString();
string text = "0.";
string test = string.Concat(text + dubTxt);
if (1 == 1)
{
MessageBox.Show(test);
return test;
}
}
You will have to develop more if statements to handle the negative numbers.
public static string test()
{
double dub = 5432;
string dubTxt = dub.ToString();
string text = "0.";
string test = string.Concat(text + dubTxt);
if (dub < 0)
{
//Do this code instead
}
}
Good Luck. Please bump me if you choose it!! I need the cred so I can do other junk. :-D
Just divide by 10 until the number is less than 1.
public static double SomeMethod(double n)
{
double d = n;
bool isNegative = (d < 0);
if(isNegative)
d = d * -1;
while(d >= 1)
d = d/10;
if(isNegative)
d = d * -1;
return d;
}
Alternative (and more precise) option:
public static double SomeMethod2(double n)
{
double d = n;
bool isNegative = (d < 0);
if(isNegative)
d = d * -1;
int numberOfDigits = ((int)d).ToString().Length;
int divisor = 1;
for(int i = 0; i < numberOfDigits; i++)
divisor = divisor * 10;
d = d/divisor;
if(isNegative)
d = d * -1;
return d;
}
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];
}
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.
The source data :
static double[] felix = new double[] { 0.003027523, 0.002012256, -0.001369238, -0.001737660, -0.001647287,
0.000275154, 0.002017238, 0.001372621, 0.000274148, -0.000913576, 0.001920263, 0.001186456, -0.000364631,
0.000638337, 0.000182266, -0.001275626, -0.000821093, 0.001186998, -0.000455996, -0.000547445, -0.000182582,
-0.000547845, 0.001279006, 0.000456204, 0.000000000, -0.001550388, 0.001552795, 0.000729594, -0.000455664,
-0.002188184, 0.000639620, 0.000091316, 0.001552228, -0.001002826, 0.000182515, -0.000091241, -0.000821243,
-0.002009132, 0.000000000, 0.000823572, 0.001920088, -0.001368863, 0.000000000, 0.002101800, 0.001094291,
0.001639643, 0.002637323, 0.000000000, -0.000172336, -0.000462665, -0.000136141 };
The variance function:
public static double Variance(double[] x)
{
if (x.Length == 0)
return 0;
double sumX = 0;
double sumXsquared = 0;
double varianceX = 0;
int dataLength = x.Length;
for (int i = 0; i < dataLength; i++)
{
sumX += x[i];
sumXsquared += x[i] * x[i];
}
varianceX = (sumXsquared / dataLength) - ((sumX / dataLength) * (sumX / dataLength));
return varianceX;
}
Excel and some online calculator says the variance is 1.56562E-06
While my function gives me 1.53492394804015E-06. I begin to doubt if the C# has accuracy problem or what. Is there anyone have this kind of problem before?
What you are seeing is the difference between sample variance and population variance and nothing to do with floating point precision or the accuracy of C#'s floating point implementation.
You are calculating population variance. Excel and that web site are calculating sample variance.
Var and VarP are distinct calculations and you do need to be careful about which one you are using. (unfortunately people often refer to them as if they are interchangeable when they are not. The same is true for standard deviation)
Sample variance for your data is 1.56562E-06, population variance is 1.53492394804015E-06.
From some code posted on codeproject awhile back:
Variance in a sample
public static double Variance(this IEnumerable<double> source)
{
double avg = source.Average();
double d = source.Aggregate(0.0, (total, next) => total += Math.Pow(next - avg, 2));
return d / (source.Count() - 1);
}
Variance in a population
public static double VarianceP(this IEnumerable<double> source)
{
double avg = source.Average();
double d = source.Aggregate(0.0, (total, next) => total += Math.Pow(next - avg, 2));
return d / source.Count();
}
Here's an alternate implementation, that is sometimes better-behaved, numerically:
mean = Average(data);
double sum2 = 0.0, sumc = 0.0;
for (int i = 0; i < data.Count; i++)
{
double dev = data[i] - mean;
sum2 += dev * dev;
sumc += dev;
}
return (sum2 - sumc * sumc / data.Count) / data.Count;