We Keep Coding

Calculating Integer Percentage

So I would like to calculate the percentage progress of my program as the nearest integer value
In my examples lets take
int FilesProcessed = 42;
int TotalFilesToProcess = 153;
So First I tried:
Int TotalProgress = ((FilesProcessed / TotalFilesToProcess) * 100)
This returned TotalProgress = 0
Then I tried
Int TotalProgress = (int)((FilesProcessed / TotalFilesToProcess) * 100)
This gives compiler error saying Cannot implicitly convert type decimal to int
Ive tried
Int TotalProgress = Math.Round((FilesProcessed / TotalFilesToProcess) * 100)
and get The call is ambiguous between decimal and double
and so now I've come here for help?

Cast to double first so it doesn't compute a division between integers:
int totalProgress = (int)((double)FilesProcessed / TotalFilesToProcess * 100);

int FilesProcessed = 42;
int TotalFilesToProcess = 153;
int TotalProgress = FilesProcessed * 100 / TotalFilesToProcess;
Console.WriteLine(TotalProgress);
https://dotnetfiddle.net/3GNlVd

If you want to be more accuracy, you can use:
int TotalProgress = Convert.ToInt32(Math.Round(((decimal)FilesProcessed / TotalFilesToProcess) * 100, 0));
If the numbers are greater you will have a difference. For example
int FilesProcessed = 42;
int TotalFilesToProcess = 1530;
The result with decimals will be: 2.74%, if you use the previous methods, you would find 2%, with the formula I am proposing you will obtain 3%. The last option has more accuracy.

Get decimals of a double?

I have a function that adds a double to another double but I need to add only to the digits after the decimal point and the number of digits varies based on the size of the number.
public double Calculate(double x, double add)
{
string xstr;
if (x >= 10)
xstr = x.ToString("00.0000", NumberFormatInfo.InvariantInfo);
if (x >= 100)
xstr = x.ToString("000.000", NumberFormatInfo.InvariantInfo);
if (x < 10)
xstr = x.ToString("0.00000", NumberFormatInfo.InvariantInfo);
string decimals = xstr.Remove(0, xstr.IndexOf(".") + 1);
decimals = (Convert.ToDouble(decimals) + add).ToString();
xstr = xstr.Substring(0, xstr.IndexOf(".") + 1) + decimals;
x = Convert.ToDouble(xstr, NumberFormatInfo.InvariantInfo);
return x;
}
I'm wondering if there isn't a simpler way to do this without having to convert the number to string first and then adding to the decimal part of it.
As you can see the number to be added to should always be a 6 digit number where ever the decimal separator is.

If you take the remainder of the object divided by 1 you'll get the fractional portion of that number:
double remainder = someDouble % 1;
To write the whole method out, it's as simple as:
public double Calculate(double x, double add)
{
return Math.Floor(x) + (x + add) % 1;
}
(This is one of those times where you're glad that % computes the remainder, rather than the modulus. This will work as is for negative numbers as well.)

A little more elegant and much more faster:
public static double Calculate(double x, double add)
{
var pow = 5 - Math.Truncate(Math.Log10(x));
var multiplier = Math.Pow(10, pow);
var decimals = Math.Truncate((x % 1)* multiplier) + add;
x = Math.Truncate(x) + Math.Truncate(decimals) / multiplier;
return x;
}

So all you want to do is add the fractional part of each value? Why don't you just do this?
public double Calculate(double x, double y)
{
double fractional_x = x - Math.Floor(x);
double fractional_y = y - Math.Floor(y);
return fractional_x + fractional_y;
}

Here is yet another version:
public double Calculate(double x, double add)
{
return (x - (int)x) + (add - (int)add);
}

Generate a random number in a Gaussian Range?

I want to use a random number generator that creates random numbers in a gaussian range where I can define the median by myself. I already asked a similar question here and now I'm using this code:
class RandomGaussian
{
private static Random random = new Random();
private static bool haveNextNextGaussian;
private static double nextNextGaussian;
public static double gaussianInRange(double from, double mean, double to)
{
if (!(from < mean && mean < to))
throw new ArgumentOutOfRangeException();
int p = Convert.ToInt32(random.NextDouble() * 100);
double retval;
if (p < (mean * Math.Abs(from - to)))
{
double interval1 = (NextGaussian() * (mean - from));
retval = from + (float)(interval1);
}
else
{
double interval2 = (NextGaussian() * (to - mean));
retval = mean + (float)(interval2);
}
while (retval < from || retval > to)
{
if (retval < from)
retval = (from - retval) + from;
if (retval > to)
retval = to - (retval - to);
}
return retval;
}
private static double NextGaussian()
{
if (haveNextNextGaussian)
{
haveNextNextGaussian = false;
return nextNextGaussian;
}
else
{
double v1, v2, s;
do
{
v1 = 2 * random.NextDouble() - 1;
v2 = 2 * random.NextDouble() - 1;
s = v1 * v1 + v2 * v2;
} while (s >= 1 || s == 0);
double multiplier = Math.Sqrt(-2 * Math.Log(s) / s);
nextNextGaussian = v2 * multiplier;
haveNextNextGaussian = true;
return v1 * multiplier;
}
}
}
Then to verify the results I plotted them with gaussianInRange(0, 0.5, 1) for n=100000000
As one can see the median is really at 0.5 but there isn't really a curve visible. So what I'm doing wrong?
EDIT
What i want is something like this where I can set the highest probability by myself by passing a value.

The simplest way to draw normal deviates conditional on them being in a particular range is with rejection sampling:
do {
retval = NextGaussian() * stdev + mean;
} while (retval < from || to < retval);
The same sort of thing is used when you draw coordinates (v1, v2) in a circle in your unconditional normal generator.
Simply folding in values outside the range doesn't produce the same distribution.
Also, if you have a good implementation of the error function and its inverse, you can calculate the values directly using an inverse CDF. The CDF of a normal distribution is
F(retval) = (1 + erf((retval-mean) / (stdev*sqrt(2)))) / 2
The CDF of a censored distribution is
C(retval) = (F(retval) - F(from)) / (F(to) - F(from)), from ≤ x < to
To draw a random number using a CDF, you draw v from a uniform distribution on [0, 1] and solve C(retval) = v. This gives
double v = random.NextDouble();
double t1 = erf((from - mean) / (stdev*sqrt(2)));
t2 = erf((to - mean) / (stdev*sqrt(2)));
double retval = mean + stdev * sqrt(2) * erf_inv(t1*(1-v) + t2*v);
You can precalculate t1 and t2 for specific parameters. The advantage of this approach is that there is no rejection sampling, so you only need a single NextDouble() per draw. If the [from, to] interval is small this will be faster.
However, it sounds like you might want the binomial distribution instead.

I have similar methods in my Graph generator (had to modify it a bit):
Returns a random floating-point number using a generator function with a specific range:
private double NextFunctional(Func<double, double> func, double from, double to, double height, out double x)
{
double halfWidth = (to - from) / 2;
double distance = halfWidth + from;
x = this.rand.NextDouble() * 2 - 1;// -1 .. 1
double y = func(x);
x = halfWidth * x + distance;
y *= height;
return y;
}
Gaussian function:
private double Gauss(double x)
{
// Graph should look better with double-x scale.
x *= 2;
double σ = 1 / Math.Sqrt(2 * Math.PI);
double variance = Math.Pow(σ, 2);
double exp = -0.5 * Math.Pow(x, 2) / variance;
double y = 1 / Math.Sqrt(2 * Math.PI * variance) * Math.Pow(Math.E, exp);
return y;
}
A method that generates a graph using the random numbers:
private void PlotGraph(Graphics g, Pen p, double from, double to, double height)
{
for (int i = 0; i < 1000; i++)
{
double x;
double y = this.NextFunctional(this.Gauss, from, to, height, out x);
this.DrawPoint(g, p, x, y);
}
}
I would rather used a cosine function - it is much faster and pretty close to the gaussian function for your needs:
double x;
double y = this.NextFunctional(a => Math.Cos(a * Math.PI), from, to, height, out x);
The out double x parameter in the NextFunctional() method is there so you can easily test it on your graphs (I use an iterator in my method).

How do I determine the standard deviation (stddev) of a set of values?

I need to know if a number compared to a set of numbers is outside of 1 stddev from the mean, etc..

While the sum of squares algorithm works fine most of the time, it can cause big trouble if you are dealing with very large numbers. You basically may end up with a negative variance...
Plus, don't never, ever, ever, compute a^2 as pow(a,2), a * a is almost certainly faster.
By far the best way of computing a standard deviation is Welford's method. My C is very rusty, but it could look something like:
public static double StandardDeviation(List<double> valueList)
{
double M = 0.0;
double S = 0.0;
int k = 1;
foreach (double value in valueList)
{
double tmpM = M;
M += (value - tmpM) / k;
S += (value - tmpM) * (value - M);
k++;
}
return Math.Sqrt(S / (k-2));
}
If you have the whole population (as opposed to a sample population), then use return Math.Sqrt(S / (k-1));.
EDIT: I've updated the code according to Jason's remarks...
EDIT: I've also updated the code according to Alex's remarks...

10 times faster solution than Jaime's, but be aware that,
as Jaime pointed out:
"While the sum of squares algorithm works fine most of the time, it
can cause big trouble if you are dealing with very large numbers. You
basically may end up with a negative variance"
If you think you are dealing with very large numbers or a very large quantity of numbers, you should calculate using both methods, if the results are equal, you know for sure that you can use "my" method for your case.
public static double StandardDeviation(double[] data)
{
double stdDev = 0;
double sumAll = 0;
double sumAllQ = 0;
//Sum of x and sum of x²
for (int i = 0; i < data.Length; i++)
{
double x = data[i];
sumAll += x;
sumAllQ += x * x;
}
//Mean (not used here)
//double mean = 0;
//mean = sumAll / (double)data.Length;
//Standard deviation
stdDev = System.Math.Sqrt(
(sumAllQ -
(sumAll * sumAll) / data.Length) *
(1.0d / (data.Length - 1))
);
return stdDev;
}

The accepted answer by Jaime is great, except you need to divide by k-2 in the last line (you need to divide by "number_of_elements-1").
Better yet, start k at 0:
public static double StandardDeviation(List<double> valueList)
{
double M = 0.0;
double S = 0.0;
int k = 0;
foreach (double value in valueList)
{
k++;
double tmpM = M;
M += (value - tmpM) / k;
S += (value - tmpM) * (value - M);
}
return Math.Sqrt(S / (k-1));
}

The Math.NET library provides this for you to of the box.
PM> Install-Package MathNet.Numerics
var populationStdDev = new List<double>(1d, 2d, 3d, 4d, 5d).PopulationStandardDeviation();
var sampleStdDev = new List<double>(2d, 3d, 4d).StandardDeviation();
See PopulationStandardDeviation for more information.

Code snippet:
public static double StandardDeviation(List<double> valueList)
{
if (valueList.Count < 2) return 0.0;
double sumOfSquares = 0.0;
double average = valueList.Average(); //.NET 3.0
foreach (double value in valueList)
{
sumOfSquares += Math.Pow((value - average), 2);
}
return Math.Sqrt(sumOfSquares / (valueList.Count - 1));
}

You can avoid making two passes over the data by accumulating the mean and mean-square
cnt = 0
mean = 0
meansqr = 0
loop over array
cnt++
mean += value
meansqr += value*value
mean /= cnt
meansqr /= cnt
and forming
sigma = sqrt(meansqr - mean^2)
A factor of cnt/(cnt-1) is often appropriate as well.
BTW-- The first pass over the data in Demi and McWafflestix answers are hidden in the calls to Average. That kind of thing is certainly trivial on a small list, but if the list exceed the size of the cache, or even the working set, this gets to be a bid deal.

I found that Rob's helpful answer didn't quite match what I was seeing using excel. To match excel, I passed the Average for valueList in to the StandardDeviation calculation.
Here is my two cents... and clearly you could calculate the moving average (ma) from valueList inside the function - but I happen to have already before needing the standardDeviation.
public double StandardDeviation(List<double> valueList, double ma)
{
double xMinusMovAvg = 0.0;
double Sigma = 0.0;
int k = valueList.Count;
foreach (double value in valueList){
xMinusMovAvg = value - ma;
Sigma = Sigma + (xMinusMovAvg * xMinusMovAvg);
}
return Math.Sqrt(Sigma / (k - 1));
}

With Extension methods.
using System;
using System.Collections.Generic;
namespace SampleApp
{
internal class Program
{
private static void Main()
{
List<double> data = new List<double> {1, 2, 3, 4, 5, 6};
double mean = data.Mean();
double variance = data.Variance();
double sd = data.StandardDeviation();
Console.WriteLine("Mean: {0}, Variance: {1}, SD: {2}", mean, variance, sd);
Console.WriteLine("Press any key to continue...");
Console.ReadKey();
}
}
public static class MyListExtensions
{
public static double Mean(this List<double> values)
{
return values.Count == 0 ? 0 : values.Mean(0, values.Count);
}
public static double Mean(this List<double> values, int start, int end)
{
double s = 0;
for (int i = start; i < end; i++)
{
s += values[i];
}
return s / (end - start);
}
public static double Variance(this List<double> values)
{
return values.Variance(values.Mean(), 0, values.Count);
}
public static double Variance(this List<double> values, double mean)
{
return values.Variance(mean, 0, values.Count);
}
public static double Variance(this List<double> values, double mean, int start, int end)
{
double variance = 0;
for (int i = start; i < end; i++)
{
variance += Math.Pow((values[i] - mean), 2);
}
int n = end - start;
if (start > 0) n -= 1;
return variance / (n);
}
public static double StandardDeviation(this List<double> values)
{
return values.Count == 0 ? 0 : values.StandardDeviation(0, values.Count);
}
public static double StandardDeviation(this List<double> values, int start, int end)
{
double mean = values.Mean(start, end);
double variance = values.Variance(mean, start, end);
return Math.Sqrt(variance);
}
}
}

/// <summary>
/// Calculates standard deviation, same as MATLAB std(X,0) function
/// <seealso cref="http://www.mathworks.co.uk/help/techdoc/ref/std.html"/>
/// </summary>
/// <param name="values">enumumerable data</param>
/// <returns>Standard deviation</returns>
public static double GetStandardDeviation(this IEnumerable<double> values)
{
//validation
if (values == null)
throw new ArgumentNullException();
int lenght = values.Count();
//saves from devision by 0
if (lenght == 0 || lenght == 1)
return 0;
double sum = 0.0, sum2 = 0.0;
for (int i = 0; i < lenght; i++)
{
double item = values.ElementAt(i);
sum += item;
sum2 += item * item;
}
return Math.Sqrt((sum2 - sum * sum / lenght) / (lenght - 1));
}

The trouble with all the other answers is that they assume you have your
data in a big array. If your data is coming in on the fly, this would be
a better approach. This class works regardless of how or if you store your data. It also gives you the choice of the Waldorf method or the sum-of-squares method. Both methods work using a single pass.
public final class StatMeasure {
private StatMeasure() {}
public interface Stats1D {
/** Add a value to the population */
void addValue(double value);
/** Get the mean of all the added values */
double getMean();
/** Get the standard deviation from a sample of the population. */
double getStDevSample();
/** Gets the standard deviation for the entire population. */
double getStDevPopulation();
}
private static class WaldorfPopulation implements Stats1D {
private double mean = 0.0;
private double sSum = 0.0;
private int count = 0;
#Override
public void addValue(double value) {
double tmpMean = mean;
double delta = value - tmpMean;
mean += delta / ++count;
sSum += delta * (value - mean);
}
#Override
public double getMean() { return mean; }
#Override
public double getStDevSample() { return Math.sqrt(sSum / (count - 1)); }
#Override
public double getStDevPopulation() { return Math.sqrt(sSum / (count)); }
}
private static class StandardPopulation implements Stats1D {
private double sum = 0.0;
private double sumOfSquares = 0.0;
private int count = 0;
#Override
public void addValue(double value) {
sum += value;
sumOfSquares += value * value;
count++;
}
#Override
public double getMean() { return sum / count; }
#Override
public double getStDevSample() {
return (float) Math.sqrt((sumOfSquares - ((sum * sum) / count)) / (count - 1));
}
#Override
public double getStDevPopulation() {
return (float) Math.sqrt((sumOfSquares - ((sum * sum) / count)) / count);
}
}
/**
* Returns a way to measure a population of data using Waldorf's method.
* This method is better if your population or values are so large that
* the sum of x-squared may overflow. It's also probably faster if you
* need to recalculate the mean and standard deviation continuously,
* for example, if you are continually updating a graphic of the data as
* it flows in.
*
* #return A Stats1D object that uses Waldorf's method.
*/
public static Stats1D getWaldorfStats() { return new WaldorfPopulation(); }
/**
* Return a way to measure the population of data using the sum-of-squares
* method. This is probably faster than Waldorf's method, but runs the
* risk of data overflow.
*
* #return A Stats1D object that uses the sum-of-squares method
*/
public static Stats1D getSumOfSquaresStats() { return new StandardPopulation(); }
}

We may be able to use statistics module in Python. It has stedev() and pstdev() commands to calculate standard deviation of sample and population respectively.
details here: https://www.geeksforgeeks.org/python-statistics-stdev/
import statistics as st
print(st.ptdev(dataframe['column name']))

This is Population standard deviation
private double calculateStdDev(List<double> values)
{
double average = values.Average();
return Math.Sqrt((values.Select(val => (val - average) * (val - average)).Sum()) / values.Count);
}
For Sample standard deviation, just change [values.Count] to [values.Count -1] in above code.
Make sure you don't have only 1 data point in your set.

Why do these division equations result in zero?

The result of all of the division equations in the below for loop is 0. How can I get it to give me a decimal e.g.:
297 / 315 = 0.30793650793650793650793650793651
Code:
using System;
namespace TestDivide
{
class Program
{
static void Main(string[] args)
{
for (int i = 0; i <= 100; i++)
{
decimal result = i / 100;
long result2 = i / 100;
double result3 = i / 100;
float result4 = i / 100;
Console.WriteLine("{0}/{1}={2} ({3},{4},{5}, {6})", i, 100, i / 100, result, result2, result3, result4);
}
Console.ReadLine();
}
}
}
Answer:
Thanks Jon and everyone, this is what I wanted to do:
using System;
namespace TestDivide
{
class Program
{
static void Main(string[] args)
{
int maximum = 300;
for (int i = 0; i <= maximum; i++)
{
float percentage = (i / (float)maximum) * 100f;
Console.WriteLine("on #{0}, {1:#}% finished.", i, percentage);
}
Console.ReadLine();
}
}
}

You're using int/int, which does everything in integer arithmetic even if you're assigning to a decimal/double/float variable.
Force one of the operands to be of the type you want to use for the arithmetic.
for (int i = 0; i <= 100; i++)
{
decimal result = i / 100m;
long result2 = i / 100;
double result3 = i / 100d;
float result4 = i / 100f;
Console.WriteLine("{0}/{1}={2} ({3},{4},{5}, {6})",
i, 100, i / 100d, result, result2, result3, result4);
}
Results:
0/100=0 (0,0,0, 0)
1/100=0.01 (0.01,0,0.01, 0.01)
2/100=0.02 (0.02,0,0.02, 0.02)
3/100=0.03 (0.03,0,0.03, 0.03)
4/100=0.04 (0.04,0,0.04, 0.04)
5/100=0.05 (0.05,0,0.05, 0.05)
(etc)
Note that that isn't showing the exact value represented by the float or the double - you can't represent 0.01 exactly as a float or double, for example. The string formatting is effectively rounding the result. See my article on .NET floating binary point for more information as well as a class which will let you see the exact value of a double.
I haven't bothered using 100L for result2 because the result would always be the same.

Try
i / 100.0

because i is an int: i / 100 performs integer division, then the result, that is always 0, is casted to the target type. You need to specify at least one non-int literal in your expression:
i / 100.0

Because i is an integer and 100 is an integer...so you have an integer division
Try (decimal)i / 100.0 instead

No matter where you store it, an integer divided by an integer will always be an integer.

You need to force a floating point operation "double / double" instead of an "int / int"
double result = (double)297 / (double)315 ;

this is integer division whatever the type of variable you storing in,
so int / int = int

double result3 = ((double)i) / 100;

Because i is a int value and you divide by an integer so the result is an integer ! and so you need to divide by 100.0 to have an implicit cast in float or specify 100f or 100d

In my case I had only vars and no int
float div = (var1 - var2) / float.Parse(var1.ToString());