I have two values, one from user input and another from DB.
var userinput = form["someInput"];
var valuefromDB = GetValue(someNumber);
public float? GetValue(int id){
return (float?) db.table.where(p=> p.id == id).select(p=> p.Value).SingleOrDefault();
}
userinput have value "1" as string, while valuefromDB havevalue 0.001 as float.
so 1 / 0.001 = 1000
but my c# code give me 999.999939 as result;
var final = float.Parse(userinput) / valuefromDB
when i have "2" as user input value, result is correct, 2000...
That's because not all decimal numbers can be accurately represented in binary (which is the representation that float uses). The solution is to format the result to the desired number of decimal places, which will cause it to be rounded and displayed "correctly" as a consequence.
Update: To format a float for display, take a look at this MSDN reference page and this page of examples.
For pure precision which is not provided by float use decimal instead.
See What is the difference between Decimal, Float and Double in C#?
Related
I have a double value: 0.314285 which I want to Round off to 5 decimal places. From a mathematical point of view my expectant result is: 0.31429. In my code I use the Math.Round with MidPointRounding.AwayFromZero parameter overload, the resultant output being: 0.31428.
Is there another way to implement to have the output result as: 0.31429??
You should read the rounding and precision article. The real representation of your number in memory can be something like 0.3142849999999999, and therefore you are getting 0.31428 result. Using a decimal type can help to solve this issue
var value = 0.314285m;
var result = Math.Round(value, 5, MidpointRounding.AwayFromZero); //0.31429
public static double RoundUp(double i, int decimalPlaces)
{
var power = Math.Pow(10, decimalPlaces);
return Math.Ceiling(i * power) / power;
}
RoundDown(0.314285, 5); //0.31429
Math.Round(decimal number to round, int number of decimal places)
I think this should work for what you are trying to do.
int trail = 14;
double mean = 14.00000587000000;
double sd = 4.47307944700000;
double zscore = double.MinValue;
zscore = (trail - mean) / sd; //zscore at this point is exponent value -1.3122950464645662E-06
zscore = Math.Round(zscore, 14); //-1.31229505E-06
Math.Round() also keeps the exponent value. should zscore.ToString("F14") be used instead of Math.Round() function to convert it to non-exponent value? Please explain.
These are completely independant concerns.
Math.Round will actually return a new value, rounded to the specified decimal (ar at least, as near as one can do with floating point).
You can reuse this result value anywhere, and show it with 16 decimals precision if you want, but it's not supposed to be the same as the original one.
The fact that it is displayed with exponent notation or not has nothing to do with Round.
When you use ToString("F14") on a number, this is a display specification only, and does not modify the underlying value in any way. The underlying value might be a number that would or would not display as exponential notation otherwise, and may or may actually have 14 significant digits.
It simply forces the number to be displayed as a full decimal without exponent notation, with the number of digits specified. So it seems to be what you actually want.
Examples :
(executable online here : http://rextester.com/PZXDES55622)
double num = 0.00000123456789;
Console.WriteLine("original :");
Console.WriteLine(num.ToString());
Console.WriteLine(num.ToString("F6"));
Console.WriteLine(num.ToString("F10"));
Console.WriteLine(num.ToString("F14"));
Console.WriteLine("rounded to 6");
double rounded6 = Math.Round(num, 6);
Console.WriteLine(rounded6.ToString());
Console.WriteLine(rounded6.ToString("F6"));
Console.WriteLine(rounded6.ToString("F10"));
Console.WriteLine(rounded6.ToString("F14"));
Console.WriteLine("rounded to 10");
double rounded10 = Math.Round(num, 10);
Console.WriteLine(rounded10.ToString());
Console.WriteLine(rounded10.ToString("F6"));
Console.WriteLine(rounded10.ToString("F10"));
Console.WriteLine(rounded10.ToString("F14"));
will output:
original :
1,23456789E-06
0,000001
0,0000012346
0,00000123456789
rounded to 6
1E-06
0,000001
0,0000010000
0,00000100000000
rounded to 10
1,2346E-06
0,000001
0,0000012346
0,00000123460000
My scenario is that if
47/15= 3.13333
i want to convert it into 4, if the result has decimal i want to increase the result by 1, right now i am doing this like
float res = ((float)(62-15) / 15);
if (res.ToString().Contains("."))
{
string digit=res.ToString().Substring(0, res.ToString().IndexOf('.'));
int incrementDigit=Convert.ToInt16(k) + 1;
}
I want to know is there any shortcut way or built in function in c# so that i can do this fast without implementing string functions.
Thanks a lot.
Do you mean you want to perform integer division, but always rounding up? I suspect you want:
public static int DivideByFifteenRoundingUp(int value) {
return (value + 14) / 15;
}
This avoids using floating point arithmetic at all - it just allows any value which isn't an exact multiple of 15 to be rounded up, due to the way that integer arithmetic truncates towards zero.
Note that this does not work for negative input - for example, if you passed in -15 this would return 0. you could fix this with:
public static int DivideByFifteenRoundingUp(int value) {
return value < 0 ? value / 15 : (value + 14) / 15;
}
Use Math.Ceiling Quoting MSDN:
Returns the smallest integral value that is greater than or equal to
the specified decimal number.
You are looking for Math.Ceiling().
Convert the value you have to a Decimal or Double and the result of that method is what you need. Like:
double number = ((double)(62-15) / (double)15);
double result = Math.Ceiling(number);
Note the fact that I cast 15 to a double, so I avoid integer division. That is most likely not what you want here.
Another way of doing what you ask is to add 0.5 to every number, then floor it (truncate the decimal places). I'm afraid I don't have access to a C# compiler right now to confirm the exact function calls!
NB: But as others have confirmed, I would think the Math.Ceiling function best communicates to others what you intend.
Something like:
float res = ((float)(62-15) / 15);
int incrementDigit = (int)Math.Ceiling(res);
or
int incrementDigit = (int)(res + 0.5f);
I'm running NUnit tests to evaluate some known test data and calculated results. The numbers are floating point doubles so I don't expect them to be exactly equal, but I'm not sure how to treat them as equal for a given precision.
In NUnit we can compare with a fixed tolerance:
double expected = 0.389842845321551d;
double actual = 0.38984284532155145d; // really comes from a data import
Expect(actual, EqualTo(expected).Within(0.000000000000001));
and that works fine for numbers below zero, but as the numbers grow the tolerance really needs to be changed so we always care about the same number of digits of precision.
Specifically, this test fails:
double expected = 1.95346834136148d;
double actual = 1.9534683413614817d; // really comes from a data import
Expect(actual, EqualTo(expected).Within(0.000000000000001));
and of course larger numbers fail with tolerance..
double expected = 1632.4587642911599d;
double actual = 1632.4587642911633d; // really comes from a data import
Expect(actual, EqualTo(expected).Within(0.000000000000001));
What's the correct way to evaluate two floating point numbers are equal with a given precision? Is there a built-in way to do this in NUnit?
From msdn:
By default, a Double value contains 15 decimal digits of precision, although a maximum of 17 digits is maintained internally.
Let's assume 15, then.
So, we could say that we want the tolerance to be to the same degree.
How many precise figures do we have after the decimal point? We need to know the distance of the most significant digit from the decimal point, right? The magnitude. We can get this with a Log10.
Then we need to divide 1 by 10 ^ precision to get a value around the precision we want.
Now, you'll need to do more test cases than I have, but this seems to work:
double expected = 1632.4587642911599d;
double actual = 1632.4587642911633d; // really comes from a data import
// Log10(100) = 2, so to get the manitude we add 1.
int magnitude = 1 + (expected == 0.0 ? -1 : Convert.ToInt32(Math.Floor(Math.Log10(expected))));
int precision = 15 - magnitude ;
double tolerance = 1.0 / Math.Pow(10, precision);
Assert.That(actual, Is.EqualTo(expected).Within(tolerance));
It's late - there could be a gotcha in here. I tested it against your three sets of test data and each passed. Changing pricision to be 16 - magnitude caused the test to fail. Setting it to 14 - magnitude obviously caused it to pass as the tolerance was greater.
This is what I came up with for The Floating-Point Guide (Java code, but should translate easily, and comes with a test suite, which you really really need):
public static boolean nearlyEqual(float a, float b, float epsilon)
{
final float absA = Math.abs(a);
final float absB = Math.abs(b);
final float diff = Math.abs(a - b);
if (a * b == 0) { // a or b or both are zero
// relative error is not meaningful here
return diff < (epsilon * epsilon);
} else { // use relative error
return diff / (absA + absB) < epsilon;
}
}
The really tricky question is what to do when one of the numbers to compare is zero. The best answer may be that such a comparison should always consider the domain meaning of the numbers being compared rather than trying to be universal.
How about converting the items each to string and comparing the strings?
string test1 = String.Format("{0:0.0##}", expected);
string test2 = String.Format("{0:0.0##}", actual);
Assert.AreEqual(test1, test2);
Assert.That(x, Is.EqualTo(y).Within(10).Percent);
is a decent option (changes it to a relative comparison, where x is required to be within 10% of y). You may want to add extra handling for 0, as otherwise you'll get an exact comparison in that case.
Update:
Another good option is
Assert.That(x, Is.EqualTo(y).Within(1).Ulps);
where Ulps means units in the last place. See https://docs.nunit.org/articles/nunit/writing-tests/constraints/EqualConstraint.html#comparing-floating-point-values.
I don't know if there's a built-in way to do it with nunit, but I would suggest multiplying each float by the 10x the precision you're seeking, storing the results as longs, and comparing the two longs to each other.
For example:
double expected = 1632.4587642911599d;
double actual = 1632.4587642911633d;
//for a precision of 4
long lActual = (long) 10000 * actual;
long lExpected = (long) 10000 * expected;
if(lActual == lExpected) { // Do comparison
// Perform desired actions
}
This is a quick idea, but how about shifting them down till they are below zero? Should be something like num/(10^ceil(log10(num))) . . . not to sure about how well it would work, but its an idea.
1632.4587642911599 / (10^ceil(log10(1632.4587642911599))) = 0.16324587642911599
How about:
const double significantFigures = 10;
Assert.AreEqual(Actual / Expected, 1.0, 1.0 / Math.Pow(10, significantFigures));
The difference between the two values should be less than either value divided by the precision.
Assert.Less(Math.Abs(firstValue - secondValue), firstValue / Math.Pow(10, precision));
open FsUnit
actual |> should (equalWithin errorMargin) expected
Im trying to write something to get my images to show correctly.
I have 2 numbers "breedtePlaatje" and "hoogtePlaatje". When i load those 2 vars with the values i get back "800" and "500" i expect "verH" to be (500 / 800) = 0,625. Tho the value of verH = 0..
This is the code:
int breedtePlaatje = Convert.ToInt32(imagefield.Width);
int hoogtePlaatje = Convert.ToInt32(imagefield.Height);
//Uitgaan van breedte plaatje
if (breedtePlaatje > hoogtePlaatje)
{
double verH = (hoogtePlaatje/breedtePlaatje);
int vHeight = Convert.ToInt32(verH * 239);
mOptsMedium.Height = vHeight;
mOptsMedium.Width = 239;
//Hij wordt te klein en je krijgt randen te zien, dus plaatje zelf instellen
if (hoogtePlaatje < 179)
{
mOptsMedium.Height = 179;
mOptsMedium.Width = 239;
}
}
Any tips regarding my approach would be lovely aswell.
Dividing int by int gives an int.
double verH = (hoogtePlaatje/breedtePlaatje);
The right hand side of the assignment is an integer value.
Change breedtePlaatje and/or hoogtePlaatje to double and you will get the answer you expect.
Integer division will result in an Integer being returned as the division result.
You need one of the parameters of the division to be a float in order for the result to be a float. You can do this by casting one of them to a float.
double verH = (double)hoogtePlaatje/breedtePlaatje;
Or
double verH = hoogtePlaatje/(double)breedtePlaatje;
See the C# spec regarding division.
When you divide two integers, C# uses integer division, where the fractional part is discarded. In your case you're getting:
500 / 800 = 0 + 5/8
Which, discarding the fractional part, gives:
500 / 800 = 0
To get floating point division, cast one of the arguments to either double, float or decimal depending on the level of precision you need, which will cause the other argument to be implicitly converted to the same type and the division carried out using floating point rules instead of integer rules, e.g.
double result = (double)breedtePlaatje / hoogtePlaatje ;
I have never used C#, but probably you will need to cast one of the variables to double, like this:
double verH = (double)hoogtePlaatje/breedtePlaatje;
Try this:
double verH = double (hoogtePlaatje) / breedtePlaateje;
If you divide an int by an int, you will get a truncated answer. Cast one of them up to a double, and the entire division will be done as double.