I have been trying to solve this formula about the wind chill, I get the result but I need more numbers after the decimal point. No matter what I tried it didn't work. Please help!
static void Main()
{
double temp = 20;
double wind = 7;
double windChill = 35.74 + 0.6215 * temp + (0.4275 * temp - 35.75) * Math.Pow(wind, 0.16);
Console.WriteLine("so wind_chill = {0}", Math.Round(windChill, 15));
}
In this I get final number 11.03490062551, but I want 11.034900625509998. What am I doing wrong?
The problem is not with Math.Round, Math.Round will actually give you 11.034900625509991 which I guess is what you want.
The problem is with Console.WriteLine, it is the method that is causing the precision loss (because internally, it calls string.Format which is actually causing the precision loss).
To fix it, use the round-trip format specifier like this:
Console.WriteLine("so wind_chill = {0:R}", Math.Round(windChill, 15));
Please note also that the value of windChill should be good enough for you. There is no need to call Math.Round.
Related
here is what I'm trying to do:
double result = Math.Pow((1 + 8), 60) - 1;
And the result variable is:
1.7970102999144311E+57 double
And trying to:
Math.Round(result, 5);
Returns same : 1.7970102999144311E+57 double
I'd like to round it to 1.79701 for example
Any solutions ?
You're misunderstanding what you're seeing.
1.7970102999144311E+57
Is scientific notation for
1797010299914431100000... (with 41 trailing zeros).
It is a whole number, thus rounding it to 5 decimal places will correctly return the same value.
What you want to do is format the output of the number
String.Format(CultureInfo.InvariantCulture, "{0:0.#####E+0}", result);
Which returns 1.79701E+57. Note that this is a very different number from 1.79701
The problem you're having is that Math.Round rounds things to the right of the decimal point. For example if you're dealing with currency and you perform an operation that leaves you with $1.5234524, you would use:
Math.Round(1.5234524,2);
// output 1.52
The number you're dealing with is actually scientific notation for a very large number with nothing to the right of the decimal point. This is why the result of Math.Round is the same as the input.
The earlier comments and answers are correct. But to get what you are trying to achieve you can use the following:
double result = Math.Pow((1 + 8), 60) - 1;
string s = String.Format("{0:E5}", result);
double d = Double.Parse(s);
I have been searching forever and I simply cannot find the answer, none of them will work properly.
I want to turn a double like 0.33333333333 into 0,33 or 0.6666666666 into 0,66
Number like 0.9999999999 should become 1 though.
I tried various methods like
value.ToString("##.##", System.Globalization.CultureInfo.InvariantCulture)
It just returns garbage or rounds the number wrongly.
Any help please?
Basically every number is divided by 9, then it needs to be displayed with 2 decimal places without any rounding.
I have found a nice function that seems to work well with numbers up to 9.999999999
Beyond that it starts to lose one decimal number. With a number like 200.33333333333
its going to just display 200 instead of 200,33. Any fix for that guys?
Here it is:
string Truncate(double value, int precision)
{
string result = value.ToString();
int dot = result.IndexOf(',');
if (dot < 0)
{
return result;
}
int newLength = dot + precision + 1;
if (newLength == dot + 1)
{
newLength--;
}
if (newLength > result.Length)
{
newLength = result.Length;
}
return result.Substring(0, newLength);
}
Have you tried
Math.Round(0.33333333333, 2);
Update*
If you don't want the decimal rounded another thing you can do is change the double to a string and then get get a substring to two decimal places and convert it back to a double.
doubleString = double.toString();
if(doubleString.IndexOf(',') > -1)
{
doubleString = doubleString.Substring(0,doubleString.IndexOf(',')+3);
}
double = Convert.ToDouble(doubleString);
You can use a if statement to check for .99 and change it to 1 for that case.
Math.Truncate(value * 100)/100
Although I make no guarantees about how the division will affect the floating point number. Decimal numbers can often not be represented exactly in floating point, because they are stored as base 2, not base 10, so if you want to guarantee reliability, use a decimal, not a double.
Math.Round((decimal)number, 2)
Casting to a decimal first will avoid the precision issues discussed on the documentation page.
Math.Floor effectively drops anything after the decimal point. If you want to save two digits, do the glitch operation - multiply then divide:
Math.Floor(100 * number) / 100
This is faster and safer than doing a culture-dependent search for a comma in a double-converted-to-string, as accepted answer suggests.
you can try one from below.there are many way for this.
1.
value=Math.Round(123.4567, 2, MidpointRounding.AwayFromZero) //"123.46"
2.
inputvalue=Math.Round(123.4567, 2) //"123.46"
3.
String.Format("{0:0.00}", 123.4567); // "123.46"
4.
string.Format("{0:F2}", 123.456789); //123.46
string.Format("{0:F3}", 123.456789); //123.457
string.Format("{0:F4}", 123.456789); //123.4568
I am calculating the average of some values. Everything works fine.
What I want to do is to round the double to the 2nd decimal place.
e.g.
I would have 0.833333333333333333 displayed as
0.83
Is there anyway to do this?
Round the double itself like:
Math.Round(0.83333, 2, MidpointRounding.AwayFromZero);
(You should define MidpointRounding.AwayAwayFromZero to get the correct results. Default this function uses bankers rounding. read more about bankers rounding: http://www.xbeat.net/vbspeed/i_BankersRounding.htm so you can see why this won't give you the right results)
Or just the display value for two decimals:
myDouble.ToString("F");
Or for any decimals determined by the number of #
myDouble.ToString("#.##")
You say displays as - so that would be:
var d = value.ToString("f2");
See Standard numeric format strings
If you actually want to adjust the value down to 2dp then you can do what #middelpat has suggested.
Use
Math.Round(decimal d,int decimals);
as
Math.Round(0.833333333,2);
This will give you the result 0.83.
double d = 0.833333333333333333;
Math.Round(d, 2).ToString();
Try
decimalVar = 0.833333333333333333;
decimalVar.ToString ("#.##");
If you want to see 0.85781.. as 0.85, the easiest way is to multiply by 100, cast to int and divide by 100.
int val = (int)(0.833333333333333333 * 100);
var result = val /100;
It should produce the result you're looking for.
string input = Console.ReadLine();
decimal sum = Convert.ToDecimal(input);
if (sum >= (decimal)500.01)
{
//40% and 8 dollars off shipping costs are taken off total amount
decimal totalprice;
totalprice = (sum - 8) * .60m;
Math.Truncate(totalprice);
Console.WriteLine("Your final cost is:${0:0.00}", totalprice);
Console.Read();
The problem is, when I enter the price 598.88 dollars into my program, I should get 354.52.
The Math:
598.88 - 8 = 590.88. 590.88 * 60% = 354.528
I actually get 354.53 because C# rounds up instead of down.
For example,
If I get an answer like 519.998, I want it to STAY at 519.99.
Another example, if I get an answer like 930.755 I want it to stay at 930.75.
I looked into some answers, but Math.Truncate obviously doesn't work for me and using the *100 / 100 trick didn't work either. Keep in mind I'm a new student, So, if an answer could be noob-safe, that would be nice. Thanks.
The * 100 / 100 works fine, you might have been using it wrong. Try this below:
decimal totalprice = TruncateToTwoDigits((sum - 8) * .60m);
Console.WriteLine("Your final cost is:${0:0.00}", totalprice);
...
private static decimal TruncateToTwoDigits(decimal Value)
{
int adjusted = (int)Math.Truncate(Value * 100m);
return adjusted / 100m;
}
As a side note, Math.Truncate returns the truncated value, it doesn't change the input parameter as your code would imply.
Math.Truncate like all the other Math function returns the value after the function is called. The function doesn't alter your variable. Actually this was not possible with doubles (please see ref parameters). So you need to do:
totalprice = Math.Truncate(totalprice);
please notice that totalprice will have just the integer part so if the value is 45.985 the result is 45 so you need to multiply by 100 and then divide. http://msdn.microsoft.com/en-us/library/7d101hyf.aspx
The rounding up that you get there is because console.Write calls String.Format that will do that. See http://www.csharp-examples.net/string-format-double/ to get your write function call.
Modulo works too. (It's hard to say which is better for safety, readability, and performance.)
Decimal totalprice = (sum - 8m) * 0.60m; // Discount $8.00 and then 40%.
totalprice -= totalprice % 0.01; // Truncate to two decimal places.
Similar question at Truncate Two decimal places without rounding
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