Question double , number of decimal places
- Thread starter andrews
- Start date
That would be Math.Round.
IanRyder
Well-known member
Hi,
I must be missing something here but, yeah, Math.Round? BTW, that syntax you posted should also be giving you an error?
Maybe you should read this:-
http://msdn.microsoft.com/en-us/library/75ks3aby(v=vs.100).aspx
Cheers,
Ian
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There is no single method to do that. Math.Truncate is similar to Math.Round and does what you want but only at the decimal point and not for a specific number of decimal places. Other than what you're doing, the only options I can think of are multiplying the number by the appropriate factor of 10, truncating and then dividing by the same factor of 10 or else converting to a String, using Substring and then converting back to a number. That second option is a real hack but you could write an extension method that used the first option:
Imports System.Runtime.CompilerServices
Friend Module DoubleExtensions
<Extension>
Public Function Truncate(source As Double, decimals As Integer) As Double
Dim wholePart = Math.Truncate(source)
Dim fractionalPart = source - wholePart
fractionalPart = fractionalPart * 10 ^ decimals
fractionalPart = Math.Truncate(fractionalPart)
fractionalPart = fractionalPart / 10 ^ decimals
Return wholePart + fractionalPart
End Function
End Module
Sample usage:
Dim number = 0.254789457845128 MessageBox.Show(number.ToString()) number = number.Truncate(8) MessageBox.Show(number.ToString())
You are not right. The Decimal type has its own Truncate method but, as I've shown, the Math.Truncate method works for Double as well as Decimal.
Solitaire
Well-known member
Truncate without rounding
Here's a short solution that seems to work:
If you don't need the original value of the variable, then you could shorten it to this:
Here's a short solution that seems to work:
VB.NET:
Sub Main()
Dim mynum As Double = 3.1234567898765
Dim newnum As Double
newnum = Math.Round(mynum, 8) 'Round changes the last digit to a 9 instead of 8
Console.WriteLine("Round to 8: " & newnum) 'not what you want
'Truncate to 8 without rounding:
newnum = Math.Round(mynum, 9) 'or newnum = mynum
newnum = newnum - 0.000000005
newnum = Math.Round(newnum, 8)
Console.WriteLine("Truncate to 8: " & newnum)
Console.ReadLine()
End SubIf you don't need the original value of the variable, then you could shorten it to this:
VB.NET:
Dim mynum As Double = 3.1234567898765
'Truncate to 8 without rounding:
mynum = mynum - 0.000000005
mynum = Math.Round(mynum, 8)
Console.WriteLine("Truncate to 8: " & mynum)
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Um, yes you're missing something. It's like you didn't even bother to read post #5. I already said myself that Math.Truncate will not truncate to a specific number of decimal places so it's no great breakthrough that you worked that out. That would also explain why I wrote a method for you that does what you want and uses Math.Truncate internally. I even provided an example of how to use it. If you're not up to copying a bit of code that I went to the trouble of writing for you then I'm wasting my time here.
Solitaire
Well-known member
andrews: I posted before I saw your short solution: round(x-0.000000005,8) is similar to what I had posted above. It works as long as you don't need the original value, or assign it to a different variable.
jmc: Thanks for your input. It wasn't a waste of time. I tried your function and it's worth keeping. (I had written something similar to this using ^ and / for a different purpose a long time ago and I don't remember offhand what it was for.)
jmc: Thanks for your input. It wasn't a waste of time. I tried your function and it's worth keeping. (I had written something similar to this using ^ and / for a different purpose a long time ago and I don't remember offhand what it was for.)
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Yes, Solitaire the same as what I propose.
And I think it is till now the best.
And not the solution Jimcilhinney which is more complicated and not so fast I think.
Imports System.Runtime.CompilerServices
Friend Module DoubleExtensions
<Extension>
Public Function Truncate(source As Double, decimals As Integer) As Double
Dim wholePart = Math.Truncate(source)
Dim fractionalPart = source - wholePart
fractionalPart = fractionalPart * 10 ^ decimals
fractionalPart = Math.Truncate(fractionalPart)
fractionalPart = fractionalPart / 10 ^ decimals
Return wholePart + fractionalPart
End Function
End Module
And I think it is till now the best.
And not the solution Jimcilhinney which is more complicated and not so fast I think.
Imports System.Runtime.CompilerServices
Friend Module DoubleExtensions
<Extension>
Public Function Truncate(source As Double, decimals As Integer) As Double
Dim wholePart = Math.Truncate(source)
Dim fractionalPart = source - wholePart
fractionalPart = fractionalPart * 10 ^ decimals
fractionalPart = Math.Truncate(fractionalPart)
fractionalPart = fractionalPart / 10 ^ decimals
Return wholePart + fractionalPart
End Function
End Module
I just used this code to test both options:
Dim source = 0.254789457845128
Dim destination As Double
Dim timer As Stopwatch
timer = Stopwatch.StartNew()
For i = 1 To 100000
destination = Math.Round(source - 0.000000005, 8)
Next
Dim time1 = timer.Elapsed
timer = Stopwatch.StartNew()
For i = 1 To 100000
destination = source.Truncate(8)
Next
Dim time2 = timer.Elapsed
MessageBox.Show(String.Format("Round: {1}{0}Truncate: {2}{0}Difference: {3}",
Environment.NewLine,
time1,
time2,
time2 - time1))
It showed that using Round was about an order of magnitude faster than using Truncate. The thing is though, Calling Truncate 100,000 times still only takes about 5/100 of a second, so it's not a difference that anyone is ever going to notice unless they're doing some serious number crunching.The advantage of my approach is that you write the method once and then you can use it as many times as you like in as many places as you like and the code to call it is much neater because all you have to do is specify the number of decimal places you want. Using your approach is less neat and not nearly as clear to anyone who may read it. Making your code as readable as possible is always something to strive for. What do I know though? I've only been doing this for 12 years. I'm sure that you and those few nanoseconds that you save will be very happy together.
Your approach is good but :
I can also make a short function with my method and then see about the time it is taken
Public Function TruncateNew(ByVal source As Double, ByVal decimals As Integer) As Double
Dim i As Integer
Dim prod, difference As Double
prod = 10
For i = 1 To decimals
prod = prod * 10
Next
difference = 5 / prod
Return Math.Round(source - difference, decimals)
End Function
a) I found that for i = 1 to decimals .... is a little faster then prod = 10^(decimals+1)
b) defining prod as integer is also a little faster but then decimals <= 8
I calculated 100000000 :
My first method : 2.4 seconds
my function : 11.5 seconds (with prod as integer 10 seconds)
your function : 21.2 seconds
Yes, I do serious number crunching, that is the reason I asked for a fast method but I agree this is very exceptional
So, the programmers can choose.
I thank you for the interest you give to my question
I can also make a short function with my method and then see about the time it is taken
Public Function TruncateNew(ByVal source As Double, ByVal decimals As Integer) As Double
Dim i As Integer
Dim prod, difference As Double
prod = 10
For i = 1 To decimals
prod = prod * 10
Next
difference = 5 / prod
Return Math.Round(source - difference, decimals)
End Function
a) I found that for i = 1 to decimals .... is a little faster then prod = 10^(decimals+1)
b) defining prod as integer is also a little faster but then decimals <= 8
I calculated 100000000 :
My first method : 2.4 seconds
my function : 11.5 seconds (with prod as integer 10 seconds)
your function : 21.2 seconds
Yes, I do serious number crunching, that is the reason I asked for a fast method but I agree this is very exceptional
So, the programmers can choose.
I thank you for the interest you give to my question
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