[RESOLVED] changing a decimal into a whole number

[dmaar]

Hey everyone.

I am attempting to take the decimal places of a given input, round it so there are only two decimal places, then change it to a whole number. So far I have been able to remove the integer part of the number, leaving only the decimal, but the other procedures are eluding me. Any help would be appreciated. Here's what I have so far: (VB 2008)

Private Function DecimalNumber(ByVal number As Double) As Double

        Dim Input As Double
        Dim SubtractNumber As Decimal
        Dim Result As Double

        Input = Math.Truncate(number)

        SubtractNumber = number - Input

        Return Result

    End Function

That code returns the decimal part of the input just fine, and I thought maybe multiplying it by 10 would work....but obviously you run into problems when there is more than one decimal place. Thanks!

[bergerkiller]

You could try using the Math.Round() function. It allows you to change the amount of decimal places a double value has.

[dmaar]

That worked great in reducing the number of places, thanks. Now all I need to do is convert it to a whole number. For example, if the value is ".24" it needs to just be "24". Here's the new code with the round function:

    Private Function DecimalNumber(ByVal number As Double) As Double

        Dim Input As Double
        Dim SubtractNumber As Decimal
        Dim Result As Double

        Input = Math.Truncate(number)

        SubtractNumber = number - Input

        Result = Math.Round(SubtractNumber, 2)

        Return Result

    End Function

[Grimfort]

You can not round a whole number, so you need to round the number first to 2 places first.

[Pradeep1210]

After what you are doing you just need to multiply with 100.

Try this:

Private Function DecimalNumber(ByVal number As Double) As Double
    Return Fix(100 * (number - Fix(number) + 0.00000000001))
End Function

[dmaar]

You're right Pradeep, that works for the particular example I posted and all decimals with 2 places, but when you get a decimal with only one place you run into problems with that equation. For example, if the decimal was ".2", .2 * 100 = 20, not 2.

Is there a way to do something conditional? So if the number of places = 2, then it multiplies by 100, and if the places is 1 then it multiplies by 10. So it would be something like this:

If (number of places after decimal point = 2) Then
DecimalNumber * 100
ElseIf (number of places after decimal point = 1) Then
DecimalNumber * 10

Thanks!

[dbasnett]

HUH? What?

        Dim foo As Double = Math.PI 'a test number with more than two decimals ;)
        Dim bar As Double = Math.Round(foo, 2) 'two decimals

        Debug.WriteLine(foo.ToString)
        Debug.WriteLine(bar.ToString)
        Stop

        Dim d As Double = bar * 100 'get rid of decimals
        Debug.WriteLine(d.ToString)
        Stop

        bar = Math.Round(foo, 1) 'one decimal
        Debug.WriteLine(bar.ToString)
        d = bar * 100 'get rid of decimals
        Debug.WriteLine(d.ToString)
        Stop

[stanav]

Try this:

Private Function FixMyNumber(ByVal input As Double) As Integer
        Dim inputAsString As String = input.ToString("#.00")
        Dim decimalPart As String = inputAsString.Substring(inputAsString.IndexOf("."c) + 1)
        Return Integer.Parse(decimalPart)
    End Function

[Pradeep1210]

Try this:

Private Function DecimalNumber(ByVal number As Double) As Double
    number = 10 * (number - Fix(number) + 0.0000000000001)
    If Math.Round(number, 9) <> Fix(number) Then number *= 10
    Return Fix(number)
End Function

[dbasnett]

If you want just the decimal part of a number

        Dim foo As Double = Math.PI 'a test number with more than two decimals ;)
        Dim dp As Double = Math.Round(foo - Math.Floor(foo), 2)

If the number can be negative then

        Dim foo As Double = -Math.PI 'a test number with more than two decimals ;)
        Dim dp As Double
        If foo > 0 Then dp = Math.Round(foo - Math.Floor(foo), 2) Else dp = Math.Round(foo - Math.Ceiling(foo), 2)

[Grimfort]

Moo:

Val(Microsoft.VisualBasic.Right((Math.Round(Foo,2)).ToString("#.0#"),2).Trim("."c))

There are 5 cases

1 = 0
1.2 = 2
1.23 = 23
1.234 = 23
0.04 = 4

[dbasnett]

Hey everyone.

I am attempting to take the decimal places of a given input, round it so there are only two decimal places, then change it to a whole number.... Thanks!

Maybe we should know why you are taking the decimal part and trying to convert it to a whole number. The answer might be different if we knew more.

[dmaar]

Thanks everybody. Solved that problem...but it created another.

What I ended up doing:
After the decimal is separated from the whole number, the length of the value is either equal to four or three, depending on whether there is one decimal place or two.

So I am just using an If statement to see whether it is 2 dec places or 1, and then using a substring(2,2) or (2,1) to remove the decimal place. Works pretty well.

The problem happens when I get a value like ".04". In that case, even though the end value shows up in the code as "04" as I want it to, it always print just "4". How could I fix this?

[Pradeep1210]

have you tried post #9. It works ok unless I have misunderstood something.

[Grimfort]

I just edited mine, should work with 0.04 as well.

[dmaar]

Thanks guys! Finally got one working the way i want it.

[Grimfort]

Show us then.

[polecat]

This should work I think

Return Cint(result)

[jmcilhinney]

I'm fairly sure that this should work for all scenarios:

Dim s = input.ToString("n2")
 
s = s.TrimEnd("0"c)
s = s.Substring(s.IndexOf(".") + 1)
 
Dim output = CInt(s)

If you want to allow for systems that use commas for decimal separators then you can get the decimal separator for the current culture.

[dmaar]

Show us then.

Here's the basic idea (this is after rounding the decimal out to two digits):

        Length = CStr(Len(DecimalString) - 2)

        If CDbl(Length) = 2 Then
            Result =  'Corresponding function Substring(2,2)
        ElseIf CDbl(Length) = 1 Then
            Result = 'Corresponding function Substring(2,1)
        End If

If there are 2 decimal places after the round, the length will always be "4". If there is 1 decimal place after the round, the length will always be "3". I subtract 2 from the length (just to make it easier to follow), then use If statements to send the decimal to it's corresponding function, depending on whether there are 2 decimal places or 1. It is sent as a substring in order to get rid of the decimal places.

Alternatively, you could use those length conditions to remove the decimal place, by multiplying it by 100 for the first and 10 for the second. However I found that the problem with this was that if you have something like ".04", you get "4" when multiplied by 100. I wanted to keep the "0" as well as the 4, and the substring method does that.

I didn't do it exactly like anyone posted, but you're responses definitely helped me understand this better and figure one out for myself. This one may seem less logical, but I'm still pretty new at this.


One last question: Does anyone know why when I subtract the truncated number from the input, the decimal it returns is often not right? For example:

DecimalNO = InputNO - TruncatedNO

Where InputNO is 8.46, TruncatedNO is 8, ALWAYS gives me "0.46000000000000085". What's up with that? It should just be 0.46.

The rounding takes care of this, but ideally I would like to throw an exception if the DecimalNO is greater than 2...this problem makes this hard to do.

[dbasnett]

        Dim iStr As String = "-8.46"

        Dim DecimalNO As Double
        Dim InputNO As Double
        Dim TruncatedNO As Double

        If Double.TryParse(iStr, InputNO) Then
            'good input value
            If InputNO > 0 Then
                TruncatedNO = Math.Floor(InputNO)
            Else
                TruncatedNO = Math.Ceiling(InputNO)
            End If
            DecimalNO = Math.Round(InputNO - TruncatedNO, 2)
        End If
        Debug.WriteLine("Input {0}, Truncated {1}, Decimal {2}", _
                        InputNO, TruncatedNO, DecimalNO)