Android Question “Solved” How to get the decimal part of a number?

[Sergey_New]

Dim latitude As Double=43.123456
Dim m As Long
'...
' need to get the value m=123456

How to do it?

 

[LucaMs]

t would crash if latitude had no decimals.

have tested it in B4J and it does not crash

 

[Mariano Ismael Castro]

View attachment 154185

This is strange, try trying it with 43123456

 

[LucaMs]

This is strange, try trying it with 43123456

Try:

    Dim latitude As Double= 43123456
    Log(latitude.As(String))

log:
4.3123456E7

 

[Mariano Ismael Castro]

Try:

    Dim latitude As Double= 43123456
    Log(latitude.As(String))

log:
4.3123456E7

Thanks for your teaching

 

[TILogistic]

?

    Dim latitude As Double= 43.123456
    Dim latitude As Double= 43
    Dim latitude As Double= .123456
    Dim m() As String = Regex.Split("\.", latitude)
    Log(IIf(m.Length = 1, 0, m(1)))

 

[ilan]

if you deal with numbers i think it is better to not convert it to string.
@Jeffrey Cameron gave you the correct answer in my opinion

“Solved” How to get the decimal part of a number?

Dim latitude As Double=43.123456 Dim m As Long '... ' need to get the value m=123456 How to do it?

www.b4x.com

it is better to stay with numbers when you deal with numbers.

 

[cklester]

Would this not work?

    Dim pVal As Double = 43.123456
    Dim pDec As Double = pVal - Floor(pVal)
    Log(pDec)

One of the problems with this approach is the rounding errors that can occur.

e.g.,

Sub AppStart (Args() As String)
    Log("Hello world!!!")
    Private num As Double = 31.987654321
    
    'inting
    Log(num.As(Int))
    Log(num-num.As(Int))

    'flooring
    Log(Floor(num))
    Log(num-Floor(num))

End Sub

resulting in

Hello world!!!
31
0.9876543210000008
31
0.9876543210000008

The best approach has to be converting it to a string first. That way, you avoid these slight rounding errors.

 

[ilan]

One of the problems with this approach is the rounding errors that can occur.

e.g.,

Sub AppStart (Args() As String)
    Log("Hello world!!!")
    Private num As Double = 31.987654321
   
    'inting
    Log(num.As(Int))
    Log(num-num.As(Int))

    'flooring
    Log(Floor(num))
    Log(num-Floor(num))

End Sub

resulting in



The best approach has to be converting it to a string first. That way, you avoid these slight rounding errors.

sorry but i dont understand.
you are dealing with doubles so why do you convert it to a int?

the best way to do it is just taking the double.
using floor to get the whole number before the decimal point and then substract the origin number with the whole number only.

as @Jeffrey Cameron did.
if you conert to string and start spliting the string you may get errors like

if there is no point in the number or if the phone instead of dot use comma "," you may get a crash.

 

[cklester]

sorry but i dont understand.
you are dealing with doubles so why do you convert it to a int?

It's the same effect as flooring. Look closely at the results again...

31
0.9876543210000008
31
0.9876543210000008

The original number is 31.987654321. Both int() and floor() achieve the same result... and both are wrong.

the best way to do it is just taking the double.
using floor to get the whole number before the decimal point and then substract the origin number with the whole number only.

That's exactly what I showed in my example. And the result is wrong.

 

[Jeffrey Cameron]

sorry but i dont understand.
you are dealing with doubles so why do you convert it to a int?

the best way to do it is just taking the double.
using floor to get the whole number before the decimal point and then substract the origin number with the whole number only.

as @Jeffrey Cameron did.
if you conert to string and start spliting the string you may get errors like

if there is no point in the number or if the phone instead of dot use comma "," you may get a crash.

To be honest, I didn't actually test this, that is why I phrased my reply as a question.

After running it in B4J it does appear that there is some conversion happening internally that is producing rounding errors. Honestly, 99% of my work is in VB.NET and I stopped using Double in favor of the Decimal datatype years ago and it does not suffer from this problem. When I ran this in B4J it came up with "0.12345599999999735" as the log result.

When I run this in VB.NET with the Decimal datatype, it produces the expected result:

        'VB.NET code
        Dim pVal As Decimal = 43.123456
        Dim pDec As Decimal = pVal - Math.Floor(pVal)
        Debug.Print("Value is " & pDec)

        'output shows "Value is 0.123456"

So, given the inherent inaccuracies of the Double datatype, perhaps a string solution (albeit less efficient) is the best answer. I would probably format the value to ensure the value is the expected type:

    Dim pVal As Double = 43.123456789
    Dim psVal As String = $"$0.9{pVal}"$
    Log(psVal.SubString(psVal.IndexOf(".")+1))

This shows the expected "123456789"

 

[Mike1970]

Did you try

dim decimal as float = my_number - my_number.As(int)

Not tested

 

[ilan]

It's the same effect as flooring. Look closely at the results again...

The original number is 31.987654321. Both int() and floor() achieve the same result... and both are wrong.




That's exactly what I showed in my example. And the result is wrong.

you are right i see it now, it is really weird

 

[epiCode]

Sub DecimalPart ( dblstr As Double ) As String
    If dblstr-Floor(dblstr) = 0 Then Return "0" Else  Return Regex.Replace("^[\d].\.*",dblstr.As(String),"")
End Sub

- Returns full String
- no rounding issues
- returns "0" when there is no decimal part
- returns leading zeroes for example 21.0023 will return 0023 as string and not 23.

 

[Cosmo]

For positive values:
Convert the decimal value 'a' to an integer 'b'
Convert b to a decimal value
subtract a-b, the result is the decimal part of your number.

HTH, regards

 

[Raart]

IndexOf(".") is not suitable, because you could have a different symbol as decimal separator (possibly you would have to use a variable in which to put the system separator).

The following code should work:

    Dim Num As Double = 2.36
'    Dim Num As Double = -2.36 ' to test negative numbers
    Dim IntPart As Int
    Dim DecimalPart As Double
   
    If Num >= 0 Then
        IntPart = Floor(Num)
    Else
        IntPart = Ceil(Num)
    End If
   
    DecimalPart = Abs(Num - IntPart)
    DecimalPart = Round2(DecimalPart, 2)
   
    Log("Original number: " & Num)
    Log("Int part: " & IntPart)
    Log("decimal part: " & DecimalPart)

Here the decimal part of a negative number results to be positive. This is incorrect.
-2 + 0.36 is not -2.36 but -1.64.

 

[LucaMs]

Here the decimal part of a negative number results to be positive. This is incorrect.
-2 + 0.36 is not -2.36 but -1.64.

The decimal part is to be considered "absolute", without sign.

 

[Raart]

Looks correct

Incorrect if the number is negative.

 

[Raart]

The decimal part is to be considered "absolute", without sign.

Not in math.

 

[LucaMs]

Not in math.

It is for the purpose.