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How to cut off decmil places (not round)

Posted on 2010-09-24
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I need to take a double variable and cut off any digits two characters to the right of the decimal point.  It cannot be rounded, must be cut off.  The number of digits both sides of the decimal will vary.  I can't seem to find a method for that.  Can someone tell me a way to do this?
Thanks, Lynn
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Question by:Lambel
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10 Comments
 
LVL 75
ID: 33755423
Int(YourNumber)

mx
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Accepted Solution

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Patrick Matthews earned 2000 total points
ID: 33755472
To remove all decimals, take MX's advice above.If you want to keep the first 2 decimal places:Int(YourNumber * 100) / 100
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LVL 75
ID: 33755508
Read that 3 times and still missed:
"two characters to the right of the decimal point."

:-(
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LVL 93

Expert Comment

by:Patrick Matthews
ID: 33755524
:)And if perchance you meant, keep the first decimal place and truncate the rest:Int(YourNumber * 10) / 10
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LVL 58
ID: 33755780
or:
Int(YourNumber *10^x)/ 10^x
where 'x' is the number of decimals you want. I've done it that way for a number of years, simply because I find it easier then counting zero's. Not as efficent of course, but I've gotten caught a couple of times by not having enough zero's on one or both sides.
FWIW,
JimD.
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LVL 49

Expert Comment

by:Dale Fye
ID: 33755931
JD,

I really like that method.  I'm going to add that to my toolkit as:

Public Function fnFix(SomeValue As Variant, Optional Decimals As Integer = 2) As Variant

    fnFix = Int(SomeValue * 10 ^ Decimals) / 10 ^ Decimals
   
End Function

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LVL 58
ID: 33756018
fyed,
Not quite as elegant as yours...this is from A2 days.
Also have attached RoundSignificant(), which I've never really seen before and ended up writting to answer a question here on EE a few years back.  You may find it useful, although I've never found an occasion to use it yet.    I'm not sure I really understand it's application (Patrick may have some input on that).
JimD.

' Accepts: a variant value
' Purpose: converts multiplace decimal numbers
' Returns: a number rounded to d decimal places
'          or a zero if the value it was called for was null
'          If d is negative or null d is set to 0 and the function is like Int()
'          In any case d is set to Int(d)!
' Author:  Marcus O. M. Grabe, CIS 100120,1405
'          Please send a message, if you like it or if you have any suggestions.
Function Round(n, d)

 On Error Resume Next
 
 If IsNull(n) Or IsNull(d) Then
    Round = 0
 Else
    If d < 0 Then
       d = 0
    Else
       d = Int(d)
    End If
    Round = CLng(n * (10 ^ d)) / (10 ^ d)
 End If

End Function


Public Function RoundSignificant(varValue As Variant, intNumSignificantDigits As Integer) As String

    Dim strPrefix As String
    
    Dim dblFactor As Double
    Dim dblABSofValue As Double
    Dim strFormatedValue As String
    
    Dim lngPos As Long
    Dim bolDecimalPoint As Boolean
    Dim strChr As String
    
    Dim lngNumOfDigits As Long
        
    
    ' Check for a prefix ('>' or '<').  If there is one,
    ' strip it off for now.
    If left(varValue, 1) = ">" Or left(varValue, 1) < "<" Then
      strPrefix = left(varValue, 1)
      varValue = Val(Mid(varValue, 2))
    Else
      strPrefix = ""
    End If
    
    ' Check for null
    varValue = Nz(varValue, 0)

    ' If value of zero, return "N/A"
    If varValue = 0 Then
      RoundSignificant = "N/A"
    Else
      
      ' Get the factor
      dblFactor = 10 ^ Int(Log(Abs(varValue)) / Log(10) - intNumSignificantDigits + 1)
      
      ' Based on the factor, get an absolute value that's rounded.
      dblABSofValue = Int(Abs(varValue) / dblFactor + 0.5) * dblFactor
      
      ' Format the value as a string.
      strFormatedValue = Format((IIf(varValue >= 0, 1, -1) * dblABSofValue), "#0.00000000000000000000")
      
      ' Do we have a decimal point?
      If InStr(strFormatedValue, ".") > 0 Then
        ' If so, chop off all zeros on the right
        While right(strFormatedValue, 1) = "0"
            strFormatedValue = left(strFormatedValue, Len(strFormatedValue) - 1)
        Wend
      End If
      
      ' Scan for the number of digits in the string
      For lngPos = 1 To Len(strFormatedValue)
        strChr = Mid(strFormatedValue, lngPos, 1)
           
        If strChr > "0" And strChr <= "9" Then
          lngNumOfDigits = lngNumOfDigits + 1
        End If
       
        If strChr = "." Then
          bolDecimalPoint = True
        End If
      Next
      
      ' Is the number of digits found less then the significance required?
      If lngNumOfDigits < intNumSignificantDigits Then
        ' If so and we have decimal point, add some zeros
        If bolDecimalPoint = True Then
          strFormatedValue = strFormatedValue & String(intNumSignificantDigits - lngNumOfDigits, "0")
        End If
      End If
          
      ' Do we have anything to the right of the decimal?
      ' If not, remove the decimal point
      If right(strFormatedValue, 1) = "." Then
        strFormatedValue = left(strFormatedValue, Len(strFormatedValue) - 1)
      End If
      
      ' Add the prefix back in if we have one
      If strPrefix <> "" Then
        RoundSignificant = strPrefix & strFormatedValue
      Else
        RoundSignificant = strFormatedValue
      End If

    End If

End Function

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Author Closing Comment

by:Lambel
ID: 33756028
Thanks much!
Lynn
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LVL 49

Expert Comment

by:Dale Fye
ID: 33756129
JD,

I would disagree, yours includes the necessary error handling to  prevent me from inadvertantly raising the value to the 2.3 power and handles NULLs, which is always a good idea when working with variants.


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LVL 52

Expert Comment

by:Gustav Brock
ID: 33757788
I see no reason why you would pass anything else than a numeric to a rounding function, indeed as:

    Int(SomeVariant * 10 ^ Decimals) / 10 ^ Decimals

would return a Double anyway. Thus:

Public Function fnFix(dblValue As Double, Optional intDecimals As Integer = 2) As Double

    fnFix = Int(dblValue * 10 ^ intDecimals) / 10 ^ intDecimals
   
End Function


As for the posted RoundSignificant function, this is quite clumsy. Ignoring that it probably is made for some specific purpose because of the filtering for "<" and ">", it is still convoluted and - worst - will fail in any international environment where the decimal point is not a dot. Except for very special cases, numbers should _always_ be handled like numbers, not strings or anything else.

Below is a proven method which demands that you recall some of the math lessons in school about logarithmics.

Rounding like this is very useful for example for converting a pricelist neatly rounded in one currency to another currency maintaining the same level of rounding.
Here are four colums:

1: Amount in EUR
2: Amount in DKK
3: Amount in DKK rounded by 100
4: Amount in DKK rounded to two significant digits

 15            111,45        100           110
 150           1114,5        1100          1100
 1500          11145         11100         11000
 15000         111450        111500        110000
 150000        1114500       1114500       1100000

And here in the other direction:

1: Amount in DKK
2: Amount in EUR
3: Amount in EUR rounded by 100
4: Amount in EUR rounded to two significant digits

 110           14,8048452220727            0             15
 1100          148,048452220727            100           150
 11000         1480,48452220727            1500          1500
 110000        14804,8452220727            14800         15000

Have fun!

/gustav
Public Function RoundSignificantCurrency( _
  ByVal curValue As Currency, _
  ByVal bytSignificantDigits As Byte, _
  Optional ByVal booInteger As Boolean) _
  As Currency

' Rounds curValue to bytSignificantDigits digits.
'
' Performs no rounding if bytSignificantDigits is zero.
' Rounds to integer if booInteger is True.
'
' Rounds correctly curValue until max/min value of currency type multiplied with
' 10 raised to the power of (the number of digits of the index of curValue) minus
' bytSignificantDigits.
' This equals roughly +/-922 * 10 ^ 12 for any value of bytSignificantDigits.
'
' Requires:
'   Function Log10.
'
' 2001-10-19. Cactus Data ApS, CPH.
' 2002-04-02. Added CDec() for correcting bit errors of reals.
' 2007-04-18. Int replaced with Fix to round negative values correctly.
'             Parameter booInteger made Optional.

  Dim dblTmp    As Double
  Dim dblFactor As Double
  Dim dblPower  As Double
  
  ' No special error handling.
  On Error Resume Next
  
  If bytSignificantDigits = 0 Or curValue = 0 Then
    ' Nothing to do.
  Else
    dblPower = Int(Log10(Abs(curValue))) + 1 - bytSignificantDigits
    If booInteger = True Then
      ' No decimals.
      If dblPower < 0 Then
        dblPower = 0
      End If
    End If
    dblFactor = 10 ^ dblPower
    dblTmp = curValue / dblFactor
    dblTmp = Fix(dblTmp + Sgn(dblTmp) / 2)
    ' Apply CDec() to correct for possible bit error when multiplying reals.
    curValue = CDec(dblTmp * dblFactor)
  End If
  
  RoundSignificantCurrency = curValue

End Function


Public Function Log10( _
  ByVal dblValue As Double) _
  As Double

' Returns Log 10 of input dblValue.

  ' No error handling as this should be handled
  ' outside this function.
  '
  ' Example:
  '
  '   If dblMyValue > 0 then
  '     dblLogMyValue = Log10(dblMyValue)
  '   Else
  '     ' Do something else ...
  '   End If

  Log10 = Log(dblValue) / Log(10)

End Function

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