How do you access the decimal part of a number in MS Access? More specifically I want only the component after the decimal point, but not including the decimal point. This must also work for all whole numbers. I've seen this answered for other SQL engines, but they don't work in Access. I can't be much more specific than this because of the sensitive nature of what I'm actually working on.
For example given the following numbers the input is on the left and the output is on the right. Output can be either text or a number.
Source Correct Incorrect1 Incorrect2
10.0 0 0.0 .0
3.14159 14159 0.14159 .14159
45.65 65 0.65 .65
173.0 0 0.0 .0
143.15 15 0.15 .15
If I was using C# the following code would give me what I want:
private string getDecimalComponent(double input)
{
String strInput = input.ToString();
if (strInput.Contains('.'))
{
return strInput.Split('.')[1];
}
else
{
return "0";
}
}
Subtract the integer portion of the value.
Example:
4.25 - Int(4.25) = 0.25
Or, as a sample SQL expression:
SELECT
[myDecimalNumber],
[myDecimalNumber] - Int([myDecimalNumber]) as [rightOfDecimal]
FROM tableA
Something like:
SELECT 3.14%1 AS mycolumn from mytable
Depends on the circumstance. If this is, for instance, in a textbox, you can use InStr to find the decimal, and then use the Mid() function to get the number after it. If it's part of an arithmetic equation, then I would use the Int() function and subtract one number from the other to get the difference.
If you can elaborate on how it's being used, and in what context, I can edit my answer to give you more specifics.
EDIT: After more info came to light, try this:
Public Function GetParts(Temp1 as Double)
Temp2 = Int(Temp1)
Temp3 = Mid(Temp1, InStr(Temp1, ".") + 1)
MsgBox Temp2
MsgBox Temp3
End Function
A stable solution producing String containing decimal places from Double.
SELECT DecimalPlaces(MyColumn) FROM MyTable
where the above user-defined function contains the following code:
Function DecimalPlaces(ByVal value As Double) As String
Dim intPartLen As String
intPartLen = Len(CStr(CInt(Abs(value))))
If Len(CStr(Abs(value))) > intPartLen Then
DecimalPlaces = Mid(CStr(Abs(value)), intPartLen + 2)
Else
DecimalPlaces = "0"
End If
End Function
It avoids common mistakes, so it is
locale-independent - works with any decimal separator (it only assumes it is a single character)
preserves precision - avoids subtraction, takes the result only from string representation
Note: those Abs() calls are really required (hint: -0.1)
Related
I am using a Math.Round function in my program as follows
Math.Round((ProductWeightByType * 4.18), 2)
This works perfect except for when the last character is already a 0 then it drops that 0 so example $1.70 becomes $1.7
How do i fix this?
When dealing with money, use Decimal. It will not suffer from precision issues as Double does. Oh, it's not clear that you're not using Decimal. Well it's also not clear that your not using strings. There are no characters in numbers, rather in strings. But here is an approach which uses proper typing
Sub Main()
Dim ProductWeightByType As Decimal = 0.4056D
Dim cost As Decimal = 4.18D
Dim formattedCost As String = $"{cost:C2}"
Dim weightedCost As Decimal = ProductWeightByType * cost
Dim formattedWeightedCost As String = $"{weightedCost:C2}"
Console.WriteLine($"Cost: {cost}")
Console.WriteLine($"Formatted Cost: {formattedCost}")
Console.WriteLine($"Weight: {ProductWeightByType}")
Console.WriteLine($"Weighted Cost: {weightedCost}")
Console.WriteLine($"Formatted Weighted Cost: {formattedWeightedCost}")
Console.ReadLine()
End Sub
Cost: 4.18
Formatted Cost: $4.18
Weight: 0.4056
Weighted Cost: 1.695408
Formatted Weighted Cost: $1.70
Actually, you should probably not use Math.Round here for money. You may start to accumulate a loss or gain of fractions of pennies if you continue to round and use that value. Of course, if you want to display the cost, then format as currency just as the other answer did (mine does it as well twice).
Note, it's important to show how if the value is $1.695408 then it is rounded to $1.70, gaining $0.004592. Keep your cost in its original unspoiled numeric format without rounding, and just use C2 format for display only.
you should try using string format, something like this could do what you need:
String.Format("{0:C2}", Math.Round((ProductWeightByType * 4.18), 2))
Vb.net has a decimal data type.
Unlike normal double or floating points, decimal data type can store values like 0.1
Now say I have a variable like precision.
Say precision is 8
So basically I want to do
Protected Overridable Sub setPairsPricesStep2(decimalPrecission As Long, Optional base As String = "", Optional quote As String = "")
If decimalPrecission = 8 Then
Return
End If
Dim price = 10D ^ (-decimalPrecission)
setPairsPriceStep1(price, base, quote)
End Sub
There is a problem there
the result of Dim price = 10D ^ (-decimalPrecission) is double, not decimal. I can convert it to decimal but then I will lost the precission.
So what is the right way to do it? Should I just use for next but that's hardly elegant.
It's simple
I want a function that given precisions give decimal value.
For example, if precision is 1 I got 0.1. If precision is 5, I got 0.00001
I ended up doing this
For i = 1 To decimalPrecission
price *= 0.1D
Next
But surely there is a better way
Update:
Per comment, I tried
Dim e = 10D ^ -5
Dim e1 = 10D ^ -5L
The type of e and e1 are both double.
I suppose I can do Cdec(e). But then it means I have lost accuracy because normal double cannot store .1 correctly.
I want a function that given precisions give decimal value.
For example, if precision is 1 I got 0.1. If precision is 5, I got 0.00001
Since you are working with the Decimal type, the simplest way to get this result is to use the Decimal constructor that allows you to specify the scale factor.
Public Sub New (lo As Integer, mid As Integer, hi As Integer, isNegative As Boolean, scale As Byte)
From the Remarks section of the above referenced documentation,
The binary representation of a Decimal number consists of a 1-bit sign, a 96-bit integer number, and a scaling factor used to divide the integer number and specify what portion of it is a decimal fraction. The scaling factor is implicitly the number 10 raised to an exponent ranging from 0 to 28.
So you can see that if take the value of one divided by 10 to the first power, the result is 0.1. Likewise, one divided by 10 to the fifth power, the result is 0.00001.
The lo, mid, and hi arguments in the constructor could be obtained by uisng the [Decimal.GetBits Method](Decimal.GetBits Method), but for this simple case, I chose to hard code the values for the value of one stored as a decimal.
To obtain a value of 0.1D:
New Decimal(1, 0, 0, False, 1)
To obtain a value of 0.00001D:
New Decimal(1, 0, 0, False, 5)
Dim stringrepresentation = "1E-" + decimalPrecission.ToString
Dim price = Decimal.Parse(stringrepresentation, System.Globalization.NumberStyles.AllowExponent)
This is what I basically did. Basically I created a string 1E-5, for example, and use decimal.parse to get the decimal
I wonder if there is a better way but I have no idea.
Actually Jimy ways may work too but rounded to a number
I have three numbers from my database and want to compare them in an if statement.
I have a simple conevert function that returns only doubles.
Public Function RetDbl(ByVal obj As Variant) As Double
On Error Resume Next
RetDbl = val(Replace(Nz(obj, 0), ",", "."))
End Function
The statement is
If RetDbl(rs.value("NumA")) + RetDbl(rs.value("NumB")) <> (RetDbl(rs.value("NumC")) * 1000) Then
'[... do some code ...]
End If
With RetDbl(rs.value("NumA")) = 0.33, RetDbl(rs.value("NumB") = 0.5 and RetDbl(rs.value("NumC")) = 0.00083
This always returns false
I also tried:
In the direct field (STRG + G): ?cdbl(0.33) + cdbl(0.50) = cdbl(0.83) returns false. When i leave out the last part it returns 0.83
How can i compare these numbers?
Comparing floating numbers is hard. Only yesterday, I've posted this question
My solution:
Public Function DblSafeCompare(ByVal Value1 As Variant, ByVal Value2 As Variant) As Boolean
'Compares two variants, dates and floats are compared at high accuracy
Const AccuracyLevel As Double = 0.00000001
'We accept an error of 0.000001% of the value
Const AccuracyLevelSingle As Single = 0.0001
'We accept an error of 0.0001 on singles
If VarType(Value1) <> VarType(Value2) Then Exit Function
Select Case VarType(Value1)
Case vbSingle
DblSafeCompare = Abs(Value1 - Value2) <= (AccuracyLevelSingle * Abs(Value1))
Case vbDouble
DblSafeCompare = Abs(Value1 - Value2) <= (AccuracyLevel * Abs(Value1))
Case vbDate
DblSafeCompare = Abs(CDbl(Value1) - CDbl(Value2)) <= (AccuracyLevel * Abs(CDbl(Value1)))
Case vbNull
DblSafeCompare = True
Case Else
DblSafeCompare = Value1 = Value2
End Select
End Function
Note that the AccuracyLevel (epsilon) could be set to a smaller value, and I'm using the same value for singles and doubles, but it did well for my purposes.
I'm using a relative epsilon, but multiplying it with the first, and not the largest value, since if there's a significant difference the comparison will fail anyway.
Note that I'm using <= and not < since else DblSafeCompare(cdbl(0) ,cdbl(0)) would fail
Note that this function checks for type equality, so comparing integers to longs, doubles to singles, etc. all fails. Comparing Null to Null passes, however.
Implement it:
?DblSafeCompare(cdbl(0.33) + cdbl(0.50) ,cdbl(0.83))
?DblSafeCompare(cdbl(0.331) + cdbl(0.50) ,cdbl(0.83))
Comparing floating point numbers is really an issue, if you try to do it without understanding of the nature of the floating numbers.
Here is a nice article about it - https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html and How dangerous is it to compare floating point values?
In general, this problem is so big, that some languages like C# have developed a specific class called Decimal which makes comparing run as it would be expected by a non-programmer. Decimal info. In VBA, a similar class is Currency. Thus
CCur(0.33) + CCur(0.50) = CCur(0.83)
Returns True. VBA supports the function CDec, which converts a double to a decimal number, but it does not support the class Decimal. Thus:
CDec(0.33) + CDec(0.50) = CDec(0.83)
would also return True. And is with some better accuracy than Currency. CDec documentation.
I've read much about rounding in Excel. I found out that VBA's Round() function uses "Bankers rounding" while Application.WorksheetFunction.Round() uses more or less "normal" rounding. But it didn't help me to understand this:
? Round(6.03499,2)
6.03
Why? I want to see 6.04, not 6.03! The trick is that
? Round(Round(6.03499,3),2)
6.04
I thought a bit and developed a subroutine like this:
Option Explicit
Function DoRound(ByVal value As Double, Optional ByVal numdigits As Integer = 0) As Double
Dim i As Integer
Dim res As Double
res = value
For i = 10 To numdigits Step -1
res = Application.Round(res, i)
Next i
DoRound = res
End Function
It works fine.
? DoRound(6.03499,2)
6.04
But it is not cool. Is there any built-in normal rounding in Excel?
If you round 6.03499 to 3 digits it will be 6.035 - which is correct.
If you round 6.03499 to 2 digits it will be 6.03 - which is correct
However - the example where you first round to 3 digits, then to 2 is also correct, by the following statement:
Round(6.03499, 3) gives 6.035
Round(6.035, 2) gives 6.04
If you want Round(6.03499, 2) to give 6.04 you have to use Application.WorksheetFunction.RoundUp
Rounding 6.0349 to two decimals is just not 6.04 hence, no, there is no such function.
Round up will round anything up. Hence, 6.0000000001 will also become 7 if you round to 0 decimals.
I'm trying to divide one number by another and then immediately ceil() the result. These would normally be variables, but for simplicity let's stick with constants.
If I try any of the following, I get 3 when I want to get 4.
double num = ceil(25/8); // 3
float num = ceil(25/8); // 3
int num = ceil(25/8); // 3
I've read through a few threads on here (tried the nextafter() suggestion from this thread) as well as other sites and I don't understand what's going on. I've checked and my variables are the numbers I expect them to be and I've in fact tried the above, using constants, and am still getting unexpected results.
Thanks in advance for the help. I'm sure it's something simple that I'm missing but I'm at a loss at this point.
This is because you are doing integer arithmetic. The value is 3 before you are calling ceil, because 25 and 8 are both integers. 25/8 is calculated first using integer arithmetic, evaluating to 3.
Try:
double value = ceil(25.0/8);
This will ensure the compiler treats the constant 25.0 as a floating point number.
You can also use an explicit cast to achieve the same result:
double value = ceil(((double)25)/8);
This is because the expressions are evaluated before being passed as an argument to the ceil function. You need to cast one of them to a double first so the result will be a decimal that will be passed to ceil.
double num = ceil((double)25/8);