A script I am making scans a 5-character code and assigns it a number based on the contents of characters within the code. The code is a randomly-generated number/letter combination. For example 7D3B5 or HH42B where any position can be any one of (26 + 10) characters.
Now, the issue I am having is I would like to figure out the number from 1-(36^5) based on the code. For example:
00000 = 0
00001 = 1
00002 = 2
0000A = 10
0000B = 11
0000Z = 36
00010 = 37
00011 = 38
So on and so forth until the final possible code which is:
ZZZZZ = 60466176 (36^5)
What I need to work out is a formula to figure out, let's say G47DU in its number form, using the examples below.
Something like this?
function getCount(s){
if (!isNaN(s))
return Number(s);
return s.charCodeAt(0) - 55;
}
function f(str){
let result = 0;
for (let i=0; i<str.length; i++)
result += Math.pow(36, str.length - i - 1) * getCount(str[i]);
return result;
}
var strs = [
'00000',
'00001',
'00002',
'0000A',
'0000B',
'0000Z',
'00010',
'00011',
'ZZZZZ'
];
for (str of strs)
console.log(str, f(str));
You are trying to create a base 36 numeric system. Since there are 5 'digits' each digit being 0 to Z, the value can go from 0 to 36^5. (If we are comparing this with hexadecimal system, in hexadecimal each 'digit' goes from 0 to F). Now to convert this to decimal, you could try use the same method used to convert from hex or binary etc... system to the decimal system.
It will be something like d4 * (36 ^ 4) + d3 * (36 ^ 3) + d2 * (36 ^ 2) + d1 * (36 ^ 1) + d0 * (36 ^ 0)
Note: Here 36 is the total number of symbols.
d0, d1, d2, d3, d4 can range from 0 to 35 in decimal (Important: Not 0 to 36).
Also, you can extend this for any number of digits or symbols and you can implement operations like addition, subtraction etc in this system itself as well. (It will be fun to implement that. :) ) But it will be easier to convert it to decimal do the operations and convert it back though.
All,
I'm intermediate for VB6. I want to get remainder from divided two doubles. I used ,
Dim a As Double, b As Double, result As Double
b = 8333.33
a = 58333.31
result = a - (b * Fix(a / b))
result should be 0. But it is not.
a/b =7 and no remain. So Fix(a / b) should be 7. But Fix(a / b)=6, Why?
If you use this as a workaround I think you will get the result you expect:
Dim a As Double, b As Double, intermediate As Double, fixed1 As Double, fixed2 As Double
b = 8333.33
a = 58333.31
intermediate = a / b
fixed1 = Fix(intermediate)
fixed2 = Fix(a / b)
Debug.Print intermediate ' 7
Debug.Print fixed1 ' 7
Debug.Print fixed2 ' 6
Documentation for Fix says:
Int, Fix functions
Returns the integer portion of a number.
... Remarks Both Int and Fix remove the fractional part of number and
return the resulting integer value.
(that is from VBA docs but should be the same for VB6).
Apparently when VB evaluates a / b within the context of the function call it results in a floating point value very slightly < 7.
BTW, another trick to get it working - depending on the floating point numbers you need to work with - is to use Currency instead of Double
Dim a As Currency, b As Currency
b = 8333.33
a = 58333.31
Debug.Print Fix(a / b) ' Returns 7
Debug.Print a - (b * Fix(a / b)) ' Returns 0
I want to take a number and convert it into lowercase a-z letters using VBScript.
For example:
1 converts to a
2 converts to b
27 converts to aa
28 converts to ab
and so on...
In particular I am having trouble converting numbers after 26 when converting to 2 letter cell names. (aa, ab, ac, etc.)
You should have a look at the Chr(n) function.
This would fit your needs from a to z:
wscript.echo Chr(number+96)
To represent multiple letters for numbers, (like excel would do it) you'll have to check your number for ranges and use the Mod operator for modulo.
EDIT:
There is a fast food Copy&Paste example on the web: How to convert Excel column numbers into alphabetical characters
Quoted example from microsoft:
For example: The column number is 30.
The column number is divided by 27: 30 / 27 = 1.1111, rounded down by the Int function to "1".
i = 1
Next Column number - (i * 26) = 30 -(1 * 26) = 30 - 26 = 4.
j = 4
Convert the values to alphabetical characters separately,
i = 1 = "A"
j = 4 = "D"
Combined together, they form the column designator "AD".
And its code:
Function ConvertToLetter(iCol As Integer) As String
Dim iAlpha As Integer
Dim iRemainder As Integer
iAlpha = Int(iCol / 27)
iRemainder = iCol - (iAlpha * 26)
If iAlpha > 0 Then
ConvertToLetter = Chr(iAlpha + 64)
End If
If iRemainder > 0 Then
ConvertToLetter = ConvertToLetter & Chr(iRemainder + 64)
End If
End Function
Neither of the solutions above work for the full Excel range from A to XFD. The first example only works up to ZZ. The second example has boundry problems explained in the code comments below.
//
Function ColumnNumberToLetter(ColumnNumber As Integer) As String
' convert a column number to the Excel letter representation
Dim Div As Double
Dim iMostSignificant As Integer
Dim iLeastSignificant As Integer
Dim Base As Integer
Base = 26
' Column letters are base 26 starting at A=1 and ending at Z=26
' For base 26 math to work we need to adjust the input value to
' base 26 starting at 0
Div = (ColumnNumber - 1) / Base
iMostSignificant = Int(Div)
' The addition of 1 is needed to restore the 0 to 25 result value to
' align with A to Z
iLeastSignificant = 1 + (Div - iMostSignificant) * Base
' convert number to letter
ColumnNumberToLetter = Chr(64 + iLeastSignificant)
' if the input number is larger than the base then the conversion we
' just did is the least significant letter
' Call the function again with the remaining most significant letters
If ColumnNumber > Base Then
ColumnNumberToLetter = ColumnNumberToLetter(iMostSignificant) & ColumnNumberToLetter
End If
End Function
//
try this
function converts(n)
Dim i, c, m
i = n
c = ""
While i > 26
m = (i mod 26)
c = Chr(m+96) & c
i = (i - m) / 26
Wend
c = Chr(i+96) & c
converts = c
end function
WScript.Echo converts(1000)
Example :
a = 1
b = 2
c = 3
..
..
z = 26
aa = 27
ab = 28
how to convert another string into an integer? for example i want to convert 'lmao' to an integer. please help me :) thank you.
in pascal :)
To convert ordinary base-10 strings into numbers, you take each character from left to right, convert it to its numeric value (between 0 and 9) and add it to the total you already have (which you initialize to zero). If there are more characters following the one you just processed, then multiply the total by 10. Repeat until you run out of characters.
For example, the number 374 is 3×102 + 7×101 + 4×100. Another way of writing that, which more closely models the conversion algorithm I described above, is (((3)×10+7)×10+4.
You can adapt that to handle any string of characters, not just numeric characters. Instead of 10, the base is 26, so multiply by that. And instead of digits, the characters are a through z. Your example string would be evaluated like this: (((l)×26+m)×26+a)×26+o. Substitute numbers for those letters, and you get 219,742.
Here's some code to do it. It doesn't check for errors; it assumes that the string will only contain valid characters and that the string won't represent a number that's too big to fit in an Integer variable.
function SpecialStrToInt(const s: string): Integer;
var
i: Integer;
subtotal: Integer;
c: Char;
charval: Integer;
begin
subtotal := 0;
for i := 1 to Length(s) do begin
c := s[i];
charval := Ord(c) - Ord('a') + 1;
subtotal := subtotal * 26;
subtotal := subtotal + charval;
end;
SpecialStrToInt := subtotal;
end;
An oddity about your format is that there's no way to represent zero.
This question already has answers here:
How to convert a column number (e.g. 127) into an Excel column (e.g. AA)
(60 answers)
Closed 9 years ago.
How would you determine the column name (e.g. "AQ" or "BH") of the nth column in Excel?
Edit: A language-agnostic algorithm to determine this is the main goal here.
I once wrote this function to perform that exact task:
public static string Column(int column)
{
column--;
if (column >= 0 && column < 26)
return ((char)('A' + column)).ToString();
else if (column > 25)
return Column(column / 26) + Column(column % 26 + 1);
else
throw new Exception("Invalid Column #" + (column + 1).ToString());
}
Here is the cleanest correct solution I could come up with (in Java, but feel free to use your favorite language):
String getNthColumnName(int n) {
String name = "";
while (n > 0) {
n--;
name = (char)('A' + n%26) + name;
n /= 26;
}
return name;
}
But please do let me know of if you find a mistake in this code, thank you.
A language agnostic algorithm would be as follows:
function getNthColumnName(int n) {
let curPower = 1
while curPower < n {
set curPower = curPower * 26
}
let result = ""
while n > 0 {
let temp = n / curPower
let result = result + char(temp)
set n = n - (curPower * temp)
set curPower = curPower / 26
}
return result
This algorithm also takes into account if Excel gets upgraded again to handle more than 16k columns. If you really wanted to go overboard, you could pass in an additional value and replace the instances of 26 with another number to accomodate alternate alphabets
Thanks, Joseph Sturtevant! Your code works perfectly - I needed it in vbscript, so figured I'd share my version:
Function ColumnLetter(ByVal intColumnNumber)
Dim sResult
intColumnNumber = intColumnNumber - 1
If (intColumnNumber >= 0 And intColumnNumber < 26) Then
sResult = Chr(65 + intColumnNumber)
ElseIf (intColumnNumber >= 26) Then
sResult = ColumnLetter(CLng(intColumnNumber \ 26)) _
& ColumnLetter(CLng(intColumnNumber Mod 26 + 1))
Else
err.Raise 8, "Column()", "Invalid Column #" & CStr(intColumnNumber + 1)
End If
ColumnLetter = sResult
End Function
Joseph's code is good but, if you don't want or need to use a VBA function, try this.
Assuming that the value of n is in cell A2
Use this function:
=MID(ADDRESS(1,A2),2,LEN(ADDRESS(1,A2))-3)
IF(COLUMN()>=26,CHAR(ROUND(COLUMN()/26,1)+64)&CHAR(MOD(COLUMN(),26)+64),CHAR(COLUMN()+64))
This works 2 letter columns (up until column ZZ). You'd have to nest another if statement for 3 letter columns.
The formula above fails on columns AY, AZ and each of the following nY and nZ columns. The corrected formula is:
=IF(COLUMN()>26,CHAR(ROUNDDOWN((COLUMN()-1)/26,0)+64)&CHAR(MOD((COLUMN()-1),26)+65),CHAR(COLUMN()+64)
Ruby one-liner:
def column_name_for(some_int)
some_int.to_s(26).split('').map {|c| (c.to_i(26) + 64).chr }.join # 703 => "AAA"
end
It converts the integer to base26 then splits it and does some math to convert each character from ascii. Finally joins 'em all back together. No division, modulus, or recursion.
Fun.
FROM wcm:
If you don't want to use VBA, you can use this
replace colnr with the number you want
=MID(ADDRESS(1,colnr),2,LEN(ADDRESS(1,colnr))-3)
Please be aware of the fact that this formula is volatile because of the usage of the ADDRESS function. Volatile functions are functions that are recalculated by excel after EVERY change.
Normally excel recalculates formula's only when their dependent references changes.
It could be a performance killer, to use this formula.
And here is a conversion from the VBScript version to SQL Server 2000+.
CREATE FUNCTION [dbo].[GetExcelColRef]
(
#col_seq_no int
)
RETURNS varchar(5)
AS
BEGIN
declare #Result varchar(5)
set #Result = ''
set #col_seq_no = #col_seq_no - 1
If (#col_seq_no >= 0 And #col_seq_no < 26)
BEGIN
set #Result = char(65 + #col_seq_no)
END
ELSE
BEGIN
set #Result = [dbo].[GetExcelColRef] (#col_seq_no / 26) + '' + [dbo].[GetExcelColRef] ((#col_seq_no % 26) + 1)
END
Return #Result
END
GO
This works fine in MS Excel 2003-2010. Should work for previous versions supporting the Cells(...).Address function:
For the 28th column - taking columnNumber=28; Cells(1, columnNumber).Address returns "$AB$1".
Doing a split on the $ sign returns the array: ["","AB","1"]
So Split(Cells(1, columnNumber).Address, "$")(1) gives you the column name "AB".
UPDATE:
Taken from How to convert Excel column numbers into alphabetical characters
' The following VBA function is just one way to convert column number
' values into their equivalent alphabetical characters:
Function ConvertToLetter(iCol As Integer) As String
Dim iAlpha As Integer
Dim iRemainder As Integer
iAlpha = Int(iCol / 27)
iRemainder = iCol - (iAlpha * 26)
If iAlpha > 0 Then
ConvertToLetter = Chr(iAlpha + 64)
End If
If iRemainder > 0 Then
ConvertToLetter = ConvertToLetter & Chr(iRemainder + 64)
End If
End Function
APPLIES TO: Microsoft Office Excel 2007 SE / 2002 SE / 2000 SE / 97 SE
I suppose you need VBA code:
Public Function GetColumnAddress(nCol As Integer) As String
Dim r As Range
Set r = Range("A1").Columns(nCol)
GetColumnAddress = r.Address
End Function
This does what you want in VBA
Function GetNthExcelColName(n As Integer) As String
Dim s As String
s = Cells(1, n).Address
GetNthExcelColName = Mid(s, 2, InStr(2, s, "$") - 2)
End Function
This seems to work in vb.net
Public Function Column(ByVal pColumn As Integer) As String
pColumn -= 1
If pColumn >= 0 AndAlso pColumn < 26 Then
Return ChrW(Asc("A"c) + pColumn).ToString
ElseIf (pColumn > 25) Then
Return Column(CInt(math.Floor(pColumn / 26))) + Column((pColumn Mod 26) + 1)
Else
stop
Throw New ArgumentException("Invalid column #" + (pColumn + 1).ToString)
End If
End Function
I took Joseph's and tested it to BH, then fed it 980-1000 and it looked good.
In VBA, assuming lCol is the column number:
function ColNum2Letter(lCol as long) as string
ColNum2Letter = Split(Cells(1, lCol).Address, "$")(0)
end function
All these code samples that these good people have posted look fine.
There is one thing to be aware of. Starting with Office 2007, Excel actually has up to 16,384 columns. That translates to XFD (the old max of 256 colums was IV). You will have to modify these methods somewhat to make them work for three characters.
Shouldn't be that hard...
Here's Gary Waters solution
Function ConvertNumberToColumnLetter2(ByVal colNum As Long) As String
Dim i As Long, x As Long
For i = 6 To 0 Step -1
x = (1 - 26 ^ (i + 1)) / (-25) - 1 ‘ Geometric Series formula
If colNum > x Then
ConvertNumberToColumnLetter2 = ConvertNumberToColumnLetter2 & Chr(((colNum - x - 1)\ 26 ^ i) Mod 26 + 65)
End If
Next i
End Function
via http://www.dailydoseofexcel.com/archives/2004/05/21/column-numbers-to-letters/
Considering the comment of wcm (top value = xfd), you can calculate it like this;
function IntToExcel(n: Integer); string;
begin
Result := '';
for i := 2 down to 0 do
begin
if ((n div 26^i)) > 0) or (i = 0) then
Result := Result + Char(Ord('A')+(n div (26^i)) - IIF(i>0;1;0));
n := n mod (26^i);
end;
end;
There are 26 characters in the alphabet and we have a number system just like hex or binary, just with an unusual character set (A..Z), representing positionally the powers of 26: (26^2)(26^1)(26^0).
FYI T-SQL to give the Excel column name given an ordinal (zero-based), as a single statement.
Anything below 0 or above 16,383 (max columns in Excel2010) returns NULL.
; WITH TestData AS ( -- Major change points
SELECT -1 AS FieldOrdinal
UNION ALL
SELECT 0
UNION ALL
SELECT 25
UNION ALL
SELECT 26
UNION ALL
SELECT 701
UNION ALL
SELECT 702
UNION ALL
SELECT 703
UNION ALL
SELECT 16383
UNION ALL
SELECT 16384
)
SELECT
FieldOrdinal
, CASE
WHEN FieldOrdinal < 0 THEN NULL
WHEN FieldOrdinal < 26 THEN ''
WHEN FieldOrdinal < 702 THEN CHAR (65 + FieldOrdinal / 26 - 1)
WHEN FieldOrdinal < 16384 THEN CHAR (65 + FieldOrdinal / 676 - 1)
+ CHAR (65 + (FieldOrdinal / 26) - (FieldOrdinal / 676) * 26 - 1)
ELSE NULL
END
+ CHAR (65 + FieldOrdinal % 26)
FROM TestData
ORDER BY FieldOrdinal
I currently use this, but I have a feeling that it can be optimized.
private String GetNthExcelColName(int n)
{
String firstLetter = "";
//if number is under 26, it has a single letter name
// otherwise, it is 'A' for 27-52, 'B' for 53-78, etc
if(n > 26)
{
//the Converts to double and back to int are just so Floor() can be used
Double value = Convert.ToDouble((n-1) / 26);
int firstLetterVal = Convert.ToInt32(Math.Floor(value))-1;
firstLetter = Convert.ToChar(firstLetterValue + 65).ToString();
}
//second letter repeats
int secondLetterValue = (n-1) % 26;
String secondLetter = Convert.ToChar(secondLetterValue+65).ToString();
return firstLetter + secondLetter;
}
=CHAR(64+COLUMN())