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)
Related
In VB6, I am trying to convert a number to binary but when the number has 10 digits i am always getting an Overflow error.
What is the data type where i can store a trillion number?
This is the code which is working when the number has less that 10 digits.
Public Function DecimalToBinary(DecimalNum As Double) As _
String
Dim tmp As String
Dim n As Double
n = DecimalNum
tmp = Trim(Str(n Mod 2))
n = n \ 2
Do While n <> 0
tmp = Trim(Str(n Mod 2)) & tmp
n = n \ 2
Loop
DecimalToBinary = tmp
End Function
One of the problems you will encounter is that the Mod operator will not work with values larger than a Long (2,147,483,647). You can rewrite a Mod function as described in this answer: VBA equivalent to Excel's mod function:
' Divide the number by 2.
' Get the integer quotient for the next iteration.
' Get the remainder for the binary digit.
' Repeat the steps until the quotient is equal to 0.
Public Function DecimalToBinary(DecimalNum As Double) As String
Dim tmp As String
Dim n As Double
n = DecimalNum
Do While n <> 0
tmp = Remainder(n, 2) & tmp
n = Int(n / 2)
Loop
DecimalToBinary = tmp
End Function
Function Remainder(Dividend As Variant, Divisor As Variant) As Variant
Remainder = Dividend - Divisor * Int(Dividend / Divisor)
End Function
You can also rewrite your function to avoid Mod altogether:
Public Function DecimalToBinary2(DecimalNum As Double) As String
Dim tmp As String
Dim n As Double
Dim iCounter As Integer
Dim iBits As Integer
Dim dblMaxSize As Double
n = DecimalNum
iBits = 1
dblMaxSize = 1
' Get number of bits
Do While dblMaxSize <= n
dblMaxSize = dblMaxSize * 2
iBits = iBits + 1
Loop
' Move back down one bit
dblMaxSize = dblMaxSize / 2
iBits = iBits - 1
' Work back down bit by bit
For iCounter = iBits To 1 Step -1
If n - dblMaxSize >= 0 Then
tmp = tmp & "1"
n = n - dblMaxSize
Else
' This bit is too large
tmp = tmp & "0"
End If
dblMaxSize = dblMaxSize / 2
Next
DecimalToBinary2 = tmp
End Function
This function finds the bit that is larger than your number and works back down, bit by bit, figuring out if the value for each bit can be subtracted from your number. It's a pretty basic approach but it does the job.
For both functions, if you want to have your binary string in groups of 8 bits, you can use a function like this to pad your string:
Public Function ConvertToBytes(p_sBits As String)
Dim iLength As Integer
Dim iBytes As Integer
iLength = Len(p_sBits)
If iLength Mod 8 > 0 Then
iBytes = Int(iLength / 8) + 1
Else
iBytes = Int(iLength / 8)
End If
ConvertToBytes = Right("00000000" & p_sBits, iBytes * 8)
End Function
I am using Excel 2007 which supports Columns upto 16,384 Columns. I would like to obtain the Column name corresponding Column Number.
Currently, I am using the following code. However this code supports upto 256 Columns. Any idea how to obtain Column Name if the column number is greater than 256.
function loc = xlcolumn(column)
if isnumeric(column)
if column>256
error('Excel is limited to 256 columns! Enter an integer number <256');
end
letters = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
count = 0;
if column-26<=0
loc = char(letters(column));
else
while column-26>0
count = count + 1;
column = column - 26;
end
loc = [char(letters(count)) char(letters(column))];
end
else
letters = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
if size(column,2)==1
loc =findstr(column,letters);
elseif size(column,2)==2
loc1 =findstr(column(1),letters);
loc2 =findstr(column(2),letters);
loc = (26 + 26*loc1)-(26-loc2);
end
end
Thanks
As a diversion, here is an all function handle example, with (almost) no file-based functions required. This is based on the dec2base function, since Excel column names are (almost) base 26 numbers, with the frustrating difference that there are no "0" characters.
Note: this is probably a terrible idea overall, but it works. Better solutions are probably found elsewhere in the file exchange.
First, the one file based function that I couldn't get around, to perform arbitrary depth function composition.
function result = compose( fnHandles )
%COMPOSE Compose a set of functions
% COMPOSE({fnHandles}) returns a function handle consisting of the
% composition of the cell array of input function handles.
%
% For example, if F, G, and H are function handles with one input and
% one output, then:
% FNCOMPOSED = COMPOSE({F,G,H});
% y = FNCOMPOSED(x);
% is equivalent to
% y = F(G(H(x)));
if isempty(fnHandles)
result = #(x)x;
elseif length(fnHandles)==1
result = fnHandles{1};
else
fnOuter = fnHandles{1};
fnRemainder = compose(fnHandles(2:end));
result = #(x)fnOuter(fnRemainder(x));
end
Then, the bizarre, contrived path to convert base26 values into the correct string
%Functions leading to "getNumeric", which creates a numeric, base26 array
remapUpper = #(rawBase)(rawBase + (rawBase>='A')*(-55)); %Map the letters 'A-P' to [10:26]
reMapLower = #(rawBase)(rawBase + (rawBase<'A')*(-48)); %Map characters '0123456789' to [0:9]
getRawBase = #(x)dec2base(x, 26);
getNumeric = #(x)remapUpper(reMapLower(getRawBase(x)));
%Functions leading to "correctNumeric"
% This replaces zeros with 26, and reduces the high values entry by 1.
% Similar to "borrowing" as we learned in longhand subtraction
borrowDownFrom = #(x, fromIndex) [x(1:(fromIndex-1)) (x(fromIndex)-1) (x(fromIndex+1)+26) (x((fromIndex+2):end))];
borrowToIfNeeded = #(x, toIndex) (x(toIndex)<=0)*borrowDownFrom(x,toIndex-1) + (x(toIndex)>0)*(x); %Ugly numeric switch
getAllConditionalBorrowFunctions = #(numeric)arrayfun(#(index)#(numeric)borrowToIfNeeded(numeric, index),(2:length(numeric)),'uniformoutput',false);
getComposedBorrowFunction = #(x)compose(getAllConditionalBorrowFunctions(x));
correctNumeric = #(x)feval(getComposedBorrowFunction(x),x);
%Function to replace numerics with letters, and remove leading '#' (leading
%zeros)
numeric2alpha = #(x)regexprep(char(x+'A'-1),'^#','');
%Compose complete function
num2ExcelName = #(x)arrayfun(#(x)numeric2alpha(correctNumeric(getNumeric(x))), x, 'uniformoutput',false)';
Now test using some stressing transitions:
>> num2ExcelName([1:5 23:28 700:704 727:729 1024:1026 1351:1355 16382:16384])
ans =
'A'
'B'
'C'
'D'
'E'
'W'
'X'
'Y'
'Z'
'AA'
'AB'
'ZX'
'ZY'
'ZZ'
'AAA'
'AAB'
'AAY'
'AAZ'
'ABA'
'AMJ'
'AMK'
'AML'
'AYY'
'AYZ'
'AZA'
'AZB'
'AZC'
'XFB'
'XFC'
'XFD'
This function I wrote works for any number of columns (until Excel runs out of columns). It just requires a column number input (e.g. 16368 will return a string 'XEN').
If the application of this concept is different than my function, it's important to note that a column of x number of A's begins every 26^(x-1) + 26^(x-2) + ... + 26^2 + 26 + 1. (e.g. 'AAA' begins on 26^2 + 26 + 1 = 703)
function [col_str] = let_loc(num_loc)
test = 2;
old = 0;
x = 0;
while test >= 1
old = 26^x + old;
test = num_loc/old;
x = x + 1;
end
num_letters = x - 1;
str_array = zeros(1,num_letters);
for i = 1:num_letters
loc = floor(num_loc/(26^(num_letters-i)));
num_loc = num_loc - (loc*26^(num_letters-i));
str_array(i) = char(65 + (loc - 1));
end
col_str = strcat(str_array(1:length(str_array)));
end
Hope this saves someone some time!
Looking for the best way to do this in VB6. Typically, I would use this approach...
' count spaces
For i = 1 To Len(text)
If Mid$(text, i, 1) = " " Then count = count + 1
Next
Not saying it's the best way, but you code do:
distinctChr = " "
count = Len(text) - Len(Replace(text, distinctChr , ""))
Use the split command like this
Dim TempS As String
TempS = " This is a split test "
Dim V As Variant
V = Split(TempS, " ")
Cls
Print UBound(V) '7
V = Split(TempS, "i")
Print UBound(V) '3
V = Split(TempS, "e")
Print UBound(V) '1
You can combine it to a single line.
Print UBound(Split(TempS, "i"))
I did some crude timing on it. On a 40,000 character string with all spaces it seems to clock in at 17 milliseconds on a 2.4 GHz Intel Core 2 processor.
A function could look like this
Function CountChar(ByVal Text As String, ByVal Char As String) As Long
Dim V As Variant
V = Split(Text, Char)
CountChar = UBound(V)
End Function
I would use a modified bucket sort:
Dim i as Integer
Dim index As Integer
Dim count as Integer
Dim FoundByAscii(0 To 255) As Boolean
For i = 1 To Len(text)
index = Asc(Mid$(text, i, 1))
FoundByAscii(index) = True
Next i
count = 0
For i = 0 To 255
If FoundByAscii(i) Then
count = count + 1
End If
Next i
...and your result is in count. The performance is O(N) - if Mid$ is O(1).
Edit:
Based on your clarification, do this:
' count spaces
Dim asciiToSearchFor As Integer
asciiToSearchFor = Asc(" ")
For i = 1 To Len(text)
If Asc(Mid$(text, i, 1)) = asciiToSearchFor Then count = count + 1
Next
As ascii compares have to be faster that string comparison. I'd profile it just in case, but I'm pretty sure.
It's not clear what you mean by the best way to do this.
If you want something very fast, but totally unmaintainable, adapt this horrible code that delves into the underlying memory of a VB6 string to count the number of words. Courtesy of VBspeed.
I want a function to calculate numerology.For example if i enter "XYZ" then my output should be 3 .
Here is how it became 3:
X = 24
Y = 25
Z = 26
on adding it becomes 75 which again adds up to 12 (7+5) which again adds up to 3(1+2) . Similarly whatever names i should pass,my output should be a single digit score.
Here you are:
Function Numerology(Str)
Dim sum, i, char
' Convert the string to upper case, so that 'X' = 'x'
Str = UCase(Str)
sum = 0
' For each character, ...
For i = 1 To Len(Str)
' Check if it's a letter and raise an exception otherwise
char = Mid(Str, i , 1)
If char < "A" Or char > "Z" Then Err.Raise 5 ' Invalid procedure call or argument
' Add the letter's index number to the sum
sum = sum + Asc(char) - 64
Next
' Calculate the result using the digital root formula (http://en.wikipedia.org/wiki/Digital_root)
Numerology = 1 + (sum - 1) Mod 9
End Function
In vbscript:
Function numerology(literal)
result = 0
for i = 1 to Len(literal)
'' // for each letter, take its ASCII value and substract 64,
'' so "A" becomes 1 and "Z" becomes 26
result = result + Asc(Mid(literal, i, 1)) - 64
next
'' // while result is bigger than 10, let's sum it's digits
while(result > 10)
partial = 0
for i = 1 to Len(CStr(result))
partial = partial + CInt(Mid(CStr(result), i, 1))
next
result = partial
wend
numerology = result
End Function
I have no idea what this could possible be used for but it was fun to write anyway.
Private Function CalcStupidNumber(ByVal s As String) As Integer
s = s.ToLower
If (s.Length = 1) Then 'End condition
Try
Return Integer.Parse(s)
Catch ex As Exception
Return 0
End Try
End If
'cover to Values
Dim x As Int32
Dim tot As Int32 = 0
For x = 0 To s.Length - 1 Step 1
Dim Val As Integer = ConvertToVal(s(x))
tot += Val
Next
Return CalcStupidNumber(tot.ToString())
End Function
Private Function ConvertToVal(ByVal c As Char) As Integer
If (Char.IsDigit(c)) Then
Return Integer.Parse(c)
End If
Return System.Convert.ToInt32(c) - 96 ' offest of a
End Function
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())