Convert column index into corresponding column letter
JavascriptGoogle Apps-ScriptIndexingGoogle SheetsJavascript Problem Overview
I need to convert a Google Spreadsheet column index into its corresponding letter value, for example, given a spreadsheet:
I need to do this (this function obviously does not exist, it's an example):
getColumnLetterByIndex(4); // this should return "D"
getColumnLetterByIndex(1); // this should return "A"
getColumnLetterByIndex(6); // this should return "F"
Now, I don't recall exactly if the index starts from 0
or from 1
, anyway the concept should be clear.
I didn't find anything about this on gas documentation.. am I blind? Any idea?
Thank you
Javascript Solutions
Solution 1 - Javascript
I wrote these a while back for various purposes (will return the double-letter column names for column numbers > 26):
function columnToLetter(column)
{
var temp, letter = '';
while (column > 0)
{
temp = (column - 1) % 26;
letter = String.fromCharCode(temp + 65) + letter;
column = (column - temp - 1) / 26;
}
return letter;
}
function letterToColumn(letter)
{
var column = 0, length = letter.length;
for (var i = 0; i < length; i++)
{
column += (letter.charCodeAt(i) - 64) * Math.pow(26, length - i - 1);
}
return column;
}
Solution 2 - Javascript
This works good
=REGEXEXTRACT(ADDRESS(ROW(); COLUMN()); "[A-Z]+")
even for columns beyond Z.
Simply replace COLUMN()
with your column number. The value of ROW()
doesn't matter.
Solution 3 - Javascript
=SUBSTITUTE(ADDRESS(1,COLUMN(),4), "1", "")
This takes your cell, gets it's address as e.g. C1, and removes the "1".
How it works
-
COLUMN()
gives the number of the column of the cell. -
ADDRESS(1, ..., <format>)
gives an address of a cell, in format speficied by<format>
parameter.4
means the address you know - e.g.C1
.- The row doesn't matter here, so we use
1
. - See
ADDRESS
docs
- The row doesn't matter here, so we use
-
Finally,
SUBSTITUTE(..., "1", "")
replaces the1
in the addressC1
, so you're left with the column letter.
Solution 4 - Javascript
No need to reinvent the wheel here, use the GAS range instead:
var column_index = 1; // your column to resolve
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sheet = ss.getSheets()[0];
var range = sheet.getRange(1, column_index, 1, 1);
Logger.log(range.getA1Notation().match(/([A-Z]+)/)[0]); // Logs "A"
Solution 5 - Javascript
This works on ranges A-Z
formula =char(64+column())
js String.fromCharCode(64+colno)
an google spreadsheet appscript code, based on @Gardener would be:
function columnName(index) {
var cname = String.fromCharCode(65 + ((index - 1) % 26));
if (index > 26)
cname = String.fromCharCode(64 + (index - 1) / 26) + cname;
return cname;
}
Solution 6 - Javascript
In javascript:
X = (n) => (a=Math.floor(n/26)) >= 0 ? X(a-1) + String.fromCharCode(65+(n%26)) : '';
console.assert (X(0) == 'A')
console.assert (X(25) == 'Z')
console.assert (X(26) == 'AA')
console.assert (X(51) == 'AZ')
console.assert (X(52) == 'BA')
Solution 7 - Javascript
Adding to @SauloAlessandre's answer, this will work for columns up from A-ZZ.
=if(column() >26,char(64+(column()-1)/26),) & char(65 + mod(column()-1,26))
I like the answers by @wronex and @Ondra Žižka. However, I really like the simplicity of @SauloAlessandre's answer.
So, I just added the obvious code to allow @SauloAlessandre's answer to work for wider spreadsheets.
As @Dave mentioned in his comment, it does help to have a programming background, particularly one in C where we added the hex value of 'A' to a number to get the nth letter of the alphabet as a standard pattern.
Answer updated to catch the error pointed out by @Sangbok Lee. Thank you!
Solution 8 - Javascript
I was looking for a solution in PHP. Maybe this will help someone.
<?php
$numberToLetter = function(int $number)
{
if ($number <= 0) return null;
$temp; $letter = '';
while ($number > 0) {
$temp = ($number - 1) % 26;
$letter = chr($temp + 65) . $letter;
$number = ($number - $temp - 1) / 26;
}
return $letter;
};
$letterToNumber = function(string $letters) {
$letters = strtoupper($letters);
$letters = preg_replace("/[^A-Z]/", '', $letters);
$column = 0;
$length = strlen($letters);
for ($i = 0; $i < $length; $i++) {
$column += (ord($letters[$i]) - 64) * pow(26, $length - $i - 1);
}
return $column;
};
var_dump($numberToLetter(-1));
var_dump($numberToLetter(26));
var_dump($numberToLetter(27));
var_dump($numberToLetter(30));
var_dump($letterToNumber('-1A!'));
var_dump($letterToNumber('A'));
var_dump($letterToNumber('B'));
var_dump($letterToNumber('Y'));
var_dump($letterToNumber('Z'));
var_dump($letterToNumber('AA'));
var_dump($letterToNumber('AB'));
Output:
NULL
string(1) "Z"
string(2) "AA"
string(2) "AD"
int(1)
int(1)
int(2)
int(25)
int(26)
int(27)
int(28)
Solution 9 - Javascript
Simple way through Google Sheet functions, A to Z.
=column(B2) : value is 2
=address(1, column(B2)) : value is $B$1
=mid(address(1, column(B2)),2,1) : value is B
It's a complicated way through Google Sheet functions, but it's also more than AA.
=mid(address(1, column(AB3)),2,len(address(1, column(AB3)))-3) : value is AB
Solution 10 - Javascript
I also was looking for a Python version here is mine which was tested on Python 3.6
def columnToLetter(column):
character = chr(ord('A') + column % 26)
remainder = column // 26
if column >= 26:
return columnToLetter(remainder-1) + character
else:
return character
Solution 11 - Javascript
A comment on my answer says you wanted a script function for it. All right, here we go:
function excelize(colNum) {
var order = 1, sub = 0, divTmp = colNum;
do {
divTmp -= order; sub += order; order *= 26;
divTmp = (divTmp - (divTmp % 26)) / 26;
} while(divTmp > 0);
var symbols = "0123456789abcdefghijklmnopqrstuvwxyz";
var tr = c => symbols[symbols.indexOf(c)+10];
return Number(colNum-sub).toString(26).split('').map(c=>tr(c)).join('');
}
This can handle any number JS can handle, I think.
Explanation:
Since this is not base26, we need to substract the base times order for each additional symbol ("digit"). So first we count the order of the resulting number, and at the same time count the number to substract. And then we convert it to base 26 and substract that, and then shift the symbols to A-Z
instead of 0-P
.
Anyway, this question is turning into a code golf :)
Solution 12 - Javascript
Here's a two liner which works beyond ZZ using recursion:
Python
def col_to_letter(n):
l = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
return col_to_letter((n-1)//26) + col_to_letter(n%26) if n > 26 else l[n-1]
Javascript
function colToLetter(n) {
l = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
return n > 26 ? colToLetter(Math.floor((n-1)/26)) + colToLetter(n%26) : l[n-1]
}
Solution 13 - Javascript
Java Apache POI
String columnLetter = CellReference.convertNumToColString(columnNumber);
Solution 14 - Javascript
This will cover you out as far as column AZ:
=iferror(if(match(A2,$A$1:$AZ$1,0)<27,char(64+(match(A2,$A$1:$AZ$1,0))),concatenate("A",char(38+(match(A2,$A$1:$AZ$1,0))))),"No match")
Solution 15 - Javascript
A function to convert a column index to letter combinations, recursively:
function lettersFromIndex(index, curResult, i) {
if (i == undefined) i = 11; //enough for Number.MAX_SAFE_INTEGER
if (curResult == undefined) curResult = "";
var factor = Math.floor(index / Math.pow(26, i)); //for the order of magnitude 26^i
if (factor > 0 && i > 0) {
curResult += String.fromCharCode(64 + factor);
curResult = lettersFromIndex(index - Math.pow(26, i) * factor, curResult, i - 1);
} else if (factor == 0 && i > 0) {
curResult = lettersFromIndex(index, curResult, i - 1);
} else {
curResult += String.fromCharCode(64 + index % 26);
}
return curResult;
}
function lettersFromIndex(index, curResult, i) {
if (i == undefined) i = 11; //enough for Number.MAX_SAFE_INTEGER
if (curResult == undefined) curResult = "";
var factor = Math.floor(index / Math.pow(26, i));
if (factor > 0 && i > 0) {
curResult += String.fromCharCode(64 + factor);
curResult = lettersFromIndex(index - Math.pow(26, i) * factor, curResult, i - 1);
} else if (factor == 0 && i > 0) {
curResult = lettersFromIndex(index, curResult, i - 1);
} else {
curResult += String.fromCharCode(64 + index % 26);
}
return curResult;
}
document.getElementById("result1").innerHTML = lettersFromIndex(32);
document.getElementById("result2").innerHTML = lettersFromIndex(6800);
document.getElementById("result3").innerHTML = lettersFromIndex(9007199254740991);
32 --> <span id="result1"></span><br> 6800 --> <span id="result2"></span><br> 9007199254740991 --> <span id="result3"></span>
Solution 16 - Javascript
In python, there is the gspread library
import gspread
column_letter = gspread.utils.rowcol_to_a1(1, <put your col number here>)[:-1]
If you cannot use python, I suggest looking the source code of rowcol_to_a1() in https://github.com/burnash/gspread/blob/master/gspread/utils.py
Solution 17 - Javascript
One-liner. No recursive calls.
const n2c = n => [...(n-1).toString(26)].map(c=>c.charCodeAt()).map((c,i,arr)=>i<arr.length-1?c-1:c).map(a=>a>96?a-22:a+17).map(a=>String.fromCharCode(a)).join('');
const n2c = n => {
// Column number to 26 radix. From 0 to p.
// Column number starts from 1. Subtract 1.
return [...(n-1).toString(26)]
// to ascii number
.map(c=>c.charCodeAt())
// Subtract 1 except last digit.
// Look at 10. This should be AA not BA.
.map((c,i,arr)=>i<arr.length-1?c-1:c)
// Convert with the ascii table. [0-9]->[A-J] and [a-p]->[K-Z]
.map(a=>a>96?a-22:a+17)
// to char
.map(a=>String.fromCharCode(a))
.join('');
};
Solution 18 - Javascript
Here is a 0-indexed JavaScript function without a maximum value, as it uses a while-loop:
function indexesToA1Notation(row, col) {
const letterCount = 'Z'.charCodeAt() - 'A'.charCodeAt() + 1;
row += 1
let colName = ''
while (col >= 0) {
let rem = col % letterCount
colName = String.fromCharCode('A'.charCodeAt() + rem)
col -= rem
col /= letterCount
}
return `${colName}${row}`
}
//Test runs:
console.log(indexesToA1Notation(0,0)) //A1
console.log(indexesToA1Notation(37,9)) //J38
console.log(indexesToA1Notation(5,747)) //ABT6
I wrote it for a web-app, so I'm not 100% sure it works in Google Apps Script, but it is normal JavaScript, so I assume it will.
For some reason I cant get the snippet to show its output, but you can copy the code to some online playground if you like
Solution 19 - Javascript
If you need a version directly in the sheet, here a solution: For the colonne 4, we can use :
=Address(1,4)
I keep the row number to 1 for simplicty. The above formula returns $D$1 which is not what you want.
By modifying the formula a little bit we can remove the dollar signs in the cell reference.
=Address(1,4,4)
Adding four as the third argument tells the formula that we are not looking for absolute cell reference. Now the returns is : D1
So you only need to remove the 1 to get the colonne lettre if you need, for example with :
=Substitute(Address(1,4,4),"1","")
That returns D.
Solution 20 - Javascript
Here is a general version written in Scala. It's for a column index start at 0 (it's simple to modify for an index start at 1):
def indexToColumnBase(n: Int, base: Int): String = {
require(n >= 0, s"Index is non-negative, n = $n")
require(2 <= base && base <= 26, s"Base in range 2...26, base = $base")
def digitFromZeroToLetter(n: BigInt): String =
('A' + n.toInt).toChar.toString
def digitFromOneToLetter(n: BigInt): String =
('A' - 1 + n.toInt).toChar.toString
def lhsConvert(n: Int): String = {
val q0: Int = n / base
val r0: Int = n % base
val q1 = if (r0 == 0) (n - base) / base else q0
val r1 = if (r0 == 0) base else r0
if (q1 == 0)
digitFromOneToLetter(r1)
else
lhsConvert(q1) + digitFromOneToLetter(r1)
}
val q: Int = n / base
val r: Int = n % base
if (q == 0)
digitFromZeroToLetter(r)
else
lhsConvert(q) + digitFromZeroToLetter(r)
}
def indexToColumnAtoZ(n: Int): String = {
val AtoZBase = 26
indexToColumnBase(n, AtoZBase)
}
Solution 21 - Javascript
In PowerShell:
function convert-IndexToColumn
{
Param
(
[Parameter(Mandatory)]
[int]$col
)
"$(if($col -gt 26){[char][int][math]::Floor(64+($col-1)/26)})$([char](65 + (($col-1) % 26)))"
}
Solution 22 - Javascript
Here's a zero-indexed version (in Python):
letters = []
while column >= 0:
letters.append(string.ascii_uppercase[column % 26])
column = column // 26 - 1
return ''.join(reversed(letters))