Reputation: 17937
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
Upvotes: 140
Views: 125870
Reputation: 731
JavaScript version of [this Python code]:
function nToAZ(n) {
return (a = Math.floor(n / 26)) >= 0
? nToAZ(a-1) + String.fromCharCode(65 + (n % 26))
: '';
}
console.log(nToAZ(0) === 'A');
console.log(nToAZ(25) === 'Z');
console.log(nToAZ(26) === 'AA');
console.log(nToAZ(51) === 'AZ');
console.log(nToAZ(52) === 'BA');
console.log(nToAZ(18_277) === 'ZZZ'); // max # of columns in Google Sheets
Upvotes: 14
Reputation: 3277
Simple typescript
functional approach
const integerToColumn = (integer: number): string => {
const base26 = (x: number): string =>
x < 26
? String.fromCharCode(65 + x)
: base26((x / 26) - 1) + String.fromCharCode(65 + x % 26)
return base26(integer)
}
console.log(integerToColumn(0)) // "A"
console.log(integerToColumn(1)) // "B"
console.log(integerToColumn(2)) // "C"
Upvotes: 0
Reputation: 2416
Don't use 26 radix. Like below.
const n2c = n => {
if (!n) return '';
// 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())
.map((c,i,arr)=> {
// last digit
if (i===arr.length-1) return c;
// 10 -> p
else if (arr.length - i > 2 && arr[i+1]===48) return c===49 ? null : c-2;
// 0 -> p
else if (c===48) return 112;
// a-1 -> 9
else if (c===97) return 57;
// Subtract 1 except last digit.
// Look at 10. This should be AA not BA.
else return c-1;
})
.filter(c=>c!==null)
// 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('');
};
const table = document.createElement('table');
table.border = 1;
table.cellPadding = 3;
for(let i=0, row; i<1380; i++) {
if (i%5===0) row = table.insertRow();
row.insertCell().textContent = i;
row.insertCell().textContent = n2c(i);
}
document.body.append(table);
td:nth-child(odd) { background: gray; color: white; }
td:nth-child(even) { background: silver; }
Upvotes: 1
Reputation: 1
This is a way to convert column letters to column numbers.
=mmult(ArrayFormula(ifna(vlookup(substitute(mid(rept(" ",3-len(filter(A:A,A:A<>"")))&filter(A:A,A:A<>""),sequence(1,3),1)," ",""),{char(64+sequence(26)),sequence(26)},2,0),0)*{676,26,1}),sequence(3,1,1,0))
Screenshot of the Google Sheet
Upvotes: 0
Reputation: 31
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$1 which is not what you want.
By modifying the formula a little bit we can remove the dollar signs in the cell reference.
=Address(1,4,4)
Adding four as the third argument tells the formula that we are not looking for absolute cell reference. Now the returns is : D1
So you only need to remove the 1 to get the colonne lettre if you need, for example with :
=Substitute(Address(1,4,4),"1","")
That returns D.
Upvotes: 0
Reputation: 31
Here is a 0-indexed JavaScript function without a maximum value, as it uses a while-loop:
function indexesToA1Notation(row, col) {
const letterCount = 'Z'.charCodeAt() - 'A'.charCodeAt() + 1;
row += 1
let colName = ''
while (col >= 0) {
let rem = col % letterCount
colName = String.fromCharCode('A'.charCodeAt() + rem)
col -= rem
col /= letterCount
}
return `${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
Upvotes: -1
Reputation: 835
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;
}
Upvotes: 30
Reputation: 59
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]
}
Upvotes: 2
Reputation: 1233
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
Upvotes: 0
Reputation: 949
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)
Upvotes: 3
Reputation: 5040
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>
Upvotes: 0
Reputation: 81
Simple way through Google Sheet functions, A to Z.
=column(B2) : value is 2
=address(1, column(B2)) : value is $B$1
=mid(address(1, column(B2)),2,1) : value is B
It's a complicated way through Google Sheet functions, but it's also more than AA.
=mid(address(1, column(AB3)),2,len(address(1, column(AB3)))-3) : value is AB
Upvotes: 2
Reputation: 1
In PowerShell:
function convert-IndexToColumn
{
Param
(
[Parameter(Mandatory)]
[int]$col
)
"$(if($col -gt 26){[char][int][math]::Floor(64+($col-1)/26)})$([char](65 + (($col-1) % 26)))"
}
Upvotes: -1
Reputation: 2660
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!
Upvotes: 3
Reputation: 46914
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 :)
Upvotes: 1
Reputation: 7
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 = $base")
def digitFromZeroToLetter(n: BigInt): String =
('A' + n.toInt).toChar.toString
def digitFromOneToLetter(n: BigInt): String =
('A' - 1 + n.toInt).toChar.toString
def lhsConvert(n: Int): String = {
val q0: Int = n / base
val r0: Int = n % base
val q1 = if (r0 == 0) (n - base) / base else q0
val r1 = if (r0 == 0) base else r0
if (q1 == 0)
digitFromOneToLetter(r1)
else
lhsConvert(q1) + digitFromOneToLetter(r1)
}
val q: Int = n / base
val r: Int = n % base
if (q == 0)
digitFromZeroToLetter(r)
else
lhsConvert(q) + digitFromZeroToLetter(r)
}
def indexToColumnAtoZ(n: Int): String = {
val AtoZBase = 26
indexToColumnBase(n, AtoZBase)
}
Upvotes: -1
Reputation: 611
No need to reinvent the wheel here, use the GAS range instead:
var column_index = 1; // your column to resolve
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sheet = ss.getSheets()[0];
var range = sheet.getRange(1, column_index, 1, 1);
Logger.log(range.getA1Notation().match(/([A-Z]+)/)[0]); // Logs "A"
Upvotes: 51
Reputation: 1721
I also was looking for a Python version here is mine which was tested on Python 3.6
def columnToLetter(column):
character = chr(ord('A') + column % 26)
remainder = column // 26
if column >= 26:
return columnToLetter(remainder-1) + character
else:
return character
Upvotes: 1
Reputation: 9
This will cover you out as far as column AZ:
=iferror(if(match(A2,$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")
Upvotes: 0
Reputation: 1984
Java Apache POI
String columnLetter = CellReference.convertNumToColString(columnNumber);
Upvotes: 0
Reputation: 737
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))
Upvotes: -2
Reputation: 46914
=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
.
1
.ADDRESS
docsSUBSTITUTE(..., "1", "")
replaces the 1
in the address C1
, so you're left with the column letter.Upvotes: 42
Reputation: 1035
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.
Upvotes: 85
Reputation: 24659
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;
}
Upvotes: 236