I need to convert a number into its column name equivalent
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I'm working on a script that generate some Excel documents and I need to convert a number into its column name equivalent. For example: 1 => A 2 => B 27 => AA 28 => AB 14558 => UMX
1 comentario
I was curious about the systematic bugs in the algorithms, which several functions shown on this page demonstrate (producing either errors or incorrect output for some specific letters). So I wrote this simple test function (attached) for checking any such conversion function. It checks all 16384 columns supported by Excel 2007 and later. Note that for simplicity's sake, it does not handle row numbers, so please first modify your function to return the column letters only.
Enjoy!
Respuestas (6)
Praveen Bulusu
el 22 de Feb. de 2020
You can use the following functions. It works for any number.
num=14558
a=num2xlcol(num) % convert number to xlcol
b=xlcol2num(a) % convert xlcol to num
function xlcol_addr=num2xlcol(col_num)
% col_num - positive integer greater than zero
n=1;
while col_num>26*(26^n-1)/25
n=n+1;
end
base_26=zeros(1,n);
tmp_var=-1+col_num-26*(26^(n-1)-1)/25;
for k=1:n
divisor=26^(n-k);
remainder=mod(tmp_var,divisor);
base_26(k)=65+(tmp_var-remainder)/divisor;
tmp_var=remainder;
end
xlcol_addr=char(base_26); % Character vector of xlcol address
end
function xlcol_num=xlcol2num(xlcol_addr)
% xlcol_addr - upper case character
if ischar(xlcol_addr) && ~any(~isstrprop(xlcol_addr,"upper"))
xlcol_num=0;
n=length(xlcol_addr);
for k=1:n
xlcol_num=xlcol_num+(double(xlcol_addr(k)-64))*26^(n-k);
end
else
error('not a valid character')
end
end
2 comentarios
John McDowell
el 6 de Jul. de 2020
Editada: John McDowell
el 6 de Jul. de 2020
These are both really useful functions! Thanks for sharing them
Frederico Pratas
el 1 de Abr. de 2021
This is useful, adding some generality to it:
function xlcol_num = xlcol2num(xlcol_addr)
if ischar(xlcol_addr)
xlcol_addr = upper(xlcol_addr);
xlcol_num=0;
n=length(xlcol_addr);
for k=1:n
xlcol_num = xlcol_num + (double(xlcol_addr(k))-double('A')+1)*(length('A':'Z')^(n-k))
end
else
error('not a valid character')
end
end
Unfortunately Andrei Bobrov's answer does not really take into account the missing zeros, which means that it leads to a kind of "off by one" bug.
Here is an alternative approach which tests correctly on all columns currently supported by Excel (A-XFD) and is limited only by floating point precision (i.e. practically unlimited columns):
d = [1, 2, 26, 27, 28, 702, 703, 704, 14558, 16384]; % A,B,Z,AA,AB,ZZ,AAA,AAB,UMX,XFD
s = arrayfun(@int2xlcol,d)
function str = int2xlcol(num)
raw = 'A':'Z';
str = raw(1+rem(num-1,26));
tmp = fix((num-1)/26);
while any(tmp)
str = [raw(1+rem(tmp-1,26)),str]; %#ok<AGROW>
tmp = fix((tmp-1)/26);
end
str = string(str);
end
You can also find several submissions on FEX which claim to make this conversion, e.g.:
1 comentario
The same approach can also convert multiple values at once:
d = [1, 2, 26, 27, 28, 702, 703, 704, 14558, 16384]; % A,B,Z,AA,AB,ZZ,AAA,AAB,UMX,XFD
z = reshape('A':'Z',[],1);
c = num2cell(z(1+rem(d-1,26)));
v = fix((d-1)/26);
while any(v)
x = v>0;
c(x) = strcat(z(1+rem(v(x)-1,26)),c(x));
v = fix((v-1)/26);
end
s = string(c)
Andrei Bobrov
el 15 de Oct. de 2015
Editada: Andrei Bobrov
el 16 de Oct. de 2015
z = 'A':'Z';
d = [1, 2, 27, 28, 14558];
ll = ceil(log(max(d(:)))/log(26));
bs = rem(floor(d(:)*26.^(1-ll:0)),26);
out = cellfun(@(x)z(x(x > 0)),num2cell(bs,2),'un',0);
or
out = arrayfun(@(x)z(rem(floor(x*26.^(1-floor(log(x)/log(26)+1):0)),26)),d(:),'un',0);
8 comentarios
Praveen Choudhury
el 15 de Oct. de 2015
Andrei Bobrov
el 16 de Oct. de 2015
corrected :)
Daniel LeBrun
el 10 de Dic. de 2019
Movida: Stephen23
el 25 de Feb. de 2025
Hey I was using this code and noticed that when it gets to "Z" it instead writes "A". Have to find a way to have the bs term not forget the 26th.
Buggy code! It does not correctly parse outputs which should contain 'Z':
>> d = [26,51,52,53,677,728]; % -> {'Z','AY','AZ','BA','ZA','AAZ'}
>> z = 'A':'Z';
>> ll = ceil(log(max(d(:)))/log(26));
>> bs = rem(floor(d(:)*26.^(1-ll:0)),26);
>> out = cellfun(@(x)z(x(x > 0)),num2cell(bs,2),'un',0)
out =
'A'
'AY'
'B'
'BA'
'AA'
'AB'
>> out = arrayfun(@(x)z(rem(floor(x*26.^(1-floor(log(x)/log(26)+1):0)),26)),d(:),'un',0)
Subscript indices must either be real positive integers or logicals.
Error in @(x)z(rem(floor(x*26.^(1-floor(log(x)/log(26)+1):0)),26))
Andrei Bobrov
el 12 de Dic. de 2019
Editada: Andrei Bobrov
el 12 de Dic. de 2019
Yes! (Thanx Daniel and Stephen!)
Other variant:
z = 'A':'Z';
d = [1, 2, 26, 27, 28, 702, 703, 704, 14558, 16384,18278];
ll = ceil(log(max(d(:)))/log(26));
bs = rem(floor(d(:)*26.^(1-ll:0)),26);
lo = bs(:,end) == 0;
bs(lo,:) = circshift(bs(lo,:),-1,2)*26;
out = cellfun(@(x)z(x(x > 0)),num2cell(bs,2),'un',0);
Xiaobing Zhou
el 22 de Nov. de 2020
Andrei Bobrov, works for 26, but not for 2*26, 3*26, ..., n*26
Andrei Bobrov
el 22 de Nov. de 2020
Editada: Andrei Bobrov
el 22 de Nov. de 2020
function ch = Int2lat(d)
function ch = finner(d)
ll = floor(log(max(d(:)))/log(26))+1;
out = rem(floor(d(:)*26.^(1-ll:0)),26);
lo = ~out;
while any(lo)
out = out + lo*26 - circshift(lo,-1);
out = out(cumsum(out,2) > 0);
lo = ~out;
end
abc = 'A':'Z';
ch = abc(out);
end
ch = arrayfun(@finner,d,'un',0);
end
>> h = Int2lat((1:20)'*26)
h =
20×1 cell array
{'Z' }
{'AZ'}
{'BZ'}
{'CZ'}
{'DZ'}
{'EZ'}
{'FZ'}
{'GZ'}
{'HZ'}
{'IZ'}
{'JZ'}
{'KZ'}
{'LZ'}
{'MZ'}
{'NZ'}
{'OZ'}
{'PZ'}
{'QZ'}
{'RZ'}
{'SZ'}
>>
@Andrei Bobrov: note that function FINNER() returns incorrect 'Z' outputs on MATLAB versions R2016a and earlier, due to the change in the default behavior of CIRCSHIFT's 2nd input argument. For example:
>> finner(26)
ans =
AY
Remco Hamoen
el 19 de Abr. de 2020
Movida: Stephen23
el 25 de Feb. de 2025
This function might belp. It converts row and column to 'A1' combinations:
function CELL = xlRC2A1(ROW,COL)
%% Returns the column characters of Excel given a certain column number
% Input COL : number of column
% Output CHAR : Character combination in Excel
if COL <= 26 % [A..Z]
CHAR = char(mod(COL-1,26)+1+64);
CELL = [CHAR num2str(ROW)];
elseif COL <= 702 % [AA..ZZ]
COL = COL-26;
CHAR1 = char(floor((COL-1)/26)+1+64);
CHAR0 = char(mod(COL-1,26)+1+64);
CHAR = [CHAR1 CHAR0];
CELL = [CHAR num2str(ROW)];
elseif COL <= 16384 % [AAA..XFD]
COL = COL-702;
CHAR2 = char(floor((COL-1)/676)+1+64);
COL=COL-(floor((COL-1)/676))*676;
CHAR1 = char(floor((COL-1)/26)+1+64);
CHAR0 = char(mod(COL-1,26)+1+64);
CHAR = [CHAR2 CHAR1 CHAR0];
CELL = [CHAR num2str(ROW)];
else
disp('Column does not exist in Excel!');
end
end
Regards,
Remco
Very good question ...
This should work whatever what ...
Xavier
These two functions have been validated with every excel column names form A (1) to XFD (16348).
Maybe some cheks for integer input is missing, but if you use it responsibly, you will find no problems.
You can call them inside arrayfun or cellfun respectively:
% Calls for arrays and cells
ColumnNumbers = 1:16348;
ColumnNames = arrayfun(@num2col, ColumnNumbers, 'un',0);
ColumnNumbers2 = cellfun(@col2num, ColumnNames);
% Tests
assert(isequal(ColumnNumbers2,ColumnNumbers)); % Check that col2num(num2col(x)) == x for every x
tic;
assert(isequal(num2col(7),'G'));
assert(isequal(num2col(21),'U'));
assert(isequal(num2col(56),'BD'));
assert(isequal(num2col(256),'IV'));
assert(isequal(num2col(539),'TS'));
assert(isequal(num2col(1000),'ALL'));
assert(isequal(num2col(16384),'XFD'));
toc;
% Fucntions
function col = num2col(num)
col = '';
if num>0
col = [num2col(floor((num-1)/26)) char('A'+mod(num-1,26))];
end
end
function num = col2num(col)
num = sum((upper(col)-'A'+1).*(26.^(length(col)-1:-1:0)));
end
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