I need to convert a number into its column name equivalent

조회 수: 122 (최근 30일)
Praveen Choudhury
Praveen Choudhury 2015년 10월 15일
편집: Stephen23 2023년 10월 11일
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
  댓글 수: 2
Remco Hamoen
Remco Hamoen 2020년 4월 19일
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
Stephen23
Stephen23 2022년 5월 18일
편집: Stephen23 2023년 10월 11일
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!

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답변 (5개)

Praveen Bulusu
Praveen Bulusu 2020년 2월 22일
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
  댓글 수: 2
John McDowell
John McDowell 2020년 7월 6일
편집: John McDowell 2020년 7월 6일
These are both really useful functions! Thanks for sharing them
Frederico Pratas
Frederico Pratas 2021년 4월 1일
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

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Stephen23
Stephen23 2019년 12월 10일
편집: Stephen23 2022년 6월 15일
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)
s = 1×10 string array
"A" "B" "Z" "AA" "AB" "ZZ" "AAA" "AAB" "UMX" "XFD"
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.:
  댓글 수: 1
Stephen23
Stephen23 2022년 6월 15일
편집: Stephen23 2022년 6월 15일
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)
s = 10×1 string array
"A" "B" "Z" "AA" "AB" "ZZ" "AAA" "AAB" "UMX" "XFD"

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Andrei Bobrov
Andrei Bobrov 2015년 10월 15일
편집: Andrei Bobrov 2015년 10월 16일
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);
  댓글 수: 7
Andrei Bobrov
Andrei Bobrov 2020년 11월 22일
편집: Andrei Bobrov 2020년 11월 22일
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'}
>>
Stephen23
Stephen23 2022년 5월 18일
편집: Stephen23 2022년 6월 15일
@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

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Daniel LeBrun
Daniel LeBrun 2019년 12월 10일
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.

Qiang Fu
Qiang Fu 2023년 10월 10일
편집: Qiang Fu 2023년 10월 10일
I had the same requirement shortly before when I used the "writematrix" function in order to convert numerical index to the corresponding column in excel file in MATLAB, and I wrote the following code based on regarding it as a geometric progression, of which the first term and common ratio were both 26.
function ac_id = idx_xlsx(num_col)
arguments
num_col {mustBeInteger}
end
% Create the geometric progression
n = 0;
f1 = @(a,q,n) a*(1-q.^n)/(1-q);
% The length of expected result
while true
if (num_col > f1(26,26,n)) && (num_col <= f1(26,26,n+1))
break
else
n = n + 1;
end
end
len_id = n + 1;
ac_id = char();
formalid = f1(26,26,n); % The sum of previous intervals
curr = num_col - formalid; % The extra length in current interval
% The 1st character from the left
if curr/26.^(len_id-1) == fix(curr/26.^(len_id-1))
ac_id(1) = char(64+fix(curr/26.^(len_id-1)));
else
ac_id(1) = char(64+fix(curr/26.^(len_id-1)+1));
end
% From the 2nd character to the last but one character
for j=2:len_id-1
% Calculate the number of secondary interval included
% in the extra length
nomin = curr-power(26,len_id-j+1)*fix(curr/power(26,len_id-j+1));
% Calculate the number of sub-interval of the secondary interval
% included in the "remaining" range
denomin = power(26,len_id-j);
idk = nomin / denomin;
if idk == 0
ac_id(j) = char(64+26);
elseif (idk == fix(idk)) && (idk ~= 0)
idk = fix(idk);
ac_id(j) = char(64+idk);
else
idk = fix(idk)+1;
ac_id(j) = char(64+idk);
end
end
% The last character
if mod(curr,26) ~= 0
ac_id(len_id) = char(64+mod(curr,26));
else
ac_id(len_id) = char(64+26);
end
end
The output is the following
>> a = idx_xlsx(1368) % Random number
a = 'AZP'
>> b = idx_xlsx(3250) % 26*n (n = 1,2,...)
b = 'DTZ'
>> c = idx_xlsx(16384) % The maximum column of excel file
c = 'XFD'
I' ve tested it with a series of numbers, and I hope this code could be reliable and we could optimize it.

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