Convert a number to Excel’s base 26
Question:
OK, I’m stuck on something seemingly simple. I am trying to convert a number to base 26 (ie. 3 = C, 27 = AA, ect.). I am guessing my problem has to do with not having a 0 in the model? Not sure. But if you run the code, you will see that numbers 52, 104 and especially numbers around 676 are really weird. Can anyone give me a hint as to what I am not seeing? I will appreciate it. (just in case to avoid wasting your time, @ is ascii char 64, A is ascii char 65)
def toBase26(x):
x = int(x)
if x == 0:
return '0'
if x < 0:
negative = True
x = abs(x)
else:
negative = False
def digit_value (val):
return str(chr(int(val)+64))
digits = 1
base26 = ""
while 26**digits < x:
digits += 1
while digits != 0:
remainder = x%(26**(digits-1))
base26 += digit_value((x-remainder)/(26**(digits-1)))
x = remainder
digits -= 1
if negative:
return '-'+base26
else:
return base26
import io
with io.open('numbers.txt','w') as f:
for i in range(1000):
f.write('{} is {}n'.format(i,toBase26(i)))
So, I found a temporary workaround by making a couple of changes to my function (the 2 if statements in the while loop). My columns are limited to 500 anyways, and the following change to the function seems to do the trick up to x = 676, so I am satisfied. However if any of you find a general solution for any x (may be my code may help), would be pretty cool!
def toBase26(x):
x = int(x)
if x == 0:
return '0'
if x < 0:
negative = True
x = abs(x)
else:
negative = False
def digit_value (val):
return str(chr(int(val)+64))
digits = 1
base26 = ""
while 26**digits < x:
digits += 1
while digits != 0:
remainder = x%(26**(digits-1))
if remainder == 0:
remainder += 26**(digits-1)
if digits == 1:
remainder -= 1
base26 += digit_value((x-remainder)/(26**(digits-1)))
x = remainder
digits -= 1
if negative:
return '-'+base26
else:
return base26
Answers:
The problem when converting to Excel’s “base 26” is that for Excel, a number ZZ is actually 26 * 26**1 + 26 * 26**0 = 702 while normal base 26 number systems would make a 1 * 26**2 + 1 * 26**1 + 0 * 26**0 = 702 (BBA) out of that. So we cannot use the usual ways here to convert these numbers.
Instead, we have to roll our own divmod_excel function:
def divmod_excel(n):
a, b = divmod(n, 26)
if b == 0:
return a - 1, b + 26
return a, b
With that, we can create a to_excel function:
import string
def to_excel(num):
chars = []
while num > 0:
num, d = divmod_excel(num)
chars.append(string.ascii_uppercase[d - 1])
return ''.join(reversed(chars))
For the other direction, this is a bit simpler
import string
from functools import reduce
def from_excel(chars):
return reduce(lambda r, x: r * 26 + x + 1, map(string.ascii_uppercase.index, chars), 0)
This set of functions does the right thing:
>>> to_excel(26)
'Z'
>>> to_excel(27)
'AA'
>>> to_excel(702)
'ZZ'
>>> to_excel(703)
'AAA'
>>> from_excel('Z')
26
>>> from_excel('AA')
27
>>> from_excel('ZZ')
702
>>> from_excel('AAA')
703
And we can actually confirm that they work correctly opposite of each other by simply checking whether we can chain them to reproduce the original number:
for i in range(100000):
if from_excel(to_excel(i)) != i:
print(i)
# (prints nothing)
Sorry, I wrote this in Pascal and know no Python
function NumeralBase26Excel(numero: Integer): string;
var
algarismo: Integer;
begin
Result := '';
numero := numero - 1;
if numero >= 0 then
begin
algarismo := numero mod 26;
if numero < 26 then
Result := Chr(Ord('A') + algarismo)
else
Result := NumeralBase26Excel(numero div 26) + Chr(Ord('A') + algarismo);
end;
end;
Simplest way, if you do not want to do it yourself:
from openpyxl.utils import get_column_letter
proper_excel_column_letter = get_column_letter(5)
# will equal "E"
You can do it in one line (with line continuations for easier reading). Written here in VBA:
Function sColumn(nColumn As Integer) As String
' Return Excel column letter for a given column number.
' 703 = 26^2 + 26^1 + 26^0
' 64 = Asc("A") - 1
sColumn = _
IIf(nColumn < 703, "", Chr(Int((Int((nColumn - 1) / 26) - 1) / 26) + 64)) & _
IIf(nColumn < 27, "", Chr( ((Int((nColumn - 1) / 26) - 1) Mod 26) + 1 + 64)) & _
Chr( ( (nColumn - 1) Mod 26) + 1 + 64)
End Function
Or you can do it in the the worksheet:
=if(<col num> < 703, "", char(floor((floor((<col num> - 1) / 26, 1) - 1) / 26, 1) + 64)) &
if(<col num> < 27, "", char(mod( floor((<col num> - 1) / 26, 1) - 1, 26) + 1 + 64)) &
char(mod( <col num> - 1 , 26) + 1 + 64)
I’ve also posted the inverse operation done similarly.
Based on @TheUltimateOptimist’s answer, I looked in the openpyxl implementation and found the "actual" algorithm used by openpyxl==3.0.10:
Be warned; it only supports values between 1 & 18278 (inclusive).
def _get_column_letter(col_idx):
"""Convert a column number into a column letter (3 -> 'C')
Right shift the column col_idx by 26 to find column letters in reverse
order. These numbers are 1-based, and can be converted to ASCII
ordinals by adding 64.
"""
# these indicies corrospond to A -> ZZZ and include all allowed
# columns
if not 1 <= col_idx <= 18278:
raise ValueError("Invalid column index {0}".format(col_idx))
letters = []
while col_idx > 0:
col_idx, remainder = divmod(col_idx, 26)
# check for exact division and borrow if needed
if remainder == 0:
remainder = 26
col_idx -= 1
letters.append(chr(remainder+64))
return ''.join(reversed(letters))
OK, I’m stuck on something seemingly simple. I am trying to convert a number to base 26 (ie. 3 = C, 27 = AA, ect.). I am guessing my problem has to do with not having a 0 in the model? Not sure. But if you run the code, you will see that numbers 52, 104 and especially numbers around 676 are really weird. Can anyone give me a hint as to what I am not seeing? I will appreciate it. (just in case to avoid wasting your time, @ is ascii char 64, A is ascii char 65)
def toBase26(x):
x = int(x)
if x == 0:
return '0'
if x < 0:
negative = True
x = abs(x)
else:
negative = False
def digit_value (val):
return str(chr(int(val)+64))
digits = 1
base26 = ""
while 26**digits < x:
digits += 1
while digits != 0:
remainder = x%(26**(digits-1))
base26 += digit_value((x-remainder)/(26**(digits-1)))
x = remainder
digits -= 1
if negative:
return '-'+base26
else:
return base26
import io
with io.open('numbers.txt','w') as f:
for i in range(1000):
f.write('{} is {}n'.format(i,toBase26(i)))
So, I found a temporary workaround by making a couple of changes to my function (the 2 if statements in the while loop). My columns are limited to 500 anyways, and the following change to the function seems to do the trick up to x = 676, so I am satisfied. However if any of you find a general solution for any x (may be my code may help), would be pretty cool!
def toBase26(x):
x = int(x)
if x == 0:
return '0'
if x < 0:
negative = True
x = abs(x)
else:
negative = False
def digit_value (val):
return str(chr(int(val)+64))
digits = 1
base26 = ""
while 26**digits < x:
digits += 1
while digits != 0:
remainder = x%(26**(digits-1))
if remainder == 0:
remainder += 26**(digits-1)
if digits == 1:
remainder -= 1
base26 += digit_value((x-remainder)/(26**(digits-1)))
x = remainder
digits -= 1
if negative:
return '-'+base26
else:
return base26
The problem when converting to Excel’s “base 26” is that for Excel, a number ZZ is actually 26 * 26**1 + 26 * 26**0 = 702 while normal base 26 number systems would make a 1 * 26**2 + 1 * 26**1 + 0 * 26**0 = 702 (BBA) out of that. So we cannot use the usual ways here to convert these numbers.
Instead, we have to roll our own divmod_excel function:
def divmod_excel(n):
a, b = divmod(n, 26)
if b == 0:
return a - 1, b + 26
return a, b
With that, we can create a to_excel function:
import string
def to_excel(num):
chars = []
while num > 0:
num, d = divmod_excel(num)
chars.append(string.ascii_uppercase[d - 1])
return ''.join(reversed(chars))
For the other direction, this is a bit simpler
import string
from functools import reduce
def from_excel(chars):
return reduce(lambda r, x: r * 26 + x + 1, map(string.ascii_uppercase.index, chars), 0)
This set of functions does the right thing:
>>> to_excel(26)
'Z'
>>> to_excel(27)
'AA'
>>> to_excel(702)
'ZZ'
>>> to_excel(703)
'AAA'
>>> from_excel('Z')
26
>>> from_excel('AA')
27
>>> from_excel('ZZ')
702
>>> from_excel('AAA')
703
And we can actually confirm that they work correctly opposite of each other by simply checking whether we can chain them to reproduce the original number:
for i in range(100000):
if from_excel(to_excel(i)) != i:
print(i)
# (prints nothing)
Sorry, I wrote this in Pascal and know no Python
function NumeralBase26Excel(numero: Integer): string;
var
algarismo: Integer;
begin
Result := '';
numero := numero - 1;
if numero >= 0 then
begin
algarismo := numero mod 26;
if numero < 26 then
Result := Chr(Ord('A') + algarismo)
else
Result := NumeralBase26Excel(numero div 26) + Chr(Ord('A') + algarismo);
end;
end;
Simplest way, if you do not want to do it yourself:
from openpyxl.utils import get_column_letter
proper_excel_column_letter = get_column_letter(5)
# will equal "E"
You can do it in one line (with line continuations for easier reading). Written here in VBA:
Function sColumn(nColumn As Integer) As String
' Return Excel column letter for a given column number.
' 703 = 26^2 + 26^1 + 26^0
' 64 = Asc("A") - 1
sColumn = _
IIf(nColumn < 703, "", Chr(Int((Int((nColumn - 1) / 26) - 1) / 26) + 64)) & _
IIf(nColumn < 27, "", Chr( ((Int((nColumn - 1) / 26) - 1) Mod 26) + 1 + 64)) & _
Chr( ( (nColumn - 1) Mod 26) + 1 + 64)
End Function
Or you can do it in the the worksheet:
=if(<col num> < 703, "", char(floor((floor((<col num> - 1) / 26, 1) - 1) / 26, 1) + 64)) &
if(<col num> < 27, "", char(mod( floor((<col num> - 1) / 26, 1) - 1, 26) + 1 + 64)) &
char(mod( <col num> - 1 , 26) + 1 + 64)
I’ve also posted the inverse operation done similarly.
Based on @TheUltimateOptimist’s answer, I looked in the openpyxl implementation and found the "actual" algorithm used by openpyxl==3.0.10:
Be warned; it only supports values between 1 & 18278 (inclusive).
def _get_column_letter(col_idx):
"""Convert a column number into a column letter (3 -> 'C')
Right shift the column col_idx by 26 to find column letters in reverse
order. These numbers are 1-based, and can be converted to ASCII
ordinals by adding 64.
"""
# these indicies corrospond to A -> ZZZ and include all allowed
# columns
if not 1 <= col_idx <= 18278:
raise ValueError("Invalid column index {0}".format(col_idx))
letters = []
while col_idx > 0:
col_idx, remainder = divmod(col_idx, 26)
# check for exact division and borrow if needed
if remainder == 0:
remainder = 26
col_idx -= 1
letters.append(chr(remainder+64))
return ''.join(reversed(letters))