Enumerate a Cell Notation

Question

So we're all hopefully familiar with Spreadsheet 'A1' cell notation.

It's simply an alphanumeric representation of the positioning of said cell within a grid. The letter(s) represent the column positioning of the cell, and the number represents the row.

The 'letter' part can consist of 1 or more letters from the 26 letter English alphabet, all of which must be capital letters. These map to numbers through the use of 26-adic bijective numeration. The 'number' part can consist of any positive, non-zero integer.

---

The challenge, write a program that given the A1 notation of any cell as a single string, can output a string containing the column position represented as a number, followed by a space and then the row number.

Sample Input/Outputs below:

A1
>>1 1
B10
>>2 10
AC4 
>>29 4
AAC753
>>705 753
F123
>>6 123
GL93
>>194 93

This is my first challenge, hence the relative simplicity and potential poorness of the criteria.

EDIT: String must be letters followed by numbers and the winning criteria is the shortest code length (if that can be a thing)

EDIT: Related to this but does the reverse process with a different starting index. Some may argue that this fact makes the linked puzzle more interesting.

Answer 1

05AB1E, 12 11 bytes

áÇ64-₂βð¹þJ

Uses the 05AB1E encoding. Try it online!

Answer 2

Microsoft Excel, 43 Bytes.

=COLUMN(INDIRECT(A1))&" "&ROW(INDIRECT(A1))

I couldn't help myself, I just had to use the right tool for the job. Takes input on A1.

Test Cases

Answer 3

Microsoft Excel, 40 bytes

Brasilian Portuguese language version.

=COL(INDIRETO(A1))&" "&LIN(INDIRETO(A1))

Translated (and therefore golfed) version of ATaco's solution.

Answer 4

Python 2, 94 91 73 bytes

^{-3 bytes thanks to Jonathan Allan
-18 bytes thanks to tsh}

s=input().strip
a=s(`3**39`);t=0
for n in a:t=t*26+ord(n)-64
print t,s(a)

Try it online!

This abuses the way that .strip works, removing all digits wtih a=s(`3**39`) where `3**39` is just a shorter way to generate the digits from 0 to 9, this will store only the chars on a that will be used to strip the chars from the numbers on s(a)

Answer 5

Google Sheets, 43 bytes

=COLUMN(INDIRECT(A1))&" "&ROW(INDIRECT(A1))

Try it online!

A1 is the input cell. Hopefully the above link works, I've never tried code golf with Google Sheets before.

Answer 6

JavaScript, 72 bytes

x=>x.replace(/\D+/,s=>[...s].reduce((n,v)=>n*26+parseInt(v,36)-9,0)+' ')

Answer 7

Proton, 113 bytes

k=>{b=(s=k.strip)('0123456789');str(sum(map((a=>26**a[1][0]*(ord(a[1][1])-64)),enumerate(b[to by-1]))))+' '+s(b)}

Try it online!

-19 bytes borrowing Rod's algorithm (note to self: add listcomps)

Answer 8

Dyalog APL, 44 bytes

(26⊥¯65+⎕AV⍳(a~⎕D)),⎕A~⍨a←⍞

Try it online!

Answer 9

Pyth - 31 bytes

First attempt, very naive.

jd,s*V.u*26NQ1_hMxLGr-QK@`UTQZK

Try it online here.

Answer 10

Jelly,  16  15 bytes

ØAɓi@€ḟ0ḅ26,⁸Kḟ

A full program accepting the string as an argument and printing the result

Try it online!

How?

ØAɓi@€ḟ0ḅ26,⁸Kḟ - Main link: list of characters, ref  e.g. "AAC753"
ØA              - uppercase alphabet yield               = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
  ɓ             - dyadic chain separation, call that B and place it on the right
   i@€          - first index for €ach (swap arguments)    [1,1,3,0,0,0]
      ḟ0        - filter discard zeros                     [1,1,3]
        ḅ26     - convert from base 26 to integer           705
            ⁸   - chain's left argument (ref)              "AAC753"
           ,    - pair                                     [705,['A','A','C','7','5','3']]
             K  - join with spaces                         [705,' ','A','A','C','7','5','3']
              ḟ - filter discard those in right (B)        [705,' ','7','5','3']
                - implicit print                        >>>705 753

Note: conversion from a list of numbers using the base conversion atom ḅ allows places to be outside of the expected range, so conversion from, say, [2,26,1] ("BZA") using ḅ26 is calculated as 2×26²+26×26¹+1×26⁰=2029 even though the "expected" representation would have been [3,0,1].

Answer 11

Perl 5, 35 + 1 (-p) = 36 bytes

map$r=$r*26-64+ord,/\D/g;s/\D*/$r /

Try it online!

The map statement treats the column like a base26 number and converts it to decimal. The substitution replaces the letters with the numbers. The input and output are implicit in the -p flag.

Answer 12

Mathematica, 108 bytes

StringReplace[c:LetterCharacter..~~r:DigitCharacter..:>ToString@FromDigits[ToCharacterCode@c-64,26]<>" "<>r]

Operator form of StringReplace which takes the alphabetic part c of the string, converts it to a list of character codes, subtracts 64 from each code point, interprets the result as a base 26 integer, converts that integer to a String, then StringJoins a space followed by the numeric part r.

Using a regex (also 108 bytes):

StringReplace[RegularExpression["([A-Z]+)(.+)"]:>ToString@FromDigits[ToCharacterCode@"$1"-64,26]<>" "<>"$2"]

Answer 13

Mathematica, 77 72 69 68 bytes

#/.{a__,b__?DigitQ}:>{f=FromDigits;f[LetterNumber@{a},26],f[b<>""]}&

Takes a list of characters.

Try it on Wolfram Sandbox

Usage

#/.{a__,b__?DigitQ}:>{f=FromDigits;f[LetterNumber@{a},26],f[b<>""]}&[{"G", "L", "9", "3"}]

{194, 93}

or

#/.{a__,b__?DigitQ}:>{f=FromDigits;f[LetterNumber@{a},26],f[b<>""]}&[Characters["GL93"]]

{194, 93}

Explanation

#/.

In the input replace...

{a__,b__?DigitQ}

A list containing two sequences a and b (with 1 or more elements), where b consists of digit characters only... (Mathematica uses lazy pattern matching, so no check is required for a)

:>

into...

f=FromDigits;

Set f to digit-to-integer conversion function.

LetterNumber@{a}

Convert a to letter numbers (a -> 1, b -> 2, etc).

{f[ ... ,26],f[b<>""]}

Convert the above from base-26 to decimal, and convert b to decimal.

Answer 14

Ruby, 50 48+1 = 51 49 bytes

Uses the -p flag. -2 bytes from m-chrzan.

sub(/\D+/){i=0;$&.bytes{|b|i=26*i+b-64};"#{i} "}

Try it online!

Answer 15

Ruby, 64 61 59 bytes

->s{s.sub(/\D+/){"#{$&.bytes.reduce(0){|t,b|t*26+b-64}} "}}

Saved 2 bytes thanks to Value Ink.

Answer 16

Excel-VBA Immediate window, 44 45 Bytes (36 35)

Set v=Range([A1]):?trim(v.Column &" "&v.Row);

1 byte added to suppress the trailing newline

---

Or for 35 with leading whitespace

Set v=Range([A1]):?v.Column""&v.Row

1 byte saved thanks to @TaylorScott!

Both take input from cell A1, output to the VBE Immediate window

Answer 17

Haskell, 67 bytes

(\(l,n)->show(foldl(\a->(a*26-64+).fromEnum)0l)++" "++n).span(>'9')

Try it online.

Answer 18

Retina, 85 bytes

[A-Z]
$& 
[T-Z]
2$&
[J-S]
1$&
T`_L`ddd
\d+
$*
{`\G1(?=.* .* )
26$*
}` (.* )
$1
1+
$.&

Try it online! Explanation:

[A-Z]
$& 

Separate the letters from each other and the final number.

[T-Z]
2$&
[J-S]
1$&
T`_L`ddd

Convert the letters to decimal.

\d+
$*

Convert everything to unary.

{`\G1(?=.* .* )
26$*
}` (.* )
$1

While there are at least three numbers, multiply the first by 26 and add it to the second.

1+
$.&

Convert everything to decimal.

Answer 19

><>, 42 Bytes

0i88*-:0(?\$2d**+!
$n88*+48*o\
oi:0(?;   >

Try It Online

Explanation:

0i88*-:0(?\$2d**+! Read in the Letters part of the input
0                  Push an initial 0 to the stack
 i                 Read a character of input
  88*-             Subtract 64 to get from the ascii code to its value ('A'=1,'B'=2 etc.)
      :0(?\        If the value is less than 0, break out of this loop
           $2d**   Multiply our accumulator by 26
                +  Add our new number to the accumulator
                 ! don't add a new zero to the stack

The above loop will read in the first number of the numbers part (eg. '3' in "AB34") before breaking, and will have already subtracted 64 from it, so the next bit of code has to deal with that

$n88*+48*o\        Output a space, and prepare to output the first number we read in during the previous loop
          \        Loop round to the left end of this line
$n                 Output the value of the letters part as a number
  88*+             Add the 64 we subtracted from the first number
      48*o         Output a space
          \        Enter the next loop

This loop starts by outputting a character which will either be the 1st character read in by the first loop, or the character read in by the previous iteration of this loop.

oi:0(?;   >        Output the numbers part, by just echoing it
          >        Loop round to the start of the line
o                  Output the character
 i                 Read in the next character
  :0(?;            If the value read in was less than 0, terminate

Answer 20

Lua, 127 Bytes

Some nice string manipulations going there, I'm using some function that are usually never used while golfing in lua :D. The best way to golf this would be to find an other way to iterate over the column part of the input while being able to retain the position of each letter. I didn't ^^'.

Try it online!

I=...l=I:gsub("%d",""):reverse()r=0
for i=1,#l
do
r=r+(l:sub(i,i):byte()-64)*math.pow(26,i-1)end
print(r.." "..I:gsub("%a",""))

Explanation

I=...                       -- shorthand for the input
l=I:gsub("%d","")           -- shorthand for the column part
   :reverse()               -- letters are put in reverse order to calculate their weight
r=0                         -- column number
for i=1,#l                  -- iterate over the letters of the input
do
  r=r+                      -- add to the column number
      (l:sub(i,i):byte()-64)-- the byte value of the current letter minus 64 (A=1 this way)
      *math.pow(26,i-1)     -- then apply its weight in base 26
end
print(r                     -- output the column number
        .." "               -- a space
        ..I:gsub("%a",""))  -- and the row number