Skip to main content

Excel column label from integer


I would like to create a function in Mathematica, which returns the equivalent Excel column label (e.g., A, Z, AA, AZ, and so on) given a column number.


I think it can be done with IntegerString[10,26], but this representation set of this chars is 0 to 9 and then a to z. However, Excel uses only A to Z.


How can this be done?



Answer



Ok, well, I only needed the inverse function so far and have implemented it as:


FromExcelCol[col_String] := FromDigits[ToCharacterCode[col] - 64, 26]


It runs fine because FromDigits does not complain about characters larger than base-1.


However, the other way round seems to be more tricky. The leading digit runs from 0 to 26 (1 to 27 if you want -> base 27), but only as long as it is leading. Then it runs as a trailing digit from 1 to 26.


I have not found an elegant, non-iterative solution so far. I was hesitating to paste my (ugly but working) piece of code, but maybe it encourages others to look for a nice solution. :)


ToExcelCol[n_Integer] := Module[{subtract, num, base},

  subtract = Accumulate[Power[26, #] & /@ (Range[7] - 1)];
  base = Position[subtract, x_ /; x > n, {1}, 1][[1, 1]] - 1; (*find largest number which is <= base ... I think there are better alg. for this but I don't have them at hand, sorry*)
  num = n - subtract[[base]];


  StringJoin@PadLeft[FromCharacterCode /@ (IntegerDigits[num, 26] + 65), base, "A"]
  ]

Basically, what you have to do and what this function does, is subtract 26^0 if n > 26^0, then subtract 26^1 if n > 26^1+26^0 and so on. Finally pad the result to x digits with the largest 26^x+26^(x-1)+....


I have tried but I could not make a satisfying solution, so I kindly invite you to improve this piece of code. I still find that it is too iterating (with the lookup table generated in the first step). I have also thought of treating part of the number as base 27 and the other part as base 26, but well... not tonight anymore. :)


EDIT: Jacob just figured out what I was looking for: You can determine the number of digits in the final column name with Log[26, 25 (n + 1)]. So, here is the simplified version:


ToExcelCol2[n_Integer] := Module[{num, base},
  base = Floor[Log[26, 25 (n + 1)]];
  num = n - Total[Power[26, #] & /@ (Range[base] - 1)];
  StringJoin@PadLeft[FromCharacterCode /@ (IntegerDigits[num, 26] + 65), base, "A"]]


You can condense this into a one-liner without Module of course, but I'll leave that up to you.


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more