Matching brackets in a string

Question

What is the most efficient or elegant method for matching brackets in a string such as:

"f @ g[h[[i[[j[2], k[[1, m[[1, n[2]]]]]]]]]] // z"

for the purpose of identifying and replacing [[ Part ]] brackets with the single character forms?

I want to get:

With everything else intact, such as the prefix @ and postfix // forms intact

---

An explanation of Mathematica syntax for those unfamiliar:

Functions use single square brackets for arguments: func[1, 2, 3]

Part indexing is done with double square brackets: list[[6]] or with single-character Unicode double brackets: list〚6〛

My intent is to identify the matching [[ ]] form in a string of ASCII text, and replace it with the Unicode characters 〚 〛

Answer 1

Ok, here is another answer, a bit shorter:

Clear[replaceDoubleBrackets];
replaceDoubleBrackets[str_String, openSym_String, closeSym_String] := 
Module[{n = 0},
  Apply[StringJoin, 
   Characters[str] /. {"[" :> {"[", ++n}, 
     "]" :> {"]", n--}} //. {left___, {"[", m_}, {"[", mp1_}, 
      middle___, {"]", mp1_}, {"]", m_}, right___} /; 
       mp1 == m + 1 :> {left, openSym, middle, 
        closeSym, right} /. {br : "[" | "]", _Integer} :> br]]

Example:

In[100]:= replaceDoubleBrackets["f[g[h[[i[[j[2], k[[1, m[[1, n[2]]]]]]]]]]]", "(", ")"]

Out[100]= "f[g[h(i(j[2], k(1, m(1, n[2]))))]]"

EDIT

You can also use Mathematica built-in facilities, if you want to replace double brackets specifically with the symbols you indicated:

Clear[replaceDoubleBracketsAlt];
replaceDoubleBracketsAlt[str_String] :=
  StringJoin @@ Cases[ToBoxes@ToExpression[str, InputForm, HoldForm],
     _String, Infinity]

In[117]:= replaceDoubleBracketsAlt["f[g[h[[i[[j[2], k[[1, m[[1, n[2]]]]]]]]]]]"]

Out[117]= f[g[h[[i[[j[2],k[[1,m[[1,n[2]]]]]]]]]]]

The result would not show here properly, but it is a Unicode string with the symbols you requested.

Answer 2

When I wrote my first solution, I hadn't noticed that you just wanted to replace the [[ with 〚 in a string, and not an expression. You can always use HoldForm or Defer as

but I think you already knew that, and you want the expression as a string, just like the input (ToString@ on the above doesn't work)

As all the answers so far focus on string manipulations, I'll take a numeric approach instead of wrestling with strings, which is more natural to me. The character code for [ is 91 and ] is 93. So doing the following

gives the locations of the brackets as a 0/1 vector. I've negated the closing brackets, just to aid the thought process and for use later on.

NOTE: I have only checked for divisibility by 91 and 93, as I certainly don't expect you to be entering any of the following characters, but if, for some reason you choose to, you can easily AND the result above with a boolean list of equality with 91 or 93.

From this, the positions of the first of Part's double bracket pair can be found as

The fact that in mma, expressions do not start with [ and that more than two [ cannot appear consecutively as [[[... has been implicitly assumed in the above calculation.

Now the closing pair is trickier to implement, but simple to understand. The idea is the following:

• For each non-zero position in closeBracket, say i, go to the corresponding position in openBracket and find the first non-zero position to the left of it (say j).
• Set doubleCloseBrackets[[i-1]]=closeBracket[[i]]+openBracket[[j]]+doubleOpenBrackets[[j]].
• You can see that doubleCloseBrackets is the counterpart of doubleOpenBrackets and is non-zero at the position of the first of Part's ]] pair.

So now we have a set of Boolean positions for the first open bracket. We simply have to replace the corresponding element in charCode with the equivalent of 〚 and similarly, with the Boolean positions for the first close bracket, we replace the corresponding element in charCode with the equivalent of 〛.

Finally, by deleting the element next to the ones that were changed, you can get your modified string with [[]] replaced by 〚 〛

NOTE 2:

A lot of my MATLAB habits have crept in the above code, and is not entirely idiomatic in Mathematica. However, I think the logic is correct, and it works. I'll leave it to you to optimize it (me thinks you can do away with Do[]) and make it a module, as it would take me a lot longer to do it.

Code as text

Clear["Global`*"]
str = "f[g[h[[i[[j[2], k[[1, m[[1, n[2]]]]]]]]]]]";
charCode = ToCharacterCode@str;
openBracket = Boole@Divisible[charCode, First@ToCharacterCode["["]];
closeBracket = -Boole@
    Divisible[charCode, First@ToCharacterCode["]"]];
doubleOpenBracket = 
  Append[Differences@Accumulate[openBracket], 0] openBracket;
posClose = Flatten@Drop[Position[closeBracket, Except@0, {1}], 1];

doubleCloseBracket = ConstantArray[0, Dimensions@doubleOpenBracket];
openBracketDupe = openBracket + doubleOpenBracket;
Do[
  tmp = Last@
    Flatten@Position[openBracketDupe[[1 ;; i]], Except@0, {1}];
  doubleCloseBracket[[i - 1]] = 
   closeBracket[[i]] + openBracketDupe[[tmp]];
  openBracketDupe[[tmp]] = 0;,
  {i, posClose}];

changeOpen = 
  Cases[Range[First@Dimensions@charCode]  doubleOpenBracket, Except@0];
changeClosed = 
  Cases[Range[First@Dimensions@charCode]  doubleCloseBracket, 
   Except@0];
charCode[[changeOpen]] = ToCharacterCode["\[LeftDoubleBracket]"];
charCode[[changeClosed]] = ToCharacterCode["\[RightDoubleBracket]"];
FromCharacterCode@
 Delete[Flatten@charCode, 
  List /@ (Riffle[changeOpen, changeClosed] + 1)]