using StringReplace and StringPattern to convert exp(...) to Exp[....]

Question

I am looking at a file generated by another CAS. I'd like to convert it to Mathematica syntax.

Only issue is how to convert exp(....) to Exp[....]. I am using StringReplace with pattern to do the replacement but it is not catching all possible cases. I am not good at regular expressions in Mathematica.

Here are 5 simple input examples to test with

ClearAll[z]
str1 = "exp(x)^2";
str2 = "exp((x+2))";
str3 = "exp( (x+2) * exp(exp(x)+2) )";
str4 = "(4*x)/exp((x-(2/5))/((3/(exp(x)+5))*4+(x-(2-x))*x))+4"
str5="(20*x^4+(-30)*x^3+12*x^2+4*x)*exp(x)^2+(200*x^4+(-300)*x^3+300*x^2+(-236)*x)*exp(x)+(500*x^4+(-750)*x^3+1500*x^2+(-1400)*x+720)"

These should become

str1 = "Exp[x]^2";
str2 = "Exp[(x+2)]";
str3 = "Exp[ (x+2) * Exp[Exp[x]+2]  ]";
str4 = "(4*x)/Exp[(x-(2/5))/((3/(Exp[x]+5))*4+(x-(2-x))*x)]+4"
str5 = "(20*x^4+(-30)*x^3+12*x^2+4*x)*Exp[x]^2+(200*x^4+(-300)*x^3+300*x^2+(-236)*x)*Exp[x]+(500*x^4+(-750)*x^3+1500*x^2+(-1400)*x+720)"

This is what I tried among many other things

StringReplace[str, "exp(" ~~ z__ ~~ ")" :> "Exp[" ~~ z ~~ "]"]

Only the first and second worked. The third one, it did not change any of the internal ones. it failed on str4 and str5 also.

Any suggestions how to do this conversion all in one command? Either using string patterns or regular expression in Mathematica.

I am using version 12.3.1

Answer 1

Update: We can convert the strings to expressions without any string processing:

ClearAll[sR0]
sR0 = ToExpression[#, TraditionalForm] &;

sR0 /@ {str1, str2, str3, str4, str5} // Column

If you need to get strings as output wrap the expression with ToString[#, InputForm]& using the third argument of ToExpression:

ClearAll[sR01]
sR01 = ToExpression[#, TraditionalForm, ToString[#, InputForm] &] &;

sR01 /@ {str1, str2, str3, str4, str5}

{"E^(2*x)", 
 "E^(2 + x)", 
 "E^(E^(2 + E^x)*(2 + x))",
 "4 + (4*x)/E^((-2/5 + x)/(12/(5 + E^x) + x*(-2 + 2*x)))",
 "720 - 1400*x + 1500*x^2 - 750*x^3 + 500*x^4 +
    E^(2*x)*(4*x + 12*x^2 - 30*x^3 + 20*x^4) + 
    E^x*(-236*x + 300*x^2 - 300*x^3 + 200*x^4)"}

Original answer:

ClearAll[sR]
sR = StringReplace[#, "exp(" ~~ Shortest[z__] ~~ ")" /; 
   Equal @@ (StringCount[z, #] & /@ {"(", ")"}) :> 
 "Exp[" <> sR @ z <> "]"] &;

sR /@ {str1, str2, str3, str4, str5} 

{"Exp[x]^2",
 "Exp[(x+2)]",
 "Exp[ (x+2) * Exp[Exp[x]+2] ]", 
 "(4*x)/Exp[(x-(2/5))/((3/(Exp[x]+5))*4+(x-(2-x))*x)]+4", 
 "(20*x^4+(-30)*x^3+12*x^2+4*x)*Exp[x]^2+(200*x^4+(-300)*x^3+300*x^2+
   (-236)*x)*Exp[x]+(500*x^4+(-750)*x^3+1500*x^2+(-1400)*x+720)"}

You can also use FixedPoint as follows:

ClearAll[sR2]
sR2 = FixedPoint[StringReplace[
     "exp(" ~~ Shortest[z__] ~~ ")" /; 
       Equal @@ (StringCount[z, #] & /@ {"(", ")"}) :> 
      "Exp[" <> z <> "]"], #] &;

sR2 /@ {str1, str2, str3, str4, str5} == sR /@ {str1, str2, str3, str4, str5}

 True

Answer 2

A simple trick is to notice that in Mathematica, Exp[x] can be written as Exp@(x) where the @ symbol has very high precedence. A replacement of the string "exp" by the string "Exp@" therefore does the necessary conversion without needing any bracket-matching logic:

convert[s_String] := ToExpression[StringReplace[s, "exp" -> "Exp@"]]

convert[str1]
(*    E^(2 x)    *)

convert[str2]
(*    E^(2 + x)    *)

convert[str3]
(*    E^(E^(2 + E^x) (2 + x))    *)

convert[str4]
(*    4 + 4 E^(-((-(2/5) + x)/(12/(5 + E^x) + x (-2 + 2 x)))) x    *)

convert[str5]
(*    720 - 1400 x + 1500 x^2 - 750 x^3 + 500 x^4 + E^(2 x) (4 x + 12 x^2 - 30 x^3 + 20 x^4) + E^x (-236 x + 300 x^2 - 300 x^3 + 200 x^4)    *)

If you want pretty string forms (instead of automatic evaluation with ToExpression as above), use an invisible application esc-@-esc to hide the @ symbols:

convert2[s_String] := StringReplace[s, "exp" -> "Exp\[InvisibleApplication]"]

convert2[str1]
(*    "Exp\[InvisibleApplication](x)^2"    *)
(*    actually looks like "Exp(x)^2" in the front-end but is
      correct Mathematica syntax                                *)

ToExpression[%]
(*    E^(2 x)    *)