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I am looking for replacement rules that will accomplish the following:

  • If the variable name of the function contains "p", append "aa" to the function name
  • If the variable name of the function does not contain "p", append "bb" to the function name

Here are some examples

myfun1[x1] ==> myfun1bb[x1]
myfun2[x1p] ==> myfun2aa[x1p]
myfun3[x4p] ==> myfun4aa[x4p]

Any help is greatly appreciated!

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  • $\begingroup$ May I ask why? :P Mixing up the name of a variable/function and the value of a variable/function is not usually considered good practice. $\endgroup$
    – Marius Ladegård Meyer
    Aug 4, 2017 at 22:27
  • $\begingroup$ Will you only consider symbols as arguments, or can they be numbers, or expressions, or... $\endgroup$
    – Marius Ladegård Meyer
    Aug 4, 2017 at 22:28
  • $\begingroup$ @MariusLadegårdMeyer Thanks for your comments, I will only use symbols as arguments. The reason why I am doing this is that I have a long expression (>1000 terms) which I need to integrate over (it is a multi dimensional integration). It happens that the individual parts can be integrated analytically, but the integral depends in on the symbolic variable (it makes a difference on whether we are integrating over x1 or x1p). Mathematica can sovle these integrals but it is rather slow. $\endgroup$
    – ftiaronsem
    Aug 6, 2017 at 14:37
  • $\begingroup$ I have therefore saved the analytically integrated parts in the variables myfun1bb and myfun2aa. So after the replacement the result is the correctly integrated function. But maybe I overlooked a simpler way of achieving this... $\endgroup$
    – ftiaronsem
    Aug 6, 2017 at 14:41

4 Answers 4

Reset to default
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Convert to strings and then back to a symbol.

rule = fun_[arg_] :> Module[
    {
     funString = ToString[fun],
     argString = ToString[arg]
     },
    If[
     StringMatchQ[argString, ___ ~~ "p" ~~ ___],
       funString = funString <> "aa",
       funString = funString <> "bb"
     ];
    ToExpression[funString <> "[" <> argString <> "]"]
    ];

Now test it on the two cases.

myfun[x1] /. rule
(* myfunbb[x1] *)

myfun[x1p] /. rule
(* myfunaa[x1p] *)
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  • $\begingroup$ This worked perfectly for what I was trying to do! Thanks so much $\endgroup$
    – ftiaronsem
    Aug 7, 2017 at 12:55
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I answer this with the caveat that I share Marius's puzzlement at why you have to do this. Nevertheless:

Operate[Function[x, Symbol[ToString[#] <>
                           If[StringFreeQ[ToString[x], "p"], "bb", "aa"]][x]] &,
        myfun1[x1]]
   myfun1bb[x1]

Operate[Function[x, Symbol[ToString[#] <>
                           If[StringFreeQ[ToString[x], "p"], "bb", "aa"]][x]] &,
        myfun3[x4p]]
   myfun3aa[x4p]
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$\begingroup$ 
f1 = # /. h_[x__] :> If[And @@ StringFreeQ[ToString /@ {x}, "p"], 
  Symbol[SymbolName[h] <> "bb"], Symbol[SymbolName[h] <> "aa"]][ x] &;


f1 /@ {myfun1[x1], myfun2[x1p], myfun3[x1, x2, x3p], myfun4[x1, x2, xyz[pqrs]]}

{myfun1bb[x1], myfun2aa[x1p], myfun3aa[x1, x2, x3p], myfun4aa[x1, x2, xyz[pqrs]]}

Or, using Replace

f2 = Replace[#, h_[x__] :> If[And @@ StringFreeQ[ToString /@ {x}, "p"], 
       Symbol[SymbolName[h] <> "bb"], Symbol[SymbolName[h] <> "aa"]][x]] &;
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Now that we have some clarification in the comments, I would propose a simple wrapper-type definition:

myfun1[arg_] := If[StringFreeQ[SymbolName[arg], "p"], myfun1bb[arg], myfun1aa[arg]]

Do you need it to be more complicated than this?

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  • $\begingroup$ Thanks for providing this answer, unfortunately it doesn't quite fit in my case since I can't redefine the function myfun1. I do need it later on in the calculations (together with the integrated function). Basically I get this very long sum of terms containing myfuns and need to integrate it. Afterwards I need access to the original sum of terms and to the integrated sum of terms. $\endgroup$
    – ftiaronsem
    Aug 7, 2017 at 3:35
  • $\begingroup$ Hmm, but thinking of it, I could make a copy of the long expression, replace all the occurances of myfun with myfuntemp and then use your approach. So it would work. I will evaluate the different answers more closely tomorrow. Thanks. $\endgroup$
    – ftiaronsem
    Aug 7, 2017 at 3:36

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