Re: Simple question

*To*: mathgroup at smc.vnet.net*Subject*: [mg132205] Re: Simple question*From*: Ulrich Arndt <ulrich.arndt at data2knowledge.de>*Date*: Thu, 16 Jan 2014 01:54:44 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <20140115091656.DEBE169F0@smc.vnet.net>

Would something like this work for you? exp = K11 Ue + (K12 - K21) Uexx + K22 Uex + (K23 + K32) Uexy + (K13 - K31) Uey + K33 Ueyy; findings = StringTrim[ StringTake[ToString[exp], StringPosition[ToString[exp], x : Shortest["Ue" ~~ ___] ~~ " " | "+" | "(" | ")" | "-" | "/" | EndOfString]]] Clear[genFun] genFun[exp_String, pat_String] := Block[{b, c, s}, If[exp === pat, StringTake[exp, 1] <> "[x,y,z,t]", b = StringTake[exp, 1]; c = StringTake[exp, {StringLength[pat] + 1, StringLength[exp]}]; s = ToString[Characters[c]]; "D[" <> StringTake[exp, 1] <> "[x,y,z,t]," <> If[StringLength[c] == 1, c, s] <> "]" ] ] genFun["Uexy", "Ue"] genFun["Uey", "Ue"] genFun["Ue", "Ue"] mapped = genFun[#, "Ue"] & /@ findings; f = ToExpression[#, TraditionalForm] & /@ findings; m = ToExpression[#, TraditionalForm] & /@ mapped; ReplaceAll[exp, Apply[Rule, Transpose[{f, m}], 1]] Am 15.01.2014 um 10:16 schrieb KFUPM: > Dear All > > I have a long expression. Below is just a short part of it: > > K11 Ue + (K12 - K21) Uex + > K22 Uexx + (K23 + K32) Uexy + (K13 - K31) Uey + K33 Ueyy > > I want to replace Ue->U[x,y,z,t]. and Uex->D[U[x,y,z,t],x] and Uexy->D[U[x,y,z,t],{x,y}] and so on. I want to make a command to scan the expression and do the conversion autmatically. I don't want to use Replace or ReplaceAll commands, because as I said, the command is very long and I have so many functions to deal with. > > Many thanks in advance for your help. > > HMQ >

**References**:**Simple question***From:*KFUPM <hussain.alqahtani@gmail.com>