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Topic: Simple question
Replies: 3   Last Post: Jan 18, 2014 5:02 AM

 Messages: [ Previous | Next ]
 Ulrich Arndt Posts: 53 Registered: 10/1/09
Re: Simple question
Posted: Jan 16, 2014 1:50 AM

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.
>
>
> HMQ
>

Date Subject Author
1/16/14 Ulrich Arndt
1/16/14 Bob Hanlon
1/18/14 Alexei Boulbitch