Hello everybody,
I stumbled across the following problem: I want to do the following substitution in some lengthy expression:
f[ x^a_ y^b_ z^c_]->g[a,b,c] including the case a,b,c ={0,1}.
Of course I could brute force code all the cases 0,1,n>1, but that would require 3^3 terms, and I will probably need to do this with a greater number of variables. So any smart idea is most welcome, and I might learn some coding trick in the process as I have relatively small experience with Mathematica. Cheers
In[154]:= exp = Table[Power[RandomChoice[{x, y, z}], i], {i, -3, 3}]
Out[154]= {1/y^3, 1/y^2, 1/y, 1, x, y^2, y^3}
Define a function that checks the form of the input expression and returns the exponent (power).
pwrChk[expr_] := Switch[expr,
(*1. Address the case of the constant*)
(*-----> If input is*)1, (*then return*)0,
(*2. Address all regular powers of the form x^y. *)
(*----->If input is*)Power[x_, y_], (*then return*) FullForm[expr][[1, 2]],
(*3. Address cases like x^1.*)
(*----->If input is*)expr, (*return*)1]
Try it on our sample list of expressions.
In[157]:= pwrChk[#] & /@ exp
Out[157]= {-3, -2, -1, 0, 1, 2, 3}
The answer is a list with the powers of the input expressions including cases where the power was a zero or one.
You could use the _. notation like in this example:
integrate[x_^n_., x_] := x^(n + 1)/(n + 1) /; FreeQ[n, x] && n != -1
here x^1 is also matched. By setting the proper defaults you can match most of them.
To match 0 and 1 exponents for monomials you could multiply the whole expression by a general monomial, e.g. :
list = Flatten[Table[x ^i y^j z^k, {k, 0, 2}, {j, 0, 2}, {i, 0, 2}]]
list x^p y^p z^p /. x^a_ y^b_ z^c_ :> f[a, b, c] /. p -> 0
I.M.