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# Pattern recognition of exponents 0 and 1.

Posted 9 years ago
 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
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Posted 9 years ago
 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.
Posted 9 years ago
 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.
Posted 9 years ago
 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.