Skip to main content

pattern matching - How can a power of 0 or 1 be replaced?


Consider this given example for ReplaceAll


1 + x^2 + x^4 /. x^p_ -> f[p]

This returns 1+f[2]+f[4] - works good (!). But if you choose


 In[175]:= 1 +x+ x^2 + x^4 /. x^p_ -> f[p]
 Out[175]= 1+x+f[2]+f[4]

It doesn't not work for 1 and x. The correct output should be f[0]+f[1]+f[2]+f[4].



What is wrong here and to cure it?


EDIT



Possible way out



One possible way out can be to not use ReplaceAll (inspired by @eldo). The coefficient and the power of x can be combined in this way:


fun[x] = a - b x + c x^2 + d x^4
nfun = Exponent[fun[x], x];
Sum[Coefficient[fun[x], x, i] f[i], {i, 0, nfun}]


And the result is a f[0] - b f[1] + c f[2] + d f[4].


However the replace rule for x0 is still a mystery!



Answer



It would be cumbersome with ReplaceAll. I suggest to use Exponent instead:


Plus @@ (f /@ Exponent[1 + x + x^2 + x^4, x, List])


f[0] + f[1] + f[2] + f[4]



Considering Öska's objection



Alternating signs are not easily handled, maybe something like this:


fun1 = 1 - x - x^2 + x^4;
fun2 = List @@ fun1;

minus = Position[fun2, Times[-1, __]] // Flatten;
plus = Complement[Range@Length@fun2, minus];

fun3 = (f /@ Exponent[fun1, x, List]);

Plus @@ Join[Part[fun3, plus], Part[fun3, minus] /. a_ :> -a]



f[0] - f[1] - f[2] + f[4]



ADDENDUM


This can be shortened to:


 Plus @@ (D[#, x] & /@ MonomialList@fun1 /. Times[a_?NumberQ, __] :> a /. 
    a_?NumberQ :> f[Abs@a]*Sign@a /. (0) -> f@0*First@fun1)



f[0] - f[1] - f[2] + f[4]



Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more