Skip to main content

evaluation - Replacing parts of a held expression with held parts of another expression



I'm applying the Trott-Strzebonski technique in order to replace inside a held expression.


x = Hold[1 + 1, 2 + 2, 3 + 3];
y = Hold[foo@bar[2]];
y /. bar[j_] :> With[{eval = x[[j]]}, eval /; True]
(* Hold[foo[4]] *)

However, I'd like the replacing expression to remain held during substitution:


(* Hold[foo[2 + 2]] *)

The unwanted evaluation of 2 + 2 seemed to be occurring during the Part statement, x[[j]]. So, I tried using Extract instead:



y /. bar[j_] :> With[{eval = Extract[x, {j}, Unevaluated]}, eval /; True]
(* Hold[foo[Unevaluated[2 + 2]]] *)

I thought Unevaluated would disappear as an argument of the Set function, but this was not so. (Perhaps I still don't quite understand Unevaluated.)


I then thought of removing that head, Unevaluated, through another replacement, but that seems rather convoluted.


Succinctly, how does one replace parts of a held expression with held parts of another expression?



Answer



One way to achieve this is to use a "vanishing" wrapper. The idea is to temporarily wrap the substituted expression with a holding symbolic head, and then remove that head in a second replacement:


Module[{h}
, SetAttributes[h, HoldAll]

; y /. bar[j_] :> RuleCondition[Extract[x, {j}, h]] /. h[x_] :> x
]

(* Hold[foo[2+2]] *)

Module is used to ensure that the vanishing symbol does not conflict with any of the symbols in the target expression. This solution is essentially equivalent to the exhibited use of Unevaluated, but forcibly removes the wrapper explicitly instead of relying upon the evaluator to do it.


This trick can be especially handy in more complex substitutions and can be scaled to use multiple wrappers, each with more elaborate semantics than simple vanishing.


The single-wrapper idiom can be made more convenient with a helper macro:


SetAttributes[vanishing, HoldAll]
vanishing[{h_}, body_] :=

Module[{h}
, SetAttributes[h, HoldAll]
; body /. h[x_] :> x
]

It is used thus:


vanishing[{h}, y /. bar[j_] :> RuleCondition[Extract[x, {j}, h]]]

(* Hold[foo[2+2]] *)

Comments

Popular posts from this blog

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...

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...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...