Skip to main content

scoping - Is there any way to implement a "sequential" With[] in Mathematica?




I want the equivalent of Scheme's let*, or basically, a sequential With that works like this:


With[{a = 0,
a = a + 1,
a = a + 1},
a]

Is there any way to implement this? Everything I tried with Hold/Unevaluated/etc. led nowhere.



Answer



This is outside the scope of With. The documentation says:




With[{x=xâ‚€, y=yâ‚€, ...}, expr]


specifies that all occurrences of the symbols x, y, ... in expr should be replaced by xâ‚€, yâ‚€, ...



So even if there was a "sequential" With, it wouldn't be able to understand a = a+1 as updating the value of a. It would always just be a replace rule.


I think you'd be best off with a Module:


Module[{a},
a = 0;
a = a+1;
a = a+1;
a]


You can write your own command which rewrites the form of the command for you, for example:


letstar[init_List, expr_] :=
With[{vars = symbols[init]},
Module[vars, CompoundExpression @@ Join[init, {expr}]]]
symbols[init_List] := Union@Hold[init][[1, All, 1]]
SetAttributes[symbols, HoldFirst]
SetAttributes[letstar, HoldFirst]

This assumes that the first argument of letstar is a list of assignments (this is not checked) and holds its form so that they are not performed. Instead, they are passed to symbols which only extracts the left-hand sides and lists unique variables appearing in them. This is passed to a Module as the local variables, the init block is converted into a compound expression and finally expr is evaluated. So if you call



letstar[{a = 0, a = a + 1, a = a + 1}, a]

this gets internally transformed to


Module[{a}, a = 0; a = a+1; a = a+1; a]

and returns



2



Comments

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...

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...

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...