Skip to main content

Problem using OptionValue with functions defined by SubValues, and the use of Options for function dispatch


I have a function that looks like:


(a : pump[_, __])["curve QH", opts : OptionsPattern[]] :=
  Plot[x, {x, 0, 2}, Frame -> True, FrameLabel -> {"a", "b"}, 

  Evaluate[FilterRules[{opts}, Options[Plot]]]

This code allows me to substitute the option of Plot, with whatever I may want (where pump acts like a very simple object).


Now suppose I want to use an option to define if a Plot or a Table should be outputed.


I imagined something like this:


(a : pump[_, __])["curve QH", opts : OptionsPattern["type" -> "plot"]] :=
 Which[
  OptionValue["type"] == "plot",
  Plot[x, {x, 0, 2}, Frame -> True, FrameLabel -> {"a", "b"}, 
   Evaluate[FilterRules[{opts}, Options[Plot]]]],

  OptionValue["type"] == "table",
  Table[x, {x, 0, 2, 0.1}]
  ]

but it doesn't work.


What is the best strategy to use Options with this kind of definition?



Answer



There are two problems I see with this code. The first one, and the one which gives you immediate problem you reported, is that, as it looks, short (magical) form of OptionValue does not work with SubValues - based definitions. So, you should replace


OptionValue["type"]


with


OptionValue[{opts}, "type"]

Or, if you wish to associate the option with pump directly and use it within the SubValue, you need to use


OptionValue[pump, {opts}, "type"]

instead.


The second, "conceptual" problem, is that you are really trying to use options for function dispatch, but in doing this your way, you give up the powerful rule-based function dispatch mechanism that is naturally available in mma. This is what I would do instead:


ClearAll[pump, dispatch];
(a : pump[_, __])["curve QH", opts : OptionsPattern["type" -> "plot"]] :=

    dispatch[pump, OptionValue[{opts}, "type"]][a, "curve QH", opts];

dispatch[pump, "plot"][_, _, opts : OptionsPattern[]] :=
   Plot[x, {x, 0, 2}, Frame -> True, FrameLabel -> {"a", "b"}, 
      Evaluate[FilterRules[{opts}, Options[Plot]]]];

dispatch[pump, "table"][_, _, opts : OptionsPattern[]] :=
   Table[x, {x, 0, 2, 0.1}]

I delegated the dispatch to a new (extra) symbol dispatch, introduced specifically for this purpose. This looks cleaner and more extensible to me than using Which statement.



EDIT


Addressing the issue raised in comments: note that, as another manifestation of OptionValue - OptionsPattern magic being lost for SubValues- based functions, the above solution, as written, will not provide the intended default behavior, when the "type" option is not explicitly given. To make that work, one can do:


(a : pump[_, __])["curve QH", opts : OptionsPattern["type" -> "plot"]] :=
    dispatch[
       pump, 
       OptionValue[{opts,"type" -> "plot"}, "type"]
    ][a, "curve QH", opts];

adding the default option explicitly in OptionValue.


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