Skip to main content

programming - How to construct custom operators with precedence?


I need to combine data structures in operations analogous to addition, subtraction, multiplication and division (and more). I need more than one operation of each type, i.e. more than one method of addition, subtraction, etc. I have made some progress. Using the Notation palette, i can for example define !/! and %/% as different operators for division. But how can I set precedence? My impression is that this is not possible. I have explored the existing symbols without in-built meanings, such as CirclePlus and CircleMinus. These symbols without in-built meanings have precedence while allowing the meaning to be defined by the user. But I can't find enough of them to meet my needs. For example, there is no "CircleDivide".


Might it be possible to use subscripts and have $/_a$ and $/_b$ for different types of division while retaining the precedence of the division operator? If I could do it for division, then I could do it for the other operators of interest too.


ADDENDUM: I've made some progress following the clue provided by Mr.Wizard. The following code creates subscripted operators for operations analogous to +,-,*,/. The conv functions are highly simplified for test purposes and in the actual application would perform operations on data structures.


In[1]:= conv[x_, y_, op_ /; op == "+" || op == "-"] := Module[{},
If[op == "-", Return[conv[x, -y, "+"]]]; x + y]


conv[x_, y_, op_ /; op == "*" || op == "/"] := Module[{},
If[op == "/", Return[conv[x, 1/y, "*"]]]; x y]

MakeExpression[RowBox[{x_, SubscriptBox["+", "i"], y_}], StandardForm] :=
MakeExpression[RowBox[{"conv", "[", x, ",", y, ",", "\"+\"", "]"}], StandardForm]
MakeExpression[RowBox[{x_, SubscriptBox["-", "i"], y_}], StandardForm] :=
MakeExpression[RowBox[{"conv", "[", x, ",", y, ",", "\"-\"", "]"}], StandardForm]
MakeExpression[RowBox[{x_, SubscriptBox["*", "i"], y_}], StandardForm] :=
MakeExpression[RowBox[{"conv", "[", x, ",", y, ",", "\"*\"", "]"}], StandardForm]

MakeExpression[RowBox[{x_, SubscriptBox["/", "i"], y_}], StandardForm] :=
MakeExpression[RowBox[{"conv", "[", x, ",", y, ",", "\"/\"", "]"}], StandardForm]

Testing the code, it works at least up to a point and the subscripted operators appear to have the same precedence as the parent operators (+,*, etc.):


test 1


However, if I remove the parentheses from the second test expression, I get an error message. Why is that occurring and how do I fix it?


test 2



Answer



The construction of custom operators with precedence is described in the Complex Patterns and Advanced Features Notation Package Tutorial. More general informations can be found in Precedence of Operators in Notations.


The following operator definitions work for the examples given in the question.

First loading the Notation Package:


Needs["Notation`"]

Now I'll add an InputAlias for each operator, although this increases the convenience only for $-_i$ significantly.


AddInputAlias["+i" -> ParsedBoxWrapper[SubscriptBox["+", "i"]]]

AddInputAlias["*i" -> ParsedBoxWrapper[SubscriptBox["*", "i"]]]

AddInputAlias["/i" -> ParsedBoxWrapper[SubscriptBox["/", "i"]]]


AddInputAlias["-i" -> ParsedBoxWrapper[
TagBox[SubscriptBox["-", "i"], Minus, SyntaxForm -> "a"]]]

Using the Notation Palette this looks like AddInputAlias First I'll define the infix operators with some not yet defined functions


InfixNotation[ParsedBoxWrapper[SubscriptBox["+", "i"]], myPlus]

InfixNotation[ParsedBoxWrapper[SubscriptBox["*", "i"]], myTimes]

InfixNotation[ParsedBoxWrapper[SubscriptBox["/", "i"]], myDivision]


InfixNotation[ParsedBoxWrapper[
TagBox[SubscriptBox["-", "i"], Minus, SyntaxForm -> "a"]], myMinus]

InfixNotation


For a first test, your testing code in a copyable format


\!\(\(9 \*TagBox[SubscriptBox["-", "i"],Minus,SyntaxForm->"a"] 2\*SubscriptBox[\(*\), \(i\)]3\)\*SubscriptBox[\(+\), \(i\)]\(4\*SubscriptBox[\(/\), \(i\)]2\)\)

\!\(4\*SubscriptBox[\(+\), \(i\)]\(6\*SubscriptBox[\(*\), \(i\)]2\*SubscriptBox[\(/\), \(i\)]4\)\)

which can be easily typed in using the InputAliases or the Subscript keyboard shortcut (Ctrl+_) for the ones without a TagBox and will look (together with their output) like



Test1


The output of the first test example nicely shows the correct parenthesization.


Finally the placeholder functions are defined as


myPlus = Plus;
myMinus[a_, b_] := Plus[a, Times[-1, b]]
myTimes[a_, b_] := Times[a, b]
myDivision = Divide;

and the examples tested again


\!\(\(9 \*TagBox[SubscriptBox["-", "i"],Minus,SyntaxForm->"a"] 2 \*SubscriptBox[\(*\), \(i\)]3\)\*SubscriptBox[\(+\), \(i\)]\(4 \*SubscriptBox[\(/\), \(i\)]2\)\)


\!\(4\*SubscriptBox[\(+\), \(i\)]\(6\*SubscriptBox[\(*\), \(i\)]2 \*SubscriptBox[\(/\), \(i\)]4\)\)

\!\(\((9 \*TagBox[SubscriptBox["-", "i"],Minus,SyntaxForm->"a"] 2 \*SubscriptBox[\(*\), \(i\)]3)\)\*SubscriptBox[\(+\), \(i\)]\(4 \*SubscriptBox[\(/\), \(i\)]2\)\)

Test2


As described in the tutorials linked to at the beginning of this answer, the SyntaxForm option of TagBox can be used to change the precedence of an operator. For the given examples this is only needed for $-_i$, because for the other operators it is determined by their corresponding operators without a subscript. For $-_i$ the precedence has to be changed, because otherwise everything following $-_i$ is parenthesized as one part. Using SyntaxForm -> "a" the precedence behavior is "group as a symbol".


Just to illustrate the wrong parenthesization, when the precedence behavior is "group as infix minus operator":


ClearAll[myPlus, myPlus, myDivision, myMinus]
RemoveInfixNotation[ParsedBoxWrapper[

TagBox[SubscriptBox["-", "i"], Minus, SyntaxForm -> "a"]], myMinus]

InfixNotation[ParsedBoxWrapper[
TagBox[SubscriptBox["-", "i"], Minus, SyntaxForm -> "a-b"]], myMinus]

\!\(9\*TagBox[SubscriptBox["-", "i"],Minus,SyntaxForm->"a-b"]\(2\*SubscriptBox[\(*\), \(i\)]3\)\*SubscriptBox[\(+\), \(i\)]\(4\*SubscriptBox[\(/\), \(i\)]2\)\)

wrongParenthesization


Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

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}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.