Skip to main content

Pattern-rewriting Pattern triggers error?



I am trying to rewrite patterns. For example, I'd like to rewrite the pattern


jFoo_Integer

which has FullForm


Pattern[jFoo,Blank[Integer]]

into just


Blank[Integer]

In other words, I want to strip out the names of patterns. I tried the following



jFoo_Integer /. {Pattern[nym_, Blank[typ_]] :> {nym, typ}}

which does not match or reduce and produces the (IMO bogus) error message


Pattern::patvar: First element in pattern Pattern[nym_,Blank[typ_]] is not a valid pattern name. >>

I also tried


    Pattern[jFoo,Blank[Integer]] /. {Pattern[nym_, Blank[typ_]] :> {nym, typ}}
Pattern[jFoo,Blank[Integer]] /. {Verbatim[Pattern][nym_, Blank[typ_]] :> {nym, typ}}
jFoo_Integer /. {Verbatim[Pattern][nym_, Blank[typ_]] :> {nym, typ}}
jFoo_Integer /. {Verbatim[Pattern][nym_, Blank[typ_]] :> {nym, typ}}


all with exactly the same (failed) results.


Any hints, please & thanks?



Answer



What you want is probably


jFoo_Integer /. Verbatim[Pattern][nym_, Verbatim[Blank][typ_]] :> {nym, typ}
(* {jFoo, Integer} *)

The usage of Verbatim points it out




Verbatim[expr] represents expr in pattern matching, requiring that expr be matched exactly as it appears, with no substitutions for blanks or other transformations.



Comments

Popular posts from this blog

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 - 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 - 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....