Skip to main content

pattern matching - Orderless and Sequence


I just ran into the following subtlety.


Let us consider a function f with attribute Orderless.


Attributes[f]={Orderless};

For pattern matching, the consequence of this attribute is that when we have an expression with head f, any ordering of the arguments is tested. That can be seen in the following result:



ReplaceList[ f[a,b,c], f[x_,y_, z_] :> {x,y,z} ]

(* {{a,b,c},{a,c,b},{b,a,c},{b,c,a},{c,a,b},{c,b,a}} *)

I would have expected the same result from the next command, where in the rule I catch the three arguments of f in a BlankSequence, thereby placing a Sequence expression in the list at the right hand side:


 ReplaceList[ f[a,b,c], f[x__] :> {x} ]

(* {{a,b,c}} *)

It only gives one result! Likely, I overlooked something simple, but I fail to see a good explanation. Why does this not work?




Answer



Here is how I make sense of this behavior. When a function that appears in a pattern has attribute Orderless, the pattern-matcher must generate all possible permutations of its argument sequence before trying to match these patterns.


Refer to a simple example expression such as a /. b -> c: in a nutshell, as Fred mentioned in his comment below, I contend that the attribute Orderless causes the system to generate possible alternatives for the b expression, rather than for a.


When the argument sequence of your orderless f function contains more than one argument, then multiple permutations are generated. The specification f[x_, y_, z_] -> {x, y, z} in the second argument of ReplaceList can be thought of as equivalent to the following "expanded form":


{f[x_, y_, z_] -> {x, y, z}, f[x_, z_, y_] -> {x, y, z}, f[y_, x_, z_] -> {x, y, z}, 
f[y_, z_, x_] -> {x, y, z}, f[z_, x_, y_] -> {x, y, z}, f[z_, y_, x_] -> {x, y, z}}

Each one of those patterns matches f[a, b, c] in the first argument of ReplaceList, hence the multiple results.


However, when the pattern specified in the second argument of ReplaceList contains only one argument, then there are no permutations to account for, so only one "equivalent pattern" is considered, which matches once.





To clarify my point, here is a helper function that approximates my vision of what the pattern matcher is doing for orderless functions. Note that here we use a regular, non-orderless g function, and simulate orderless behavior explicitly.


Clear[generateOrderlessPatterns]
Attributes[g] = {};

generateOrderlessPatterns[functiontoapply_, list_, patterntype_] :=
Table[
functiontoapply[Sequence @@ (Pattern[#, patterntype] & /@ i)] -> list,
{i, Permutations[list]}
]


We can then generate "orderless-style" patterns for the non-orderless g function:


generateOrderlessPatterns[g, {x, y, z}, Blank[]]

(* Out:
{g[x_, y_, z_] -> {x, y, z}, g[x_, z_, y_] -> {x, y, z}, g[y_, x_, z_] -> {x, y, z},
g[y_, z_, x_] -> {x, y, z}, g[z_, x_, y_] -> {x, y, z}, g[z_, y_, x_] -> {x, y, z}}
*)

On the other hand, if we use a BlankSequence pattern, we obtain:


generateOrderlessPatterns[g, {x}, BlankSequence[]]


(* Out: {g[x__] -> {x}} *)

Using these patterns in ReplaceList emulates the Orderless behavior of f:


ReplaceList[g[a, b, c], generateOrderlessPatterns[g, {x, y, z}, Blank[]]]

(* Out:
{{a, b, c}, {a, c, b}, {b, a, c}, {c, a, b}, {b, c, a}, {c, b, a}}
*)


ReplaceList[g[a, b, c], generateOrderlessPatterns[g, {x}, BlankSequence[]]]

(* Out: {{a, b, c}} *)

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....