Skip to main content

replacement - Using PatternSequence on its own in ReplaceAll?



Say I have this list:


list = {0, 1, 2, 0, 1, 3, 0, 1, 4}

and say I want to remove consecutive sequences of 0 and 1. I'd normally go for this:


list //. {before___, 0, 1, after___} :> {before, after}


{2,3,4}

But a more concise way would be the following:



list /. PatternSequence[0, 1] -> Sequence[]

Alas, this doesn't do anything. The examples in the PatternSequence documentation are all of the form


list //. {before___, PatternSequence[0, 1], after___} :> {before, after}

which kind of defeat the purpose. (It's pretty useful if you want to name the PatternSequence on the LHS of the rule, but that's not my aim).


So the question is, is there a way to use constructs like PatternSequence[0, 1] -> Sequence[] in ReplaceAll?



Answer



You cannot do what you are trying to, unfortunately.


As Leonid says, PatternSequence is for grouping purposes. And in general the WL pattern matcher is not geared for "sequence-based patterns" for a whole variety of reasons. At a fundamental level the pattern matcher uses a 'cursor' that is always rooted at the level of an entire expression, and sequences are matched within that expression, rather than the 'cursor' itself being a sequence.



For example, expr //. {a___, 0, 1, b___} :> {a, b} takes O(n2) time in the length of the list, when it really could be O(n), if we had better ways of describing and using sequence-based patterns.


What we really need is one or two functions with the same semantics as the String* functions, except that they operate on lists of expressions rather than strings of characters.


I do think that PatternSequence could be re-used in this context.


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