Skip to main content

performance tuning - MapThread and Compile


After discovering that MapIndexed cannot be used with compile (see MapIndexed and Compile) I am now trying do implement similar functionality using a combination of MapThread and Map. Unfortunately I am running into problems even when only using MapThread.


Here is a test-scenario:


Create Lists first:


A = {{1, 2}, {2, 3}};
B = {{{1, 2, 3}, {2, 3, 4}}, {{2, 3, 4}, {3, 4, 5}}};


The I simply run MapThread with function List[] to obtain a suitable combination of those lists and put the whole thing inside a Compile-statement:


test = Compile[{{B, _Real, 3}, {A, _Real, 2}}, MapThread[List[#1, #2, #3] &, {B, A, A}, 2]];

But when running test[B,A] I get the following compile error:



CompiledFunction::cfex: Could not complete external evaluation at instruction 15; proceeding with uncompiled evaluation.



When using CompilePrint[test] instruction 15 shows a call to MainEvaluate.


Unfortunately I currently have no clue how to fix that problem. Any hint is appreciated.




Answer



Incorporating the comments into an answer:



I don't think you are allowed to have lists of that irregular shape, i.e. {{1, 2, 3}, 1, 1}. Note that it works as expected with Plus[#1,#2,#3]&. See for instance Compile[{}, Block[{a = {1, 2}, b = 3}, {a, b}]][] – ssch Oct 15 '13 at 14:20


indeed, Compile only works with rectangular arrays of consistent type -- because otherwise they cannot be packed. Arrays in Compile are packed arrays, and ragged arrays destroy packing, and hence, compilation. – Andreas Lauschke Oct 15 '13 at 14:39



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