我在Mathematica中有點新意,我的代碼中有很多appendto,我認爲它們佔用了一些時間。我知道還有其他一些優化方法,但我無法真正知道如何實現。我認爲getBucketShocks可以提高很多?任何人?Mathematica優化在循環中追加表格
getBucketShocks[BucketPivots_,BucketShock_,parallelOffset_:0]:=
Module[{shocks,pivotsNb},
shocks={};
pivotsNb=Length[BucketPivots];
If[pivotsNb>1,
AppendTo[shocks,LinearFunction[{0,BucketShock},{BucketPivots[[1]],BucketShock},{BucketPivots[[2]],0},BucketPivots[[2]],0},parallelOffset]];
Do[AppendTo[shocks,LinearFunction[{BucketPivots[[i-1]],0},{BucketPivots[[i]],BucketShock},{BucketPivots[[i+1]],0},{BucketPivots[[i+1]],0},parallelOffset]],{i,2,pivotsNb-1}];
AppendTo[shocks,LinearFunction[{BucketPivots[[pivotsNb-1]],0},{BucketPivots[[pivotsNb]],BucketShock},{BucketPivots[[pivotsNb]],BucketShock},{BucketPivots[[pivotsNb]],BucketShock},parallelOffset]],
If[pivotsNb==1,AppendTo[shocks,BucketShock+parallelOffset&]];
];
shocks];
LinearInterpolation[x_,{x1_,y1_},{x2_,y2_},parallelOffset_:0]:=parallelOffset+y1+(y2-y1)/(x2-x1)*(x-x1);
LinearFunction[p1_,p2_,p3_,p4_,parallelOffset_:0]:=Which[
#<=p1[[1]],parallelOffset+p1[[2]],
#<=p2[[1]],LinearInterpolation[#,p1,p2,parallelOffset],
#<=p3[[1]],LinearInterpolation[#,p2,p3,parallelOffset],
#<=p4[[1]],LinearInterpolation[#,p3,p4,parallelOffset],
#>p4[[1]],parallelOffset+p4[[2]]]&;
這裏有幾個想法http://stackoverflow.com/q/39599232/1004168。 – agentp