Re: Efficient Replacement Rules to Matrix?

*To*: mathgroup at smc.vnet.net*Subject*: [mg21779] Re: [mg21743] Efficient Replacement Rules to Matrix?*From*: "Tomas Garza" <tgarza at mail.internet.com.mx>*Date*: Thu, 27 Jan 2000 22:57:01 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Roger Jones [rmj at leland.stanford.edu] wrote: > What is the most efficient (in terms of time) method to transform a set > of replacement rules to a matrix. For example, I have: > > matrix = ZeroMatrix[5]; > repmat = {{1, 1} -> 4., {5, 5} -> 3,{4, 4} -> 10,{2, 2} -> 2 + I 6, {3, > 3} -> 40.}; > > and I transfor to a matrix thus: > > matrix = ReplacePart[matrix, Sequence @@ #]) & /@ ( > {Last[#], #[[1]]} & /@ matrix); > > But for large matrices this is quite slow! Is there a more efficient > method? Perhaps there are a couple of misprints in the expression for matrix above? I guess (correct me if I'm wrong) it should read matrix = (ReplacePart[matrix, Sequence @@ #]) & /@ ( {Last[#], #[[1]]} & /@ repmat); Notice the left parenthesis before ReplacePart, and repmat instead of matrix at the end of the expression. If so, I think the problem is - partly - that you end up with a huge matrix of dimensions {5, 5, 5}. From this, one has to select those elements which really correspond to the desired matrix. I would suggest a slightly different approach: In[1]:= f[m_, j_] := ReplacePart[m, Sequence[repmat[[j, 2]]], repmat[[j, 1]]] In[2]:= matrix = Fold[f, matrix, {1, 2, 3, 4, 5}] Out[2]= {{4., 0, 0, 0, 0}, {0, 2 + 6 I, 0, 0, 0}, {0, 0, 40., 0, 0}, {0, 0, 0, 10, 0}, {0, 0, 0, 0, 3}} No doubt this is much faster (I was too lazy to perform actual numeric experiments with larger matrices), since it always remains within the {5, 5} dimensions of the original matrix. Tomas Garza Mexico City