Re: Replacing list elements while retaining structure

*To*: mathgroup at smc.vnet.net*Subject*: [mg73996] Re: Replacing list elements while retaining structure*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Mon, 5 Mar 2007 04:49:35 -0500 (EST)*References*: <esdj70$olc$1@smc.vnet.net>

Hi, with structurize[expr_, repl_] := Block[{replaceList = repl, doReplace}, doReplace[a_List] := doReplace /@ a; doReplace[a_] /; replaceList =!= {} := Module[{f}, f = First[replaceList]; replaceList = Rest[replaceList]; f]; doReplace[{a_List, b__}] := {doReplace[a], Sequence @@ doReplace[{b}]}; doReplace[a_] := a; doReplace[expr] ] structurize[{x, x, {{x}, x}, x}, {1, 2, 3, 4, 5}] gives {1, 2, {{3}, 4}, 5} Regards Jens D. Grady wrote: > Okay, so I have two lists, X and Y. X has some complex structure > which is important, but the values contained in X are not important. > Y has values that are important, but the structure of Y is not. I was > to sequentially replace all the elements in X with the elements of Y. > > It's assumed that X and Y have the same total number of elements, i.e. > Length[Flatten[X]] == Length[Flatten[Y]] > > Here is an example: > > In[8]:= > structurize[{x,x,{{x},x},x},{1,2,3,4,5}] > > Out[8]= > {1,2,{{3},4},5} > > This is what I have so far: > > structurize[X_List, Y_List] := ReplacePart[X, Flatten[Y], Position[X, > a_ /; \ > (Head[a] =!= List), Heads -> False], > Table[{i}, {i, 1, Length[ > Flatten[Y]]}]] /; Length[Flatten[X]] == Length[Flatten[Y]] > > This works fine so long as the elements of X are atomic expressions; > however, if there is an element of X which is a more complicated > expression, like x^2, then this function does not work as desired > because the pattern in Position[] matches x^2 as well as x and 2. Is > there a way to avoid matching parts of a subexpression? Is there a > better way to approach the problem from the get-go? > > Thanks in advance! > >