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Re: Replacing list elements while retaining structure

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73990] Re: [mg73967] Replacing list elements while retaining structure
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Mon, 5 Mar 2007 04:46:15 -0500 (EST)
  • References: <200703040450.XAA25252@smc.vnet.net>

On 4 Mar 2007, at 05:50, 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!
>
>


Since it does not matter what happens to your values in X you can  
simply replace them first by some atomic expression. A simple way to  
do that is to multiply X by 0; it will produce a message if X  
contains any "infinite" symbols like Ininity, ComplexInfinity, but  
will work anyway. After that you can eiter use your method or  
something like this:

structurize[values_, structure_] := Module[{st = structure*0, i = 0, g},
    g[_] := values[[++i]]; Map[g, st, {-1}]]

In[2]:=
structurize[Range[5],{x,x,{{x^2},x},x}]

Out[2]=
{1,2,{{3},4},5}

Andrzej Kozlowski


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