Re: Insulating data from code

*To*: mathgroup at smc.vnet.net*Subject*: [mg66595] Re: Insulating data from code*From*: "Pratik Desai" <pratikd at wolfram.com>*Date*: Sun, 21 May 2006 22:30:41 -0400 (EDT)*References*: <e4ekai$9av$1@smc.vnet.net><e4jtnn$d02$1@smc.vnet.net> <e4mn63$6o7$1@smc.vnet.net> <e4or6i$6cr$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Peter Pein wrote: > Paul Abbott schrieb: > > In article <e4jtnn$d02$1 at smc.vnet.net>, Peter Pein <petsie at dordos.net> > > wrote: > > > >> Ray Koopman schrieb: > >>> >From time to time I've wanted to partition the first level of one > >>> list, say A, the same way that another list, say B, is partitioned. > >>> One way to do this is > >>> > >>> copyPartition[A_List, B_List] /; Length@A >= Length@Flatten@B := > >>> Module[{i = 0}, Map[A[[++i]]&,B,{-1}]] > >>> > >>> But all the methods I've thought of have a pointer that functions > >>> something like i in the above code. I'd like to eliminate the pointer, > >>> because in the unlikely event that A contains an unevaluated symbol > >>> that is the same as the name of the pointer with $ appended -- e.g., > >>> i$, if the pointer is i -- then in the returned list that symbol will > >>> have a numeric value assigned to it. Unique[i] doesn't help. The > >>> only solution I see is the probabilistic one of giving the pointer a > >>> strange (random?) name that hopefully would be very unlikely to show > >>> up as data. But that would be giving up. Does anyone have any ideas? > >>> > >> Hi Ray, > >> > >> use Replace[]: > >> > >> A={a,b,c,d,e}; > >> B={{1},{2,3},{{{4}},5}}; > >> > >> Ap=B/.Thread[Flatten[B]\[Rule]A] > >> --> {{a},{b,c},{{{d}},e}} > > > > No, that won't work. Try > > > > A={a,b,c,d,a}; > > B={{1},{2,1},{{{3}},2}}; > > > > Ap=B/.Thread[Flatten[B] -> A] > > > > You get > > > > {{a}, {b, a}, {{{d}}, b}} > > > > whereas I think the OP wanted > > > > {{a}, {b, c}, {{{d}}, a}} > > > > Cheers, > > Paul > > > > _______________________________________________________________________ > > Paul Abbott Phone: 61 8 6488 2734 > > School of Physics, M013 Fax: +61 8 6488 1014 > > The University of Western Australia (CRICOS Provider No 00126G) > > AUSTRALIA http://physics.uwa.edu.au/~paul > > > Hi Paul, > > well, I recognized this and came to a solution similar to J. Siehler's: > > A={a,{b,{c,{d,{e}}}},f}; > B={x,{x},{{x,x},x},x}; > > cpStruct=ReplacePart[##, > Sequence@@(Position[#,_,{-1},Heads->False]&/@{##})]&; > > cpStruct[B,A] > --> {a,{b},{{c,d},e},f} > > Peter Maybe something like this ?? In[1]:= Clear[list,skeletor] A={a,b,c,d,e}; B={{1},{3,5},{{{5}},5}}; skeletor[list1_? ListQ,list2_?ListQ] /;Length@list2=Length@Flatten@list1:=list1/.{ x_?NumberQ\[RuleDelayed]Hold[Part[list1,x]]}/.{list1\[RuleDelayed]list2} skeletor[B,A]//ReleaseHold Out[5]= {{a},{c,e},{{{e}},e}} Pratik Desai Wolfram Research, Inc Technical Support