RE: Replacement Rule

*To*: mathgroup at smc.vnet.net*Subject*: [mg32075] RE: [mg32064] Replacement Rule*From*: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.de>*Date*: Sat, 22 Dec 2001 04:22:48 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

> -----Original Message----- > From: bghiggins at ucdavis.edu [mailto:bghiggins at ucdavis.edu] To: mathgroup at smc.vnet.net > Sent: Friday, December 21, 2001 9:58 AM > To: mathgroup at smc.vnet.net > Subject: [mg32075] [mg32064] Replacement Rule > > > I have the following list > > l1={{1,2},{3,4},{5,6}}; > > and then I operate on it with the following pattern/replacement rule > > In[3]:= l1 /. {x_, y_} -> x > > Out[3]= {1, 3, 5} > > This is what I would expect. Now consider the following list > > In[4]:= l2 = {{3, 4}, {5, 6}}; > > Then I use the same pattern/replacement rule > > In[5]:=l2 /. {x_, y_} -> x > > Out[5]= {3, 4} > > I was hoping to get > > {3,5} > > What am I missing? The FullForm of l2 is basically the same structure > as l1, as far as I can tell.... > > Brian > There is no mystery here, but a match you didn't reckon with. To see all possible matches look at In[2]:= Position[{{3, 4}, {5, 6}}, {x_, _}] Out[2]= {{1}, {2}, {}} at positions 1 and 2 of the list (the ones you liked), plus one additional, indicated by {} which means the total expression. If you compare with In[3]:= Position[{{3, 4}, {5, 6}}, {x_Integer, _}] Out[3]= {{1}, {2}} the pattern now cannot math the total expression any longer. These are the ways out of your dilemma: be more specific with your patter, or direct the matching process. Some examples In[4]:= {{3, 4}, {5, 6}} /. {x_Integer, _} :> x Out[4]= {3, 5} In[5]:= anotherHead[{3, 4}, {5, 6}] /. {x_, _} :> x /. anotherHead -> List Out[5]= {3, 5} In[7]:= Replace[{{3, 4}, {5, 6}}, {x_, _} :> x, {1}] Out[7]= {3, 5} compare with In[6]:= (# /. {x_, _} :> x &) /@ {{3, 4}, {5, 6}} Out[6]= {3, 5} In[8]:= Replace[{{3, 4}, {5, 6}}, {x_, _} :> x, Infinity] Out[8]= {3, 5} However In[9]:= Replace[{{3, 4}, {5, 6}}, {x_, _} :> x, {0, Infinity}] Out[9]= 3 compare with In[10]:= Replace[#, {x_, _} :> x] & //@ {{3, 4}, {5, 6}} Out[10]= 3 So we might say ReplaceAll works top down, Replace bottom up. Make shure you understand these cases. Hartmut