Re: Applying "Replace" to subsets of lists

*To*: mathgroup at smc.vnet.net*Subject*: [mg129192] Re: Applying "Replace" to subsets of lists*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Thu, 20 Dec 2012 03:24:21 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

On 12/19/12 at 4:55 AM, abed.alnaif at gmail.com wrote: >Hello, Say I have the following list, and I'd like to replace the >second 'b' with the value 1, leaving the first b untouched: >MagicFunction[{{b, 2}, b}] = {{b, 2}, 1} >How do I do this? I've tried the following: >This doesn't work since it replaces both 'b' In: {{b, 2}, b} /. b -> >1 Out: {{1, 2}, 1} >This works: In: {{b, 2}, b} /. {{x_, y_}, b} -> {{x, y}, 1} Out: >{{b, 2}, 1} >However, 'b' may appear in different forms, in which case the >previous approach fails: In: {{b, 2}, b^2} /. {{x_, y_}, b} -> {{x, >y}, 1} Out: {{b, 2}, b^2} Here is one approach In[5]:= {{b, 2}, b} /. a_ :> {First[a], Last[a] /. b -> 1} Out[5]= {{b,2},1} In[6]:= {{b, 2}, b - 1} /. a_ :> {First[a], Last[a] /. b -> 1} Out[6]= {{b,2},0} In[7]:= {{b, 2}, b^2} /. a_ :> {First[a], Last[a] /. b -> 1} Out[7]= {{b,2},1}