```Date: Dec 20, 2012 3:19 AM
Author: Simons, F.H.
Subject: Re: Applying "Replace" to subsets of lists

Since you want to replace the second b, no matter on what level it occurs, I think you need a counter:In[3]:= n=0; {{b,2},b} /. {b :> (n=n+1;If[n==2,1,b])}Out[3]= {{b,2},1}In[4]:= n = 0; {{b, 2},  3 + b + b^2} /. {b :> (n = n + 1; If[n == 2, 1, b])}Out[4]= {{b, 2}, 4 + b^2}Regards,Fred SimonsEindhoven University of TechnologyOp 19-12-2012 10:55, abed.alnaif@gmail.com schreef:> 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 doesn't work (I'm not sure why):> In: {{b, 2}, b} /. {{x_, y_}, f_[b]} -> {{x, y}, 1, f[1]}> Out: {{b, 2}, b}>> 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}>> Using the 'levelspec' argument of 'Replace' also fails since 'b' can appear in different forms:> In: Replace[{{b, 2}, b^2}, b -> 1, 1]> Out: {{b, 2}, b^2}> In: Replace[{{b, 2}, b^2}, b -> 1, 2]> Out: {{1, 2}, 1}>> Thank you,>> Abed>
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