
Re: Applying "Replace" to subsets of lists
Posted:
Dec 20, 2012 3:19 AM


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 Simons Eindhoven University of Technology
Op 19122012 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 >

