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


magicFunction[list_List, part_, rule_Rule] := ReplacePart[list, part > (list[[part]] /. rule)]
magicFunction[{{b, 2}, b}, 2, b > 1]
{{b, 2}, 1}
magicFunction[{{b, 2}, b^2}, 2, b > 1]
{{b, 2}, 1}
magicFunction[{{b, 2}, b^2}, 1, b > 1]
{{1, 2}, b^2}
Bob Hanlon
On Wed, Dec 19, 2012 at 4:55 AM, <abed.alnaif@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 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 >

