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Topic: Applying "Replace" to subsets of lists
Replies: 2   Last Post: Dec 20, 2012 3:19 AM

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 Bob Hanlon Posts: 906 Registered: 10/29/11
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
>

Date Subject Author
12/20/12 Bob Hanlon
12/20/12 Simons, F.H.