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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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 Simons, F.H. Posts: 107 Registered: 12/7/04
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 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
>

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