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Applying "Replace" to subsets of lists

• To: mathgroup at smc.vnet.net
• Subject: [mg129161] Applying "Replace" to subsets of lists
• From: abed.alnaif at gmail.com
• Date: Wed, 19 Dec 2012 04:55:27 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• Delivered-to: l-mathgroup@wolfram.com
• Delivered-to: mathgroup-newout@smc.vnet.net
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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

• Follow-Ups:
  • Re: Applying "Replace" to subsets of lists
    • From: Bob Hanlon <hanlonr357@gmail.com>
  • Re: Applying "Replace" to subsets of lists
    • From: Fred Simons <f.h.simons@tue.nl>