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Re: Replacement question
*To*: mathgroup at smc.vnet.net
*Subject*: [mg35132] Re: [mg35114] Replacement question
*From*: Sseziwa Mukasa <mukasa at jeol.com>
*Date*: Tue, 25 Jun 2002 19:55:06 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
On Tuesday, June 25, 2002, at 03:41 AM, ginak wrote:
> Suppose I have something like
>
> In [100] := h[3]/f[2.33342]
>
> Out[100] := 24.12711
>
> and now I want to evaluate h[5]/f[2.33342], i.e. the same as in
> In[100] but replaceing h[3] with h[5]. This won't work
>
> In [101] := In[100] /. h[3]->h[5]
>
> Out[101] := 24.12711
>
> because In[100] is fully evaluated to 24.12711 before the rule is
> applied. (Generally, the expressions I'm interested in are more of a
> pain to type than h[5]/f[2.33342]). How do I tell Mathematica to
> evaluate
> In[100] only enough to apply the given substition rule, apply the
> substitution rule, and only then proceed with the evaluation?
>
> In fact, I don't even know how to do a replacement like
>
> In [102] := h[3]/f[2.33342] /. h[3]->h[5]
>
> for the same reason: the LHS is evaluated before the rule can be
> applied. (Of course, in this case this replacment task is pointless,
> since it is so easy to type out the desired expression, but there are
> situations in which one can obtain a complicated expression by cutting
> and pasting, and wants to apply a substitution rule to the complicated
> expression before Mathematica evaluates it.)
>
>
As far as I can tell the two situations are slightly different. In the
case of the expression In[100]/.h[3]->h[5], In[100] evaluates using the
value of h[3] before the rule gets applied, so the expression that the
rule is being applied to is not h[3]/f[2.33342] but the value of that
expression. The following block removes the downvalues of h so that
h[3] does not evaluate, then restores the downvalues and evaluates the
resulting expression:
Block[{dv=DownValues[h]},h[x_]=.;expr=In[401]/.h
[3]->h[5];DownValues[h]=dv;expr]
In the other case you simply need to use HoldPattern to prevent the left
hand side of the rule from evaluating
h[3]/f[2.33342]/.HoldPattern[h[3]]->h[5]
There may be a simpler way to handle the first case but I can't think of
it right now.
Regards,
Ssezi
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