Re: Replacement question
- To: mathgroup at smc.vnet.net
- Subject: [mg35143] Re: [mg35114] Replacement question
- From: Andrzej Kozlowski <andrzej at lineone.net>
- Date: Tue, 25 Jun 2002 19:55:57 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
How about something like:
In[1]:=
h[x_] := x^3; f[x_] := x^4;
In[2]:=
h[3]/f[2]
Out[2]=
27/16
In[3]:=
Block[{h, f}, In[2] /. h[3] -> h[5]]
Out[3]=
125/16
Andrzej Kozlowski
Toyama International University
JAPAN
htt
On Tuesday, June 25, 2002, at 04:41 PM, 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.)
>
> Thanks!
>
> G.
>
>
>
>
p://platon.c.u-tokyo.ac.jp/andrzej/