Re: ReplaceAll feature
- To: mathgroup at yoda.physics.unc.edu
- Subject: Re: ReplaceAll feature
- From: withoff
- Date: Thu, 8 Oct 92 10:21:05 CDT
> In the following, why is Out[1] different from Out[2] and Out[3]:
>
> Mathematica 2.0 for SGI Iris
> Copyright 1988-91 Wolfram Research, Inc.
> -- Terminal graphics initialized --
>
> In[1]:= (a+b+c+d)/.{a+b->x,c+d->y}
>
> Out[1]= c + d + x
>
> In[2]:= (a+b+c+d)/.a+b->x/.c+d->y
>
> Out[2]= x + y
>
> In[3]:= (a+b+c+d)//.{a+b->x,c+d->y}
>
> Out[3]= x + y
>
> I guess it has something to do with Flat. This is not a big problem, but IF
> it can be easily fixed, it should be fixed. But maybe there are some "deep"
> reasons why it behaves like this ?
> -------------------------------------------------------------------
> Pekka Janhunen tel (+358) 0 1929 535
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Yes, this does have something to do with Flat.
ReplaceAll[e, rules] (or e /. rules) stops making replacements in
a particular subexpression as soon as a replacement causes the
subexpression to change.
ReplaceAll starts here with the entire expression, a+b+c+d, and applies
the first rule, a+b->x. Since Plus has attribute Flat, the pattern
matcher looks for combinations of terms in the sum that match the pattern
a+b. When a match is found, application of the rule causes the expression
to change, so ReplaceAll stops without ever looking at the second rule:
In[10]:= a+b+c+d /. {a+b->x, c+d->y}
Out[10]= c + d + x
If there are no matches with the entire expression for either rule,
ReplaceAll starts looking at subexpressions. In the example below,
the first rule (a+b->x) matches the first subexpression (a+b), and no
further rules are applied to that subexpression. The second rule
(c+d->y) is applied in the same manner to the second subexpression:
In[11]:= {a+b, c+d} /. {a+b->x, c+d->y}
Out[11]= {x, y}
If Replace all is nested, as in ReplaceAll[ReplaceAll[e, r1], r2]
(or e /. r1 /. r2), the inner evaluation, ReplaceAll[e, r1], applies
the first rule, and the outer evaluation applies the second.
Dave Withoff
withoff at wri.com