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RE: ? about Rule (rewritten)

  • To: mathgroup at
  • Subject: [mg67212] RE: [mg67153] ? about Rule (rewritten)
  • From: "David Park" <djmp at>
  • Date: Sun, 11 Jun 2006 23:08:55 -0400 (EDT)
  • Sender: owner-wri-mathgroup at


On my system, Version 5.2 it appears to work.

{y == z, t == 7, 1 <= x <= 2, u >= -4} /.
  (a_) <= (w_) <= (b_) -> (a + b)/2

{y == z, t == 7, 3/2, u >= -4}

David Park
djmp at

From: jackgoldberg at [mailto:jackgoldberg at]
To: mathgroup at


My post on this topic was garbled.  Here I hope is a clearer version :-).

I want to replace an inequality of the type  1 <=  x <= 2 which appears in a
list some of whose entries are  y==3 or z >= 1 and so forth.  (This arose in
using Reduce on a set of linear inequalities.)  My life is simpler if  1 <=
x <=2  is replace by an interior point of the interval, say  x ==1.5.   It
has frustrated me for about a week since I thought I knew the ins and outs
of Rule and ReplaceAll.  So, 2 questions.  The first and most important is
how to do it (keep in mind that the list I am speaking of is constructed by
programming and I do not want to examine it and make the replacement that
way).  The second question is, why doesn't the following technique work?
   (1)       {y==z,  t==7, 1 <=x<=2, u>=-4}/. (a_<=w_<=b_)->(a+b)/2
To save space, I choose not to write all my variations of trying this
including the obvious one of finding the FullForm of  1 <=x<=2  and using it
in  (1).  I might mention also that every now and then, in isolation from my
module, I have managed to make it work, not often but enough to delay my
plea for help.  I am guessing that there is a pattern matching issue here
(sort of obvious comment) but I am at a loss anyway.

Incidentally I have designed a work around that is so combersome that if
there were a prize for the klutzest Mathematica code, this would be a
candidate for top prize.


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