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MathGroup Archive 2004

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Re: conditionnal rule

  • To: mathgroup at smc.vnet.net
  • Subject: [mg51536] Re: conditionnal rule
  • From: Curt Fischer <crf3 at po.cwru.edu>
  • Date: Thu, 21 Oct 2004 22:21:01 -0400 (EDT)
  • References: <cl4t4i$8ee$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Wishmaster7 wrote:
> hi all !!
> 
> I want to apply a rule under a certain condition. here is the example : 
> 
> myList = {{a, b}, {c, d}};
> 
> myList //. {a,x_}->{a,d}
> 
> apply this rule if Length[x]===1 (this means that x can not be a list)

Here is one solution to your present example.  You will also want to 
investigate the Condition operator /; and the pattern test operator ?. 
Note that in Mathematica, a list can have length one, or zero, so your 
Length criterion is not the same as testing whether something is a list 
or not.

In[1]:=
?ListQ

ListQ[expr] gives True if expr is a list, and False otherwise.

In[2]:=
?Not

!expr is the logical NOT function. It gives False if expr is \
True, and True if it is False.

In[3]:=
myList={{a,b},{c,d}}

Out[3]=
{{a,b},{c,d}}

In[4]:=
myRule={a,x_?(Not[ListQ[#]]&)}\[Rule] {a,d}

Out[4]=
{a,x_?(!ListQ[#1]&)}\[Rule]{a,d}

In[5]:=
myList//.myRule

Out[5]=
{{a,d},{c,d}}


> I tried something like : 
> 
> myList //. {a_,x_}->{c,d}/;Length[x]===1
> 
> or
> 
> myList //. {a_,x_?Length[x]===1}->{c,d}/
> 
> but it does not work.
> 
> can someone help me ?
> 


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