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Re: Passing arguments and pattern matching

  • To: mathgroup at
  • Subject: [mg33828] Re: [mg33792] Passing arguments and pattern matching
  • From: BobHanlon at
  • Date: Fri, 19 Apr 2002 02:28:02 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

In a message dated 4/16/02 7:37:53 AM, mark.coleman at writes:

>I have a question on pattern matching that I have been struggling with.
>First I'll describe the issue broadly, and then attempt to illustrate it
>with a specific example. My intution is that is a relatively simple answer,
>but I have been unable to find it.
>I would like to write a general function that accepts these two lists as
>arguments, i.e., myPattern[a_List,b_List]. The function would act something
>like the built-in Mathematica Postion[] pattern function and return the
>positions of any elements of b that occur in List a. This function will
>be called by other functions, so the user will not know in advance the
>values of lists a or b. In my attempts to build this function I have used
>the Position[] function.  The difficulty is that Position treats the 
>I pass it as 'literal', and not as references to what the arguement 
>For instance, let's say I define two lists as follows:
>   a = {x, Log[x], y, x y, Log[y] };
>   b = {x};
>Now say I wish to compute the position of x or various functions of x in
>a, then of course
>In[24]:= Position[a,x]
>Out[24]= {{1},{2,1},{4,1}}
>In[26]:= Position[a,Log[x]]
>Out[26]= {{2}}
>and so on. But now assume I wish to define a function
>   myPattern[lista_List,listb_List] := Position[lista,listsb]
>If I call myPatten[a,b], this is the same as executing
>Out[27]= {}
>My question is, how do I have Mathematica treat the second argument not
>as a literal ("b"), but as representing a the value of the underlying 
>For instance
>  myPattern[a,{Log[x]}] would return  {{2}}

This may be what you want.

myPattern[lista_List,listb_List]:=Position[lista,#]& /@ listb;

a={x,Log[x],y,x y,Log[y]};


{{{1}, {2, 1}, {4, 1}}}


{{{1}, {2, 1}, {4, 1}}, {{2}}, {{3}, {4, 2}, {5, 1}}, {{4}}, {{5}}}

Bob Hanlon
Chantilly, VA  USA

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