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Re: Finding variables in a very long expression
*To*: mathgroup at smc.vnet.net
*Subject*: [mg31410] Re: Finding variables in a very long expression
*From*: "Allan Hayes" <hay at haystack.demon.co.uk>
*Date*: Sat, 3 Nov 2001 05:29:16 -0500 (EST)
*References*: <9rq7ia$ill$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Ted:
You say
>If I had my way removing a symbol would make the kernel work as if the
symbol
> never had any values, but that's not the way it works.
Some thoughts on this:
We see Remove apparently wrapped round occurences of removed symbols in
stored values
a = A[b];
Remove[b];
a
A[Removed[b]]
Something like this is needed to preserve the stucture of the value, and to
indicate the removal.
But this is more than just wrapping
FullForm[%]
A[Removed["b"]]
AtomQ[%]
False
and replacement does not work:
a/."b"->3
A[Removed[b]]
Moreover, if I simply type and evaluate the result is not the same:
AtomQ[Removed["b"]]
False
So something more is going on that I haven't sorted out yet.
b does not now exist
?b
Information::notfound: Symbol b not found.
And we can re-create and set a value to b:
b= 5
5
without affecting the value of a:
a
A[Removed[b]]
If we had just "make the kernel work as if the symbol never had any values"
(which would seem to be what ClearAll[b] would have done) then this would
not have been been possible.
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
"Ersek, Ted R" <ErsekTR at navair.navy.mil> wrote in message
news:9rq7ia$ill$1 at smc.vnet.net...
> Jose Flanigan wanted to know how to find all variables in an expression.
> Consider expr below.
>
> In[1]:=
> expr=Sqrt[x*y]+x*z+w Sqrt[x]z^(1/n)+Pi/3;
> w=v;
>
> The simple line below almost works, but the list includes Pi which isn't a
> variable. The third argument of Cases here is a level specification. You
> can read about Cases and level specification at my web site (URL below).
>
> In[3]:=
> Union[Cases[expr, _Symbol, {0, -1}]]
>
> Out[3]=
> {n, Pi, v, x, y, z}
>
>
> The next line eliminates symbols such as Pi.
>
> In[4]:=
> Union[Cases[expr, _Symbol?( !NumericQ[#]& ), {0, -1} ]]
>
> Out[4]=
> {n, v, x, y, z}
>
>
> --- The Plot Thickens ---
> At the 1999 Developer Conference Robby Villegas explained that this sort
of
> thing gets tricky when you are working with symbols that were previously
> removed. Consider the lines below where the symbol (v) is removed. If I
> had my way removing a symbol would make the kernel work as if the symbol
> never had any values, but that's not the way it works.
>
> In[5]:=
> Remove[v]
>
> In[6]:=
> Union[Cases[expr, _Symbol?( !NumericQ[#]& ), {0, -1} ]]
>
> Out[6]=
> {n, Removed[v], x, y, z}
>
>
> If you want to make sure your list of variables is free of Removed Symbols
> use the next line (based on Robby Villegas talk).
>
> In[7]:=
> Union[Cases[expr, _Symbol?( !NumericQ[#] && NameQ[ToString[#]]& ), {0,
> -1} ]]
>
> Out[7]=
> {b, x, y, z}
>
>
> ---
> Regards,
> Ted Ersek
> Check Mathematica Tips, Tricks at
> http://www.verbeia.com/mathematica/tips/Tricks.html
>
>
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