Re: How to localize a symbol when using Manipulate?
- To: mathgroup at smc.vnet.net
- Subject: [mg112620] Re: How to localize a symbol when using Manipulate?
- From: Docente Sergio Miguel Terrazas Porras <sterraza at uacj.mx>
- Date: Thu, 23 Sep 2010 04:22:19 -0400 (EDT)
Hi Nasser
I am not one of the experts in the group, but I think that the Manipulate should be inside Module, not the other way around.
Module[{var1,var2,var3},
Manipulate[ ,{ }]
]
is supposed to make the variables var1, var2, var3 etc. be local variables.
Am I correct?
Sergio Terrazas
________________________________________
De: Nasser M. Abbasi [nma at 12000.org]
Enviado el: martes, 21 de septiembre de 2010 11:58 p.m.
Para: mathgroup at smc.vnet.net
Asunto: [mg112602] How to localize a symbol when using Manipulate?
Mathematica experts:
I came up against this problem when writing a demo.
I simplified to it to the following:
I want to make a call to a lower level function which makes an
expression (say build a polynomial in s) and returns it back to be
displayed.
I want this function to use a symbol, say s, to build the polynomial.
The problem is how to localize the symbol itself so it is not global.
Since one dose not declare things in Mathematica, (sometime I wish I
could) I can't declare the s to be a symbol, but must just use it.
Here is a simple example to show the problem:
If one does not pass a symbol for a lower level function during the call
to be used in constructing an expression, then the expression will have
$nnnn added to it.
Manipulate[process[n], {n, 2, 10},
Initialization :>
(process[n_] :== Module[{s}, s^n])
]
If you run the above, the result shown will be s$nnn^2 where nnn is
some number.
To solve the above problem, typical solution is to pass the symbol name
to be used. Example
Manipulate[process[s, n], {n, 2, 10},
Initialization :>
(process[s_, n_] :== Module[{}, s^n])
]
If you run the above, the result shown will be s$^2
There is a problem with the above solution. s in the above is global
variable. Hence s needs to be localized. Since 's' is not a control
variable (like 'n' is in this example), it needs to be explicitly
localized (control variables are localized by default).
The problem is how to localize 's'?
One can not write the following, since 's' becomes Null
Manipulate[process[s, n], {n, 2, 10},
{s, ControlType -> None},
Initialization :> (process[s_, n_] :== Module[{}, s^n])
]
If you run the above, the result shown will be Null^2
I also can't use the Symbol[] trick
Manipulate[process[Symbol["s"], n], {n, 2, 10},
Initialization :> (
process[s_, n_] :== Module[{}, s^n])
]
One way is not not pass 's' at all, and on return, strip the $$$n from
the symbol, as follows, However using ToExpression is not allowed in a
demo (for same reason as Symbol is not allowed, security). (someone
helped me long time ago with providing the following nice strip function
here in this newsgroup)
Manipulate[ strip[process[n]], {n, 2, 10},
Initialization :> (
process[n_] :== Module[{s}, s^n];
strip[expr_] :==
Module[{}, ToExpression[StringReplace[ToString[expr,
FormatType -> TraditionalForm],
c:LetterCharacter~~"$"~~DigitCharacter.. :> c]]]; )
]
I have to use ToExpression, else the result will be String, which makes
it hard to use later on, but is OK for just printing and display.
Any one has other ideas I could try?
thanks
--Nasser