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Re: Interaction of Remove and Global variables in a Module

It depends what you really want to do. If you want to clear all the 
definitions involving just use  Clear[Subscript]


In[1]:= Subscript[a,0]=1;Subscript[a,n_]:=Subscript[a,n]=2 
Subscript[a,n-1];
In[2]:= Subscript[a,2]
Out[2]= 4

In[3]:= DownValues[Subscript]

Out[3]= {HoldPattern[Subscript[a, 0]]:>1,HoldPattern[Subscript[a, 
1]]:>2,HoldPattern[Subscript[a, 2]]:>4,HoldPattern[Subscript[a, 
n_]]:>(Subscript[a, n]=2 Subscript[a, n-1])}

Clear[Subscript]

In[5]:= DownValues[Subscript]
Out[5]= {}


Al cleared. The caveat is that all DownValues associated with Subscript 
will be cleared, not just those involving a, and you may not wish to do 
that. In that case you need to proceed in a different way:


In[1]:= Subscript[a, 0] = 1;
Subscript[a, n_] := Subscript[a, n] = 2 Subscript[a, n - 1];

In[2]:= Subscript[a, 2]

Out[2]= 4

In[3]:= DownValues[Subscript]
Out[3]= {HoldPattern[Subscript[a, 0]]:>1,HoldPattern[Subscript[a, 
1]]:>2,HoldPattern[Subscript[a, 2]]:>4,HoldPattern[Subscript[a, 
n_]]:>(Subscript[a, n]=2 Subscript[a, n-1])}

Now:

Subscript[a, i_] =.

In[5]:= DownValues[Subscript]
Out[5]= {HoldPattern[Subscript[a, 0]]:>1,HoldPattern[Subscript[a, 
1]]:>2,HoldPattern[Subscript[a, 2]]:>4}

In[5]:= Do[Subscript[a, i] =., {i, 0, 2}]

In[6]:= DownValues[Subscript]

Out[6]= {}

So we all the values have been cleared.

On the other hand, the method you use (Remove) only seems to work (which 
may be all you want in practice):

In[1]:= Subscript[a, 0] = 1;
Subscript[a, n_] := Subscript[a, n] = 2 Subscript[a, n - 1];

In[2]:= Subscript[a, 2]

Out[2]= 4

Remove[a]

In[4]:= DownValues[Subscript]

Out[4]= {HoldPattern[Subscript[Removed[a], 
0]]:>1,HoldPattern[Subscript[Removed[a], 
1]]:>2,HoldPattern[Subscript[Removed[a], 
2]]:>4,HoldPattern[Subscript[Removed[a], n_]]:>(Subscript[Removed[a], 
n]=2 Subscript[Removed[a], n-1])}


They are all still there, just that a got replaced by Removed[a], so

In[5]:= Subscript[Removed[a],2]
Out[5]= Subscript[Removed[a], 2]


Andrzej Kozlowski










On 12 Jul 2011, at 12:59, Gianluca Gorni wrote:

>
> Suppose I define a recursive sequence
>
> a[0] = 1; a[n_] := a[n] = 2 a[n - 1]
>
> If I evaluate a[2], then its value will be kept in memory.
> When I need to clear and reuse the symbol a, I use Clear[a].
>
> Suppose now that I do it more traditional-looking:
>
> Subscript[a, 0] = 1; Subscript[a, n_] :=
> Subscript[a, n] = 2 Subscript[a, n - 1];
>
> If I call Subscript[a, 2], its value will be remembered.
> What can I do to clear the values? Clear[a] and ClearAll[a]
> do not work. Remove[a] does work.
>
> Is there a better alternative?
>
> Best regards,
> Gianluca
>
> On 09/lug/2011, at 13.32, DrMajorBob wrote:
>
>> So far, I haven't heard a purpose for Remove that ClearAll does not 
serve
>> just as well... without the side-effect of "spoiling" other symbols.
>>
>> Bobby
>>
>> On Fri, 08 Jul 2011 03:53:55 -0500, Leonid Shifrin <lshifr at gmail.com>
>> wrote:
>>
>>> Brian,
>>>
>>> The problem you encountered is actually more subtle. Remember that when
>>> you
>>> use Remove (as compared to Clear or ClearAll), the symbol is completely
>>> removed
>>> from the system. This is a pretty disruptive operation. Now, what should
>>> the system do if the symbol you are removing is referenced by some other
>>> symbols?
>>> Keeping it there unchanged would mean that the symbol has not been
>>> really removed
>>> from the system. The solution used in Mathematica is to change a
>>> reference to a
>>> symbol
>>> (say "a") to Removed[a]. In practice, this means that, even when you
>>> re-introduce
>>> the symbol, those definitions that were involving it are still "spoiled"
>>> and
>>> can not
>>> be used.  IMO, this is a very sensible design, but this is what leads to
>>> the
>>> behavior
>>> that puzzled you. Have a look:
>>>
>>> f[b_]:=Module[{t},a[t_]=b*t^2;]
>>> Remove[a];
>>> ?f
>>>
>>> Global`f
>>> f[b_]:=Module[{t},Removed[a][t_]=b t^2;]
>>>
>>> What you have to do in your approach is to re-run the definition for
>>> "f", to
>>> be able to
>>> use it. This is however pretty error-prone. You can cure this by calling
>>> Clear or ClearAll
>>> instead of Remove, but even this approach I don't consider a good
>>> practice.
>>> Even though
>>> you use "a"  as a function, calling Clear or Remove on it means that in a
>>> way, you use
>>> it as a global variable. In this post:
>>>
>>> 
> 
http://stackoverflow.com/questions/6236458/plot-using-with-versus-plot-using-block-mathematica/6236808#6236808
>>>
>>> there is a lengthy discussion why making implicit dependencies on global
>>> variables is a
>>> bad practice.
>>>
>>> Here are a few ways out. What you seem to want is to generate a function
>>> with embedded
>>> parameters (a closure). One way is to generate a pure function and return
>>> it:
>>>
>>> In[31]:= Clear[f];
>>> f[b_]:=Function[t,b*t^2];
>>> a = f[3];
>>> a[t]
>>> a=f[4];
>>> a[t]
>>> Remove[a];
>>> a = f[5];
>>> a[t]
>>>
>>> Out[34]= 3 t^2
>>> Out[36]= 4 t^2
>>> Out[39]= 5 t^2
>>>
>>> Another way is to explicitly pass to "f"  the symbol to which you want to
>>> assign
>>> the definition:
>>>
>>> In[40]:= Clear[ff,a];
>>> ff[fname_Symbol,b_]:=Module[{t},fname[t_]=b*t^2;];
>>> ff[a,3];
>>> a[t]
>>> ff[a,4];
>>> a[t]
>>> Remove[a];
>>> ff[a,5];
>>> a[t]
>>>
>>> Out[43]= 3 t^2
>>> Out[45]= 4 t^2
>>> Out[48]= 5 t^2
>>>
>>> By making the function name an explicit parameter to "ff", you make the
>>> problematic
>>> situation above impossible to happen.
>>>
>>> HTH
>>>
>>> Regards,
>>> Leonid
>>>
>>>
>>>
>>>
>>>
>>> On Thu, Jul 7, 2011 at 3:28 PM, blamm64 <blamm64 at charter.net> wrote:
>>>
>>>> This is what I get for querying SetDelayed
>>>>
>>>> In[1]:= ?SetDelayed
>>>> lhs:=rhs assigns rhs to be the delayed value of lhs. rhs is maintained
>>>> in an unevaluated form. When lhs appears, it is replaced by rhs,
>>>> evaluated afresh each time.  >>
>>>>
>>>> Note particularly the above reads AFRESH EACH time.  It appears then
>>>> the following is inconsistent behavior based on the above description:
>>>>
>>>> In[2]:= f[b_]:=Module[{t},a[t_]=b*t^2;]
>>>> In[3]:= a[t]
>>>> Out[3]= a[t]
>>>> In[4]:= f[3]
>>>> In[5]:= a[t]//InputForm
>>>> Out[5]//InputForm=
>>>> 3*t^2
>>>> In[6]:= f[5]
>>>> In[7]:= a[t]//InputForm
>>>> Out[7]//InputForm=
>>>> 5*t^2
>>>> In[8]:= Remove[a]
>>>> In[9]:= f[4]
>>>> In[10]:= a[t]//InputForm
>>>> Out[10]//InputForm=
>>>> a[t]
>>>>
>>>> Apparently AFRESH is not an accurate description of how SetDelayed
>>>> operates in this case, or I am missing something about this particular
>>>> interaction of Module, Remove, and global variables inside Modules.
>>>>
>>>> However, if I go back, after executing the last line above (<a> has
>>>> been Removed), and place the cursor in the input line where <f> is
>>>> defined and hit Enter, which I thought would be identical to just
>>>> evaluating <f> AFRESH again, and then execute the <f[4]> line again,
>>>> then the global <a> definition is re-constituted.
>>>>
>>>> The documentation for Remove reads the name is no longer recognized by
>>>> Mathematica.  My understanding is that if the same name is defined
>>>> AFRESH, it will once again be recognized.
>>>>
>>>> So if anyone would let me know what I am missing, regarding why the
>>>> definition of <a> is not created AFRESH each time <f> is evaluated, I
>>>> would appreciate it.
>>>>
>>>> Please don't construe the definition of <f> as my way of
>>>> 'parameterizing' a function definition, I just use that definition to
>>>> convey the apparent inconsistency.
>>>>
>>>> -Brian L.
>>>>
>>>>
>>
>>
>> --
>> DrMajorBob at yahoo.com
>>
>
>
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