Re: [Maxima-discuss] substitution problem

[Eli D. <pu...@el...>]

That's exactly what I want. Thank you. Eli

On 7/7/20 3:02 AM, Stavros Macrakis wrote:
> I think you want
>
>        block ( [ n:3 ], ... )
>
> not sublis.
>
>
> On Tue, Jul 7, 2020, 02:32 Eli Damon <pu...@el...> wrote:
>
>> Hi, Robert. Thanks. I feel stupid now. Maybe it's because I was writing
>> R code recently, where you have "quoting functions".
>>
>> Anyway, I was just using print to demonstrate. The function I was
>> actually using was makelist. I was trying to do something like
>>
>> sublis([ n = a.bunch.of.stuff ], makelist(
>> a.bunch.of.stuff.with.an.n.in.it, i, 0, n-1 ));
>>
>> and I was getting an error because n-1 didn't evaluate to a number. I
>> suppose I could use a temporary variable or a utility function, but this
>> seemed like the tidiest way to do it.
>>
>> Eli
>>
>> On 7/7/20 12:53 AM, Robert Dodier wrote:
>>
>>> On Mon, Jul 6, 2020 at 8:43 PM Eli Damon <pu...@el...> wrote:
>>>
>>>> (%i1) sublis([n=3],print("f(n)","=",f(n)));
>>>> f(n) = f(n)
>>>> (%o1) f(3)
>>> Eli, it turns out this is exactly the behavior one should expect.
>>> sublist is an ordinary function, i.e. one which evaluates its
>>> arguments. So print("f(n)","=",f(n)) is evaluated, and it evaluates to
>>> f(n), and has the side effect of printing "f(n) = f(n)". Now sublis is
>>> called with arguments [n = 3] and f(n), so it returns f(3), which you
>>> see as %o1.
>>>
>>> I guess that you want to print an expression without evaluating it and
>>> also print it evaluated. The built-in functions display and ldisplay
>>> are a little bit like that, but not exactly. Here is an attempt to do
>>> just that.
>>>
>>> mydisplay (e, [vv]) ::= buildq ([e, e1: sublis (vv, e)], print ('e =
>>> 'e1));
>>>
>>> It's a little delicate to get just the right amount of evaluation ...
>>> note that mydisplay is a macro, defined by ::= , which is a function
>>> which quotes its arguments, and its return value is evaluated by the
>>> caller. Note also the presence of single quotes in print('e = 'e1),
>>> that prevents e and e1 from being evaluated by the caller. Here's what
>>> I get for the above input.
>>>
>>> (%i35) mydisplay (f(n), n = 3);
>>> f(n) = f(3)
>>> (%o35) f(n) = f(3)
>>>
>>> mydisplay collects any arguments aside from e and puts them all in the
>>> list vv.
>>>
>>> (%i36) mydisplay (g(m, n), n = a, m = 2*a);
>>> g(m, n) = g(2 a, a)
>>> (%o36) g(m, n) = g(2 a, a)
>>>
>>> I don't know whether mydisplay should print e = e1 or just return it.
>>> I guess which one is more useful depends on what you need to do. Maybe
>>> you can say more about what your goal is.
>>>
>>> There are probably other variations which are useful in some context.
>>> E.g. f(n) = f(3) = 1234 or whatever depending on the definition of f.
>>>
>>> Hope this helps,
>>>
>>> Robert Dodier
>>
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