On 04/07/2015 8:21 AM, David Winsemius wrote:
>
>> On Jul 3, 2015, at 11:05 PM, Duncan Murdoch <murdoch.dun...@gmail.com> wrote:
>>
>> On 04/07/2015 3:45 AM, David Winsemius wrote:
>>>
>>>> On Jul 3, 2015, at 5:08 PM, David Winsemius <dwinsem...@comcast.net> wrote:
>>>>
>>>>
>>>>
>>>> It doesn’t appear to me that mpfr was ever designed to handle expressions
>>>> as the first argument.
>>>
>>> This could be a start. Obviously one would wnat to include code to do other
>>> substitutions probably using the all.vars function to pull out the other
>>> “constants” and ’numeric’ values to make them of equivalent precision. I’m
>>> guessing you want to follow the parse-tree and then testing the numbers for
>>> integer-ness and then replacing by paste0( “mpfr(“, val, “L, “, prec,”)” )
>>>
>>> Pre <- function(expr, prec){ sexpr <- deparse(substitute(expr) )
>>
>> Why deparse? That's almost never a good idea. I can't try your code (I
>> don't have mpfr available), but it would be much better to modify the
>> expression than the text representation of it. For example, I think
>> your code would modify strings containing "pi", or variables with those
>> letters in them, etc. If you used substitute(expr) without the
>> deparse(), you could replace the symbol "pi" with the call to the Const
>> function, and be more robust.
>>
>
> Really? I did try. I was fairly sure that someone could do better but I
> don’t see an open path along the lines you suggest. I’m pretty sure I tried
> `substitute(expr, list(pi= pi))` when `expr` had been the formal argument and
> got disappointed because there is no `pi` in the expression `expr`. I
> _thought_ the problem was that `substitute` does not evaluate its first
> argument, but I do admit to be pretty much of a klutz with this sort of
> programming. I don’t think you need to have mpfr installed in order to
> demonstrate this.
The substitute() function really does two different things.
substitute(expr) (with no second argument) grabs the underlying
expression out of a promise. substitute(expr, list(pi = pi)) tries to
make the substitution in the expression "expr", so it doesn't see "pi".
This should work:
do.call(substitute, list(expr = substitute(expr), env=list(pi =
Const(pi, 120))))
(but I can't evaluate the Const function to test it).
This works:
do.call(substitute, list(expr = substitute(expr), env = list(pi =
quote(Const(pi, 120)))))
because it never evaluates the Const function, it just returns some code
that you could modify more if you liked.
Duncan Murdoch
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