6
$\begingroup$

I'm confused with Mathematica's way of parsing expressions. I've been struggling with this for a while and never found an exhaustive answer, sometimes things don't parse the way I think they would and I don't really understand why.

As an example, with Mathematica 8:

(* A works *) Manipulate[ Plot[(y/a) /. {y -> (x - b)}, {x, 0, 2}, PlotRange -> {0, 1}], {a, 1, 2}, {b, 0, 1}] (* B doesn't work *) Manipulate[ Plot[(x/a) /. {x -> (x - b)}, {x, 0, 2}, PlotRange -> {0, 1}], {a, 1, 2}, {b, 0, 1}] (* C works *) (x/a) /. {x -> (x - b)} Manipulate[ Plot[%, {x, 0, 2}, PlotRange -> {0, 1}], {a, 1, 2}, {b, 0, 1}] (* D doesn't work *) test = (x/a) /. {x -> (x - b)} Manipulate[ Plot[test, {x, 0, 2}, PlotRange -> {0, 1}] , {a, 1, 2}, {b, 0, 1}] (* E works *) test2[x_, a_, b_] = (x/a) /. {x -> (x - b)} Manipulate[ Plot[test2[x, a, b], {x, 0, 2}, PlotRange -> {0, 1}], {a, 1, 2}, {b, 0, 1}]

Case A works, so the substitution is performed fine inside Plot and Manipulate.

But B doesn't, which I could understand as an issue trying to substitute a variable with an expression containing itself, but then, if you evaluate it beforehand, as in C, everything works again, so it has to be a problem with x being part of Plot, I guess.

Then if you assign the result of the substitution to a variable, you can't directly plot it, so it seems that variables are not evaluated if they are not functions (as in E, with pattern matching) inside Plot. But % is, so % is "special" as it gets evaluated inside plot while a standard symbol assigned to a value does not.

Can someone explain me all this? I guess it's related to the Hold attributes a function can have?

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
4
$\begingroup$

Plot has Attributes HoldAll, so one possibility to get what you expect is to do just

SetOptions[Plot, Evaluated -> True];

at the beginning.

Another possibility (better documented) would be to use Evaluate inside Plot:

Plot[Evaluate[...], {x, 0, 2}]
Improve this answer
$\endgroup$
7
  • $\begingroup$ I see, but why A works? I guess it's because when Plot comes to Evaluate the expression it first replaces x with a numeric value, and that replacement goes "inside" my /.{x->x-b} which then starts making very little sense... Right? $\endgroup$ Commented Feb 12, 2013 at 1:14
  • $\begingroup$ Also, why % behaves in that "special" way, i.e. see C versus D $\endgroup$ Commented Feb 12, 2013 at 1:15
  • $\begingroup$ @user61865. % is a shortcut for a function call. Evaluate FullForm@Hold@%; you will get Hold[Out[]]. $\endgroup$ Commented Feb 12, 2013 at 2:00
  • $\begingroup$ "Evaluate inside Plot" better documented?! Now that's an understatement if I ever read one. The use of Evaluated ->True is essentially undocumented. $\endgroup$ Commented Feb 12, 2013 at 2:05
  • 1
    $\begingroup$ @user61865. Your last comment is really another question. Don't ask a new question in a comment. Post it in a new question. $\endgroup$ Commented Feb 12, 2013 at 15:29

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.