How to set the properties of a plot that's already been plotted?

Question

Let's say I have a function that I've plotted, like so:

a = Plot[f, {x, 0, 5}, Background -> Blue]

So I don't really have access to the function anymore, but I want to replot it with a changed property of the graph, for example, I want to make the color of the line Black, or the background Red.

How can I do that without replotting it, or not having access to the function again?

I know I can access the original color using Option[], like:

Last@Last@Options[a, Background]

But SetOptions and SetProperty don't really work. It seems like SetOptions sets the general property for plots in general, and SetProperty isn't for this. (SetProperty[a, Background -> Red] gives the error that my plot isn't an object with properties. SetOptions[a, Background -> Red] says that my argument plot isn't a symbol or stream.)

Is there a way to do this?

Answer 1

If you have the plot in Mathematica format (for instance saved in an unexecuted Mathematica notebook) but not the code that created it, then copy it into a new line of code, like

to obtain the same plot with a blue background. The point is, without the code to generate the plot, execute the instructions directly on the plot itself.

Addendum

It is natural to ask whether the approach give by kgir would work with the more complicate plot used in this answer.

The line styles are changed, but the background is not. The reason can be seen from

(* {} *)

Background does not appear in the internal code for the plot, so it cannot be changed in this way. Instead, Append the command to the plot. (Using Show, as above, also would work in this case.)

More generally, for complicated plots it is necessary to peer inside the plot by

and tailor the needed commands to change or add to the internal instructions of the plot. For instance, one might then use

which both removes the old line styling commands and adds the new ones, to obtain the plot just above.

Second Addendum

As requested in a comment below, the code used to create the original plot is

Plot[Flatten@Table[{MathieuCharacteristicA[r, q], 
    MathieuCharacteristicB[r + 1, q], -MathieuCharacteristicA[r, q], 
   -MathieuCharacteristicB[r + 1, q],}, {r, 0, 1}], {q, 0, 8}, 
   Evaluated -> True, PlotRange -> {All, {0, 4}}, 
   PlotStyle -> {Directive[Thick, Blue], Directive[Thick, Red], 
       Directive[Thick, Dashed, Blue], Directive[Thick, Dashed, Red]}, 
   Filling -> Table[{2 n + 1 -> {{2 n + 2}, Directive[Opacity[1/2], Purple]}}, {n, 0, 3}]]

which is based on a recent question.

Answer 2

Perhaps:

a = Plot[{x Sin[1/x], x Cos[1/x]}, {x, -1/2, 1/2}, Background -> Blue]

a /. {HoldPattern[Background->_] :> Rule[Background, Red], 
    ll_Line :> {Thick, Dashed, ll}}

Note: a /. {Rule[Background, _] :> Rule[Background, Red], ll_Line :> {Thick, Dashed, ll}} also works.

Update: If the input plot does not have an explicit Background option, as in

b = Plot[{x Sin[1/x], x Cos[1/x]}, {x, -1/2, 1/2}]

then the replacement rule above would not work. You could use

Show[b, Background->Red]/. { ll_Line :> {Thick, Dashed, ll}}

to get

See this answer by Mr.Wizard for methods to modify plot styles in more general ways.