4
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I can set the Options for Plot as:

SetOptions[
  Plot, 
  AxesStyle -> 
 {{Black, 14, Italic, FontFamily -> "Times"}, 
 {Black, 14, Italic, FontFamily -> "Times"}}];

But it also changes the style on the labels themselves (which I want in a different style), e.g., "x" and "y" in this example:

Plot[Cos[x],
 {x, -5, 5},
 AxesLabel -> {"x", "y"}]

How can I set the Options for the added labels to be separate from the rest of the AxesStyle? As an example, suppose I want the "x" and "y" to be Red, Bold, and 28 points?

Of course I can hand code those styles each time into the AxesLabel specifications, but I'd like to do it with Options.

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7
  • $\begingroup$ Add the options FontColor -> Red, FontWeight -> Bold when you set AxesStyle? $\endgroup$
    – kglr
    Dec 19, 2019 at 2:41
  • $\begingroup$ @kglr: Nope. Doesn't work: It makes all the text red. And note that I want the size of the axes ticks to be different from the "x" and "y" labels too. $\endgroup$ Dec 19, 2019 at 2:43
  • $\begingroup$ sorry; i see... $\endgroup$
    – kglr
    Dec 19, 2019 at 2:45
  • $\begingroup$ maybe add the option TicksStyle -> FontColor -> Black to override the FontColor ->Red in set in AxesStyle? $\endgroup$
    – kglr
    Dec 19, 2019 at 2:49
  • $\begingroup$ @kglr: You set me on the right track; override was the key. SetOptions[Plot, Sequence[ AxesStyle -> {{28, Italic, FontFamily -> "Times", FontColor -> Red}, {28, Italic, FontFamily -> "Times", FontColor -> Red}}], TicksStyle -> {{14, FontColor -> Black}, {14, FontColor -> Black}}]; Post that and I'll accept it. $\endgroup$ Dec 19, 2019 at 2:53

1 Answer 1

4
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You can override the font settings given in AxesStyle using TicksStyle:

SetOptions[Plot, 
  {AxesStyle -> Directive[28, Italic, FontFamily -> "Times", FontColor -> Red], 
   TicksStyle -> Directive[14, FontColor -> Black]}]

Plot[Sin[x], {x, 0, 2 Pi}, AxesLabel -> {"X", "Y"}]

enter image description here

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1
  • 2
    $\begingroup$ I really do dislike how TicksStyle is misnamed... it's really more like TickLabelStyle. $\endgroup$
    – b3m2a1
    Dec 19, 2019 at 8:32

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