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I wish to show 20 circles in a vertical rectangular whose m circles are filled. n circles for example colored by green and m-n colored by red. The radius of circles are the same and can be r. The circles are not filled are empty. How do I reach this aim? I wish to specify the position of green circles and red circles manually. For example: the first be green, the second be red, the third be green and the fourth, the remainder be empty. AND NOT Random.

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    $\begingroup$ You can do that with a combination of Disk and Graphics. $\endgroup$
    – anderstood
    Commented Feb 19, 2018 at 16:43

2 Answers 2

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Here's another way:

colors = {Green, Red, Green, Sequence @@ ConstantArray[White, 17]};
r = 
  Table[Rectangle[{m, n}, RoundingRadius -> .5], {m, 1}, {n, 20}];

Graphics[{EdgeForm[Blue], Riffle[Reverse@colors, Flatten@r]},
  Frame -> True, FrameStyle -> Directive[Blue, Bold], 
 FrameTicks -> None]

enter image description here

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  • $\begingroup$ You just need to specify your desired order in colors. I was doing something similar with Rectangles, so that's why I use them here instead of explicit circles (RoundingRadius->0.5 of a Rectangle[] equals a circle). $\endgroup$
    – TumbiSapichu
    Commented Feb 19, 2018 at 23:21
  • $\begingroup$ Thank you so much, your approach is very near to what I was searching, $\endgroup$
    – Unbelievable
    Commented Feb 20, 2018 at 5:45
  • $\begingroup$ @Irreversible, glad I could help! $\endgroup$
    – TumbiSapichu
    Commented Feb 20, 2018 at 7:29
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Here's one way:

GraphicsColumn[Table[Graphics[{RandomChoice[{Green, Red, White}],
     Disk[{i, 0}, 0.4], Black, Circle[{i, 0}, 0.4]}], {i, 20}]]

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  • $\begingroup$ Thank you so much. But I wish to specify the position of green circles and red circles manually. For example: the first be green, the second be red, the third be green and the fourth, the remainder be empty. AND NOT Random. $\endgroup$
    – Unbelievable
    Commented Feb 19, 2018 at 19:37
  • $\begingroup$ @Irreversible What have you tried and where did you get stuck? This answer shows the basics. You can just as easily pull the colour from another list indexed by i. $\endgroup$
    – Szabolcs
    Commented Feb 19, 2018 at 20:34
  • $\begingroup$ @Szabolcs, I have exploited GraphicsColumn[{Graphics[Disk[]], Graphics[Disk[]], Graphics[Disk[]]}, (which is in the help Mathematica) Frame -> All], But unfortunately I had to create twenty copies of the line for specifying the position of colored circles. $\endgroup$
    – Unbelievable
    Commented Feb 20, 2018 at 5:44

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