How to plot an integer with a specific color for each digit?

Question

I want to plot something like this, the first 100,000 digits of Pi:

... in a rectangular or circular form.

Answer 1

Before asking for help you should always try your own solution and post the code that you have tried (see docs and EIWL), even if you cannot finish it, it's a good etiquette :-) (please do that in future). But you are new user and this problem have a nice minimal solution in Wolfram Language, so I post my take on it.

Your colors look like Hue (or HSB), I will use that and transfer to RGB:

colors = ColorConvert[Hue /@ Range[0, .9, .1], "RGB"]

Build rules for replacement of a digit by color:

rules = Dispatch[Thread[Range[0, 9] -> List @@@ colors]]

Build image. You are basically building a matrix partitioning Pi digits list into a matrix, and then replacing in that matrix digits by RGB values. And that is the array structure of an Image which is efficient in this case, I think, more than Graphics. You can partition at a different than 400 width to achieve different aspect ratio.

Image[Partition[First[RealDigits[Pi, 10, 100000]], 400] /. rules]

Answer 2

You mentioned plotting the digits of Pi in a "circular form". Perhaps a "sunflower" plot is of interest. Digits are plotted in a spiral from the centre outwards, and coloured by the chosen colour scheme. As the rational approximation to Pi gets better, that is, as exponent k increases, patterns in the plot disappear. A visual confirmation of the randomness of the digits.

This function plots a sunflower with each disk having the property accorded by the function f.

SunflowerPlot[n_, c_, d_, phi_, f_, cs_String] :=
   Graphics[
      Table[{
         ColorData[cs, f[k]],
         Disk[c Sqrt[k] {Cos[k phi Degree], Sin[k phi Degree]}, d]},
         {k, 1, n}]]

The plot may be manipulated with the following.

Manipulate[
   SunflowerPlot[n, c, d, phi,
                 ((RealDigits[Rationalize[Pi, 10.^-k], 10, 2500][[1]][[#]]*0.1)^e &),
                 cs],
   {{n, 1000, "Number of Florets"}, 10, 2500, 10, Appearance -> "Labeled"},
   {{c, 0.2, "Scaling Parameter"}, 0.1, 0.5, Appearance -> "Labeled"},
   {{d, 0.2, "Floret Radius"}, 0.1, 0.5, Appearance -> "Labeled"},
   {{k, 5., "Rationalize Exponent"}, 1., 15., Appearance -> "Labeled"},
   {{phi, 137.50776405, "Spiral Angle"}, 10, 360, Appearance -> "Labeled"},
   {{cs, "SolarColors", "Colour Scheme"}, ColorData["Gradients"]},
   {{e, 1.0, "Colour Exponent"}, 0.1, 3.0, Appearance -> "Labeled"}]

Answer 3

You can also use MatrixPlot:

legend = Graphics[{Hue[#/10], Disk[{#, 0}, .5], Black, 
      Text[Style[#, 16], {#, 0}]} & /@ Range[0, 9], ImageSize -> 300];
MatrixPlot[Partition[First[RealDigits[Pi, 10, 100000]], 400], 
 ColorFunction -> ( Hue[#/10] &), ColorFunctionScaling -> False, 
 Frame -> False, ImageSize -> 2 -> 3,  PlotLegends -> Placed[legend, Below]]

Alternatively, use an array of colors as the first argument of MatrixPlot:

MatrixPlot[Map[Hue[#/10] &, Partition[First[RealDigits[Pi, 10, 100000]], 400], {-1}],
 Frame -> False, ImageSize -> 2 -> 3, 
 PlotLegends -> Placed[legend, Below]]

same picture

Answer 4

Small 10 x 5 version showing the first 50 digits of Pi

colours = Transpose@Reverse@Partition[Characters[
      "3" <> StringDrop[ToString[N[Pi, 50]], 2]], 10];
insets = Table[Inset[Style[Text[colours[[i + 1, j + 1]]],
     Black, FontSize -> 20],
    {10 i, 10 j}, {0, 0}, {10, 10}], {i, 0, 10 - 1}, {j, 0, 5 - 1}];
Graphics[{Yellow, Rectangle[{0, 0}, {100, 50}], insets}]

Same again with colours

colours = Transpose@Reverse@Partition[Characters[
       "3" <> StringDrop[ToString[N[Pi, 50]], 2]] /. Thread[
       ToString /@ Range[0, 9] -> ColorData[3, "ColorList"]], 10];
insets = Table[Inset[Graphics[{colours[[i + 1, j + 1]],
      Rectangle[{0, 0}, {10, 10}]},
     PlotRange -> {{0, 10}, {0, 10}}, ImageSize -> 10],
    {10 i, 10 j}, {0, 0}, {10, 10}], {i, 0, 10 - 1}, {j, 0, 5 - 1}];
Graphics[{Yellow, Rectangle[{0, 0}, {100, 50}],
  insets}, PlotRange -> {{0, 100}, {0, 50}}]

Similar version showing the first 5000 characters of Pi

colours = Transpose@Reverse@Partition[Characters[
       "3" <> StringDrop[ToString[N[Pi, 5000]], 2]] /. Thread[
       ToString /@ Range[0, 9] -> ColorData[3, "ColorList"]], 100];
insets = Table[Inset[Graphics[{colours[[i + 1, j + 1]],
      Rectangle[{0, 0}, {10, 10}]},
     PlotRange -> {{0, 10}, {0, 10}}, ImageSize -> 10],
    {10 i, 10 j}, {0, 0}, {10, 10}], {i, 0, 100 - 1}, {j, 0, 50 - 1}];
Graphics[{Yellow, Rectangle[{0, 0}, {1000, 500}],
  insets}, PlotRange -> {{0, 1000}, {0, 500}}]