Mixing
How to plot an integer with a specific color for each digit?
Clash Royale CLAN TAG#URR8PPP
up vote
8
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I want to plot something like this, the first 100,000 digits of Pi: 
... in a rectangular or circular form.
plotting
edited Sep 5 at 18:41
Johu
2,636829
asked Sep 5 at 18:30
mhd.math
1434
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mhd.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
8
down vote
favorite
I want to plot something like this, the first 100,000 digits of Pi:
... in a rectangular or circular form.
plotting
edited Sep 5 at 18:41
Johu
2,636829
asked Sep 5 at 18:30
mhd.math
1434
New contributor
mhd.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Welcome to Mathematica.SE! You should watch the introductory Tour to get the basics of the site.
– Vitaliy Kaurov
Sep 5 at 19:21
@VitaliyKaurov Thank You for the welcoming and your answer, I appreciate your efforts.
– mhd.math
Sep 5 at 19:25
For plotting PI, the circular form should be logical. But it would be interesting to find out if such a plot for other irrational numbers (e, sqrt(2)) pr just random numbers, would look different
– Roland
Sep 6 at 11:02
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
I want to plot something like this, the first 100,000 digits of Pi:
... in a rectangular or circular form.
plotting
edited Sep 5 at 18:41
Johu
2,636829
asked Sep 5 at 18:30
mhd.math
1434
New contributor
mhd.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I want to plot something like this, the first 100,000 digits of Pi:
... in a rectangular or circular form.
plotting
edited Sep 5 at 18:41
Johu
2,636829
asked Sep 5 at 18:30
mhd.math
1434
New contributor
mhd.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 5 at 18:41
Johu
2,636829
edited Sep 5 at 18:41
Johu
2,636829
edited Sep 5 at 18:41
Johu
2,636829
2,636829
asked Sep 5 at 18:30
mhd.math
1434
New contributor
mhd.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 5 at 18:30
mhd.math
1434
asked Sep 5 at 18:30
mhd.math
1434
1434
New contributor
mhd.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mhd.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mhd.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Welcome to Mathematica.SE! You should watch the introductory Tour to get the basics of the site.
– Vitaliy Kaurov
Sep 5 at 19:21
@VitaliyKaurov Thank You for the welcoming and your answer, I appreciate your efforts.
– mhd.math
Sep 5 at 19:25
For plotting PI, the circular form should be logical. But it would be interesting to find out if such a plot for other irrational numbers (e, sqrt(2)) pr just random numbers, would look different
– Roland
Sep 6 at 11:02
add a comment |Â
1
Welcome to Mathematica.SE! You should watch the introductory Tour to get the basics of the site.
– Vitaliy Kaurov
Sep 5 at 19:21
@VitaliyKaurov Thank You for the welcoming and your answer, I appreciate your efforts.
– mhd.math
Sep 5 at 19:25
For plotting PI, the circular form should be logical. But it would be interesting to find out if such a plot for other irrational numbers (e, sqrt(2)) pr just random numbers, would look different
– Roland
Sep 6 at 11:02
1
1
Welcome to Mathematica.SE! You should watch the introductory Tour to get the basics of the site.
– Vitaliy Kaurov
Sep 5 at 19:21
Welcome to Mathematica.SE! You should watch the introductory Tour to get the basics of the site.
– Vitaliy Kaurov
Sep 5 at 19:21
@VitaliyKaurov Thank You for the welcoming and your answer, I appreciate your efforts.
– mhd.math
Sep 5 at 19:25
@VitaliyKaurov Thank You for the welcoming and your answer, I appreciate your efforts.
– mhd.math
Sep 5 at 19:25
For plotting PI, the circular form should be logical. But it would be interesting to find out if such a plot for other irrational numbers (e, sqrt(2)) pr just random numbers, would look different
– Roland
Sep 6 at 11:02
For plotting PI, the circular form should be logical. But it would be interesting to find out if such a plot for other irrational numbers (e, sqrt(2)) pr just random numbers, would look different
– Roland
Sep 6 at 11:02
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
9
down vote
accepted
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]
answered Sep 5 at 19:04
Vitaliy Kaurov
55.4k6155269
add a comment |Â
up vote
8
down vote
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"]
answered Sep 6 at 3:20
KennyColnago
11.6k1753
add a comment |Â
up vote
6
down vote
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
answered Sep 5 at 20:03
kglr
159k8183383
add a comment |Â
up vote
4
down vote
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]
answered Sep 5 at 19:32
Chris Degnen
21.4k23281
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
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]
answered Sep 5 at 19:04
Vitaliy Kaurov
55.4k6155269
add a comment |Â
up vote
9
down vote
accepted
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]
answered Sep 5 at 19:04
Vitaliy Kaurov
55.4k6155269
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
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]
answered Sep 5 at 19:04
Vitaliy Kaurov
55.4k6155269
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]
answered Sep 5 at 19:04
Vitaliy Kaurov
55.4k6155269
edited Sep 5 at 19:16
answered Sep 5 at 19:04
Vitaliy Kaurov
55.4k6155269
answered Sep 5 at 19:04
Vitaliy Kaurov
55.4k6155269
answered Sep 5 at 19:04
Vitaliy Kaurov
55.4k6155269
55.4k6155269
add a comment |Â
add a comment |Â
up vote
8
down vote
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"]
answered Sep 6 at 3:20
KennyColnago
11.6k1753
add a comment |Â
up vote
8
down vote
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"]
answered Sep 6 at 3:20
KennyColnago
11.6k1753
add a comment |Â
up vote
8
down vote
up vote
8
down vote
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"]
answered Sep 6 at 3:20
KennyColnago
11.6k1753
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"]
answered Sep 6 at 3:20
KennyColnago
11.6k1753
answered Sep 6 at 3:20
KennyColnago
11.6k1753
answered Sep 6 at 3:20
KennyColnago
11.6k1753
answered Sep 6 at 3:20
KennyColnago
11.6k1753
11.6k1753
add a comment |Â
add a comment |Â
up vote
6
down vote
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
answered Sep 5 at 20:03
kglr
159k8183383
add a comment |Â
up vote
6
down vote
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
answered Sep 5 at 20:03
kglr
159k8183383
add a comment |Â
up vote
6
down vote
up vote
6
down vote
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
answered Sep 5 at 20:03
kglr
159k8183383
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
answered Sep 5 at 20:03
kglr
159k8183383
edited Sep 5 at 23:15
answered Sep 5 at 20:03
kglr
159k8183383
answered Sep 5 at 20:03
kglr
159k8183383
answered Sep 5 at 20:03
kglr
159k8183383
159k8183383
add a comment |Â
add a comment |Â
up vote
4
down vote
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]
answered Sep 5 at 19:32
Chris Degnen
21.4k23281
add a comment |Â
up vote
4
down vote
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]
answered Sep 5 at 19:32
Chris Degnen
21.4k23281
add a comment |Â
up vote
4
down vote
up vote
4
down vote
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]
answered Sep 5 at 19:32
Chris Degnen
21.4k23281
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]
answered Sep 5 at 19:32
Chris Degnen
21.4k23281
edited Sep 5 at 19:40
answered Sep 5 at 19:32
Chris Degnen
21.4k23281
answered Sep 5 at 19:32
Chris Degnen
21.4k23281
answered Sep 5 at 19:32
Chris Degnen
21.4k23281
21.4k23281
add a comment |Â
add a comment |Â
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mhd.math is a new contributor. Be nice, and check out our Code of Conduct.
mhd.math is a new contributor. Be nice, and check out our Code of Conduct.
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1
Welcome to Mathematica.SE! You should watch the introductory Tour to get the basics of the site.
– Vitaliy Kaurov
Sep 5 at 19:21
@VitaliyKaurov Thank You for the welcoming and your answer, I appreciate your efforts.
– mhd.math
Sep 5 at 19:25
For plotting PI, the circular form should be logical. But it would be interesting to find out if such a plot for other irrational numbers (e, sqrt(2)) pr just random numbers, would look different
– Roland
Sep 6 at 11:02