How to get the coordinates of the extrema from a plot?
Clash Royale CLAN TAG#URR8PPP
1
$begingroup$
I generated a number of plots. Let us assume that we now have these plots simply as an image. How do I get the coordinates of the extrema in this image?
I am interested in getting the coordinates as pairs of x and y, for each color in the plot.
Here is an example of such an image:
plotting
asked Feb 1 at 15:14
user3318424user3318424
874
$endgroup$
add a comment |
1
$begingroup$
I generated a number of plots. Let us assume that we now have these plots simply as an image. How do I get the coordinates of the extrema in this image?
I am interested in getting the coordinates as pairs of x and y, for each color in the plot.
Here is an example of such an image:
plotting
asked Feb 1 at 15:14
user3318424user3318424
874
$endgroup$
$begingroup$
You mean aGraphics-object or a pixel image?
$endgroup$
– Ulrich Neumann
Feb 1 at 16:07
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
Feb 1 at 16:45
add a comment |
1
1
1
$begingroup$
I generated a number of plots. Let us assume that we now have these plots simply as an image. How do I get the coordinates of the extrema in this image?
I am interested in getting the coordinates as pairs of x and y, for each color in the plot.
Here is an example of such an image:
plotting
asked Feb 1 at 15:14
user3318424user3318424
874
$endgroup$
I generated a number of plots. Let us assume that we now have these plots simply as an image. How do I get the coordinates of the extrema in this image?
I am interested in getting the coordinates as pairs of x and y, for each color in the plot.
Here is an example of such an image:
plotting
plotting
asked Feb 1 at 15:14
user3318424user3318424
874
asked Feb 1 at 15:14
user3318424user3318424
874
asked Feb 1 at 15:14
user3318424user3318424
874
asked Feb 1 at 15:14
user3318424user3318424
874
asked Feb 1 at 15:14
user3318424user3318424
874
874
$begingroup$
You mean aGraphics-object or a pixel image?
$endgroup$
– Ulrich Neumann
Feb 1 at 16:07
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
Feb 1 at 16:45
add a comment |
$begingroup$
You mean aGraphics-object or a pixel image?
$endgroup$
– Ulrich Neumann
Feb 1 at 16:07
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
Feb 1 at 16:45
$begingroup$
You mean a
Graphics-object or a pixel image?$endgroup$
– Ulrich Neumann
Feb 1 at 16:07
$begingroup$
You mean a
Graphics-object or a pixel image?$endgroup$
– Ulrich Neumann
Feb 1 at 16:07
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
Feb 1 at 16:45
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
Feb 1 at 16:45
add a comment |
3 Answers
3
active
oldest
votes
4
$begingroup$
If "image" is a Graphics-object try
pic = Plot[Sin[x]/x, Exp[-.1 x] Sin[x], x, 0, 20] (*two functions*)
lines = Cases[pic, Line[p_] -> p, Infinity] (*get the points*)
Evaluate all extrema and plot
extrema = Map[Cases[
Partition[#, 3,1], _, a_, p : _, b_, _, c_ /; a < b && c < b || a > b && c > b -> p] &, lines]
Show[pic, Graphics[Point[extrema ]]]
answered Feb 1 at 15:37
Ulrich NeumannUlrich Neumann
9,001516
$endgroup$
$begingroup$
Thank you for your answer. However, I assume that I no longer have or know the function that resulted in the plots. I only have the "image".
$endgroup$
– user3318424
Feb 1 at 15:41
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
Feb 1 at 15:42
$begingroup$
All the extrema.
$endgroup$
– user3318424
Feb 1 at 15:46
add a comment |
4
$begingroup$
If "image" is a pixel image
(named pic , sorry, don't know how to include pic="image" in the coding ) try:
dc = Rest@DominantColors[pic] (* dominant colors without white*)
curves = Map[ListPlot[PixelValuePositions[pic, #, .1 ],Axes -> False, PlotStyle -> #] &, dc] (* three colored curves *)
Get the points of the different curves
points = Cases[curves , Point[pi_] -> pi, Infinity];
...see my first answer
answered Feb 1 at 17:06
Ulrich NeumannUlrich Neumann
9,001516
$endgroup$
add a comment |
3
$begingroup$
Ulrich's approach for a Graphics-object misses the end points. You can use FindPeaks to also catch the end points.
pic = Plot[Sin[x]/x, Exp[-.1 x] Sin[x], x, 0, 20]; (*two functions*)
lines = Cases[pic, Line[p_] -> p, Infinity]; (*get the points*)
max = Flatten[(#[[FindPeaks[#[[All, 2]]][[All, 1]]]] & /@ lines), 1];
min = Flatten[(#[[FindPeaks[(# /. x_?NumericQ, y_?NumericQ :> x, -y)[[All,
2]]][[All, 1]]]] & /@ lines), 1];
Show[pic, Epilog -> AbsolutePointSize[4],
Red, Point[max],
Blue, Point[min]]
answered Feb 1 at 17:45
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
$begingroup$
@ BobHanlon Thanks, I didn't knowFindPeaks
$endgroup$
– Ulrich Neumann
Feb 2 at 9:35
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
$begingroup$
If "image" is a Graphics-object try
pic = Plot[Sin[x]/x, Exp[-.1 x] Sin[x], x, 0, 20] (*two functions*)
lines = Cases[pic, Line[p_] -> p, Infinity] (*get the points*)
Evaluate all extrema and plot
extrema = Map[Cases[
Partition[#, 3,1], _, a_, p : _, b_, _, c_ /; a < b && c < b || a > b && c > b -> p] &, lines]
Show[pic, Graphics[Point[extrema ]]]
answered Feb 1 at 15:37
Ulrich NeumannUlrich Neumann
9,001516
$endgroup$
$begingroup$
Thank you for your answer. However, I assume that I no longer have or know the function that resulted in the plots. I only have the "image".
$endgroup$
– user3318424
Feb 1 at 15:41
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
Feb 1 at 15:42
$begingroup$
All the extrema.
$endgroup$
– user3318424
Feb 1 at 15:46
add a comment |
4
$begingroup$
If "image" is a Graphics-object try
pic = Plot[Sin[x]/x, Exp[-.1 x] Sin[x], x, 0, 20] (*two functions*)
lines = Cases[pic, Line[p_] -> p, Infinity] (*get the points*)
Evaluate all extrema and plot
extrema = Map[Cases[
Partition[#, 3,1], _, a_, p : _, b_, _, c_ /; a < b && c < b || a > b && c > b -> p] &, lines]
Show[pic, Graphics[Point[extrema ]]]
answered Feb 1 at 15:37
Ulrich NeumannUlrich Neumann
9,001516
$endgroup$
$begingroup$
Thank you for your answer. However, I assume that I no longer have or know the function that resulted in the plots. I only have the "image".
$endgroup$
– user3318424
Feb 1 at 15:41
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
Feb 1 at 15:42
$begingroup$
All the extrema.
$endgroup$
– user3318424
Feb 1 at 15:46
add a comment |
4
4
4
$begingroup$
If "image" is a Graphics-object try
pic = Plot[Sin[x]/x, Exp[-.1 x] Sin[x], x, 0, 20] (*two functions*)
lines = Cases[pic, Line[p_] -> p, Infinity] (*get the points*)
Evaluate all extrema and plot
extrema = Map[Cases[
Partition[#, 3,1], _, a_, p : _, b_, _, c_ /; a < b && c < b || a > b && c > b -> p] &, lines]
Show[pic, Graphics[Point[extrema ]]]
answered Feb 1 at 15:37
Ulrich NeumannUlrich Neumann
9,001516
$endgroup$
If "image" is a Graphics-object try
pic = Plot[Sin[x]/x, Exp[-.1 x] Sin[x], x, 0, 20] (*two functions*)
lines = Cases[pic, Line[p_] -> p, Infinity] (*get the points*)
Evaluate all extrema and plot
extrema = Map[Cases[
Partition[#, 3,1], _, a_, p : _, b_, _, c_ /; a < b && c < b || a > b && c > b -> p] &, lines]
Show[pic, Graphics[Point[extrema ]]]
answered Feb 1 at 15:37
Ulrich NeumannUlrich Neumann
9,001516
edited Feb 1 at 16:33
answered Feb 1 at 15:37
Ulrich NeumannUlrich Neumann
9,001516
answered Feb 1 at 15:37
Ulrich NeumannUlrich Neumann
9,001516
answered Feb 1 at 15:37
Ulrich NeumannUlrich Neumann
9,001516
9,001516
$begingroup$
Thank you for your answer. However, I assume that I no longer have or know the function that resulted in the plots. I only have the "image".
$endgroup$
– user3318424
Feb 1 at 15:41
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
Feb 1 at 15:42
$begingroup$
All the extrema.
$endgroup$
– user3318424
Feb 1 at 15:46
add a comment |
$begingroup$
Thank you for your answer. However, I assume that I no longer have or know the function that resulted in the plots. I only have the "image".
$endgroup$
– user3318424
Feb 1 at 15:41
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
Feb 1 at 15:42
$begingroup$
All the extrema.
$endgroup$
– user3318424
Feb 1 at 15:46
$begingroup$
Thank you for your answer. However, I assume that I no longer have or know the function that resulted in the plots. I only have the "image".
$endgroup$
– user3318424
Feb 1 at 15:41
$begingroup$
Thank you for your answer. However, I assume that I no longer have or know the function that resulted in the plots. I only have the "image".
$endgroup$
– user3318424
Feb 1 at 15:41
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
Feb 1 at 15:42
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
Feb 1 at 15:42
$begingroup$
All the extrema.
$endgroup$
– user3318424
Feb 1 at 15:46
$begingroup$
All the extrema.
$endgroup$
– user3318424
Feb 1 at 15:46
add a comment |
4
$begingroup$
If "image" is a pixel image
(named pic , sorry, don't know how to include pic="image" in the coding ) try:
dc = Rest@DominantColors[pic] (* dominant colors without white*)
curves = Map[ListPlot[PixelValuePositions[pic, #, .1 ],Axes -> False, PlotStyle -> #] &, dc] (* three colored curves *)
Get the points of the different curves
points = Cases[curves , Point[pi_] -> pi, Infinity];
...see my first answer
answered Feb 1 at 17:06
Ulrich NeumannUlrich Neumann
9,001516
$endgroup$
add a comment |
4
$begingroup$
If "image" is a pixel image
(named pic , sorry, don't know how to include pic="image" in the coding ) try:
dc = Rest@DominantColors[pic] (* dominant colors without white*)
curves = Map[ListPlot[PixelValuePositions[pic, #, .1 ],Axes -> False, PlotStyle -> #] &, dc] (* three colored curves *)
Get the points of the different curves
points = Cases[curves , Point[pi_] -> pi, Infinity];
...see my first answer
answered Feb 1 at 17:06
Ulrich NeumannUlrich Neumann
9,001516
$endgroup$
add a comment |
4
4
4
$begingroup$
If "image" is a pixel image
(named pic , sorry, don't know how to include pic="image" in the coding ) try:
dc = Rest@DominantColors[pic] (* dominant colors without white*)
curves = Map[ListPlot[PixelValuePositions[pic, #, .1 ],Axes -> False, PlotStyle -> #] &, dc] (* three colored curves *)
Get the points of the different curves
points = Cases[curves , Point[pi_] -> pi, Infinity];
...see my first answer
answered Feb 1 at 17:06
Ulrich NeumannUlrich Neumann
9,001516
$endgroup$
If "image" is a pixel image
(named pic , sorry, don't know how to include pic="image" in the coding ) try:
dc = Rest@DominantColors[pic] (* dominant colors without white*)
curves = Map[ListPlot[PixelValuePositions[pic, #, .1 ],Axes -> False, PlotStyle -> #] &, dc] (* three colored curves *)
Get the points of the different curves
points = Cases[curves , Point[pi_] -> pi, Infinity];
...see my first answer
answered Feb 1 at 17:06
Ulrich NeumannUlrich Neumann
9,001516
answered Feb 1 at 17:06
Ulrich NeumannUlrich Neumann
9,001516
answered Feb 1 at 17:06
Ulrich NeumannUlrich Neumann
9,001516
answered Feb 1 at 17:06
Ulrich NeumannUlrich Neumann
9,001516
9,001516
add a comment |
add a comment |
3
$begingroup$
Ulrich's approach for a Graphics-object misses the end points. You can use FindPeaks to also catch the end points.
pic = Plot[Sin[x]/x, Exp[-.1 x] Sin[x], x, 0, 20]; (*two functions*)
lines = Cases[pic, Line[p_] -> p, Infinity]; (*get the points*)
max = Flatten[(#[[FindPeaks[#[[All, 2]]][[All, 1]]]] & /@ lines), 1];
min = Flatten[(#[[FindPeaks[(# /. x_?NumericQ, y_?NumericQ :> x, -y)[[All,
2]]][[All, 1]]]] & /@ lines), 1];
Show[pic, Epilog -> AbsolutePointSize[4],
Red, Point[max],
Blue, Point[min]]
answered Feb 1 at 17:45
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
$begingroup$
@ BobHanlon Thanks, I didn't knowFindPeaks
$endgroup$
– Ulrich Neumann
Feb 2 at 9:35
add a comment |
3
$begingroup$
Ulrich's approach for a Graphics-object misses the end points. You can use FindPeaks to also catch the end points.
pic = Plot[Sin[x]/x, Exp[-.1 x] Sin[x], x, 0, 20]; (*two functions*)
lines = Cases[pic, Line[p_] -> p, Infinity]; (*get the points*)
max = Flatten[(#[[FindPeaks[#[[All, 2]]][[All, 1]]]] & /@ lines), 1];
min = Flatten[(#[[FindPeaks[(# /. x_?NumericQ, y_?NumericQ :> x, -y)[[All,
2]]][[All, 1]]]] & /@ lines), 1];
Show[pic, Epilog -> AbsolutePointSize[4],
Red, Point[max],
Blue, Point[min]]
answered Feb 1 at 17:45
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
$begingroup$
@ BobHanlon Thanks, I didn't knowFindPeaks
$endgroup$
– Ulrich Neumann
Feb 2 at 9:35
add a comment |
3
3
3
$begingroup$
Ulrich's approach for a Graphics-object misses the end points. You can use FindPeaks to also catch the end points.
pic = Plot[Sin[x]/x, Exp[-.1 x] Sin[x], x, 0, 20]; (*two functions*)
lines = Cases[pic, Line[p_] -> p, Infinity]; (*get the points*)
max = Flatten[(#[[FindPeaks[#[[All, 2]]][[All, 1]]]] & /@ lines), 1];
min = Flatten[(#[[FindPeaks[(# /. x_?NumericQ, y_?NumericQ :> x, -y)[[All,
2]]][[All, 1]]]] & /@ lines), 1];
Show[pic, Epilog -> AbsolutePointSize[4],
Red, Point[max],
Blue, Point[min]]
answered Feb 1 at 17:45
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
Ulrich's approach for a Graphics-object misses the end points. You can use FindPeaks to also catch the end points.
pic = Plot[Sin[x]/x, Exp[-.1 x] Sin[x], x, 0, 20]; (*two functions*)
lines = Cases[pic, Line[p_] -> p, Infinity]; (*get the points*)
max = Flatten[(#[[FindPeaks[#[[All, 2]]][[All, 1]]]] & /@ lines), 1];
min = Flatten[(#[[FindPeaks[(# /. x_?NumericQ, y_?NumericQ :> x, -y)[[All,
2]]][[All, 1]]]] & /@ lines), 1];
Show[pic, Epilog -> AbsolutePointSize[4],
Red, Point[max],
Blue, Point[min]]
answered Feb 1 at 17:45
Bob HanlonBob Hanlon
60.6k33597
answered Feb 1 at 17:45
Bob HanlonBob Hanlon
60.6k33597
answered Feb 1 at 17:45
Bob HanlonBob Hanlon
60.6k33597
answered Feb 1 at 17:45
Bob HanlonBob Hanlon
60.6k33597
60.6k33597
$begingroup$
@ BobHanlon Thanks, I didn't knowFindPeaks
$endgroup$
– Ulrich Neumann
Feb 2 at 9:35
add a comment |
$begingroup$
@ BobHanlon Thanks, I didn't knowFindPeaks
$endgroup$
– Ulrich Neumann
Feb 2 at 9:35
$begingroup$
@ BobHanlon Thanks, I didn't know
FindPeaks$endgroup$
– Ulrich Neumann
Feb 2 at 9:35
$begingroup$
@ BobHanlon Thanks, I didn't know
FindPeaks$endgroup$
– Ulrich Neumann
Feb 2 at 9:35
add a comment |
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$begingroup$
You mean a
Graphics-object or a pixel image?$endgroup$
– Ulrich Neumann
Feb 1 at 16:07
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
Feb 1 at 16:45