How to get the coordinates of the extrema from a plot?
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 13 hours ago
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 13 hours ago
user3318424user3318424
874
$endgroup$
$begingroup$
You mean aGraphics-object or a pixel image?
$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
11 hours ago
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 13 hours ago
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 13 hours ago
user3318424user3318424
874
asked 13 hours ago
user3318424user3318424
874
asked 13 hours ago
user3318424user3318424
874
asked 13 hours ago
user3318424user3318424
874
asked 13 hours ago
user3318424user3318424
874
874
$begingroup$
You mean aGraphics-object or a pixel image?
$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
11 hours ago
add a comment |
$begingroup$
You mean aGraphics-object or a pixel image?
$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
11 hours ago
$begingroup$
You mean a
Graphics-object or a pixel image?$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
You mean a
Graphics-object or a pixel image?$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
11 hours ago
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
11 hours ago
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 12 hours ago
Ulrich NeumannUlrich Neumann
8,380516
$endgroup$
$begingroup$
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
12 hours ago
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
All the extrema.
$endgroup$
– user3318424
12 hours ago
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 11 hours ago
Ulrich NeumannUlrich Neumann
8,380516
$endgroup$
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 10 hours ago
Bob HanlonBob Hanlon
59.5k33596
$endgroup$
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 12 hours ago
Ulrich NeumannUlrich Neumann
8,380516
$endgroup$
$begingroup$
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
12 hours ago
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
All the extrema.
$endgroup$
– user3318424
12 hours ago
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 12 hours ago
Ulrich NeumannUlrich Neumann
8,380516
$endgroup$
$begingroup$
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
12 hours ago
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
All the extrema.
$endgroup$
– user3318424
12 hours ago
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 12 hours ago
Ulrich NeumannUlrich Neumann
8,380516
$endgroup$
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 12 hours ago
Ulrich NeumannUlrich Neumann
8,380516
edited 11 hours ago
answered 12 hours ago
Ulrich NeumannUlrich Neumann
8,380516
answered 12 hours ago
Ulrich NeumannUlrich Neumann
8,380516
answered 12 hours ago
Ulrich NeumannUlrich Neumann
8,380516
8,380516
$begingroup$
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
12 hours ago
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
All the extrema.
$endgroup$
– user3318424
12 hours ago
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
12 hours ago
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
All the extrema.
$endgroup$
– user3318424
12 hours ago
$begingroup$
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
12 hours ago
$begingroup$
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
12 hours ago
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
I didn't use the knowledge of the function! One questiom : Are you looking for all extrema or only the global?
$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
All the extrema.
$endgroup$
– user3318424
12 hours ago
$begingroup$
All the extrema.
$endgroup$
– user3318424
12 hours ago
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 11 hours ago
Ulrich NeumannUlrich Neumann
8,380516
$endgroup$
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 11 hours ago
Ulrich NeumannUlrich Neumann
8,380516
$endgroup$
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 11 hours ago
Ulrich NeumannUlrich Neumann
8,380516
$endgroup$
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 11 hours ago
Ulrich NeumannUlrich Neumann
8,380516
answered 11 hours ago
Ulrich NeumannUlrich Neumann
8,380516
answered 11 hours ago
Ulrich NeumannUlrich Neumann
8,380516
answered 11 hours ago
Ulrich NeumannUlrich Neumann
8,380516
8,380516
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 10 hours ago
Bob HanlonBob Hanlon
59.5k33596
$endgroup$
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 10 hours ago
Bob HanlonBob Hanlon
59.5k33596
$endgroup$
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 10 hours ago
Bob HanlonBob Hanlon
59.5k33596
$endgroup$
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 10 hours ago
Bob HanlonBob Hanlon
59.5k33596
answered 10 hours ago
Bob HanlonBob Hanlon
59.5k33596
answered 10 hours ago
Bob HanlonBob Hanlon
59.5k33596
answered 10 hours ago
Bob HanlonBob Hanlon
59.5k33596
59.5k33596
add a comment |
add a comment |
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$begingroup$
You mean a
Graphics-object or a pixel image?$endgroup$
– Ulrich Neumann
12 hours ago
$begingroup$
@UlrichNeumann I'm pretty sure they mean they have just a pixel image, which would complicate things a lot.
$endgroup$
– MassDefect
11 hours ago