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:



enter image description here






plotting







share|improve this question









$endgroup$












  • $begingroup$
    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
















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:



enter image description here






plotting







share|improve this question









$endgroup$












  • $begingroup$
    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














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:



enter image description here






plotting







share|improve this question









$endgroup$




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:



enter image description here






plotting




plotting






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 13 hours ago









user3318424user3318424

874




874












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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










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 ]]]


enter image description here






share|improve this answer











$endgroup$













  • $begingroup$
    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



















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






share|improve this answer









$endgroup$





















    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]}]


    enter image description here






    share|improve this answer









    $endgroup$













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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 ]]]


      enter image description here






      share|improve this answer











      $endgroup$













      • $begingroup$
        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
















      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 ]]]


      enter image description here






      share|improve this answer











      $endgroup$













      • $begingroup$
        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














      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 ]]]


      enter image description here






      share|improve this answer











      $endgroup$



      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 ]]]


      enter image description here







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 11 hours ago

























      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


















      • $begingroup$
        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











      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






      share|improve this answer









      $endgroup$


















        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






        share|improve this answer









        $endgroup$
















          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






          share|improve this answer









          $endgroup$



          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







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 11 hours ago









          Ulrich NeumannUlrich Neumann

          8,380516




          8,380516























              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]}]


              enter image description here






              share|improve this answer









              $endgroup$


















                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]}]


                enter image description here






                share|improve this answer









                $endgroup$
















                  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]}]


                  enter image description here






                  share|improve this answer









                  $endgroup$



                  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]}]


                  enter image description here







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 10 hours ago









                  Bob HanlonBob Hanlon

                  59.5k33596




                  59.5k33596






























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                      格利澤436b

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                      格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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