Finding minimum from ListPlot

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2












$begingroup$


I have Table with 10000 elements in it and I plot a graphic using ListPlot.
enter image description here



I need find minimum in graph.
enter image description here



Here is a .txt file with data.



I tried to use



Min[Data[[All,1]]]


the result is 0., cuz it search the first point of x but I don't need the first one.










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$endgroup$











  • $begingroup$
    Data[[Ordering[Data[[All,2]],1][[1]],1]]?
    $endgroup$
    – Henrik Schumacher
    Feb 10 at 15:56















2












$begingroup$


I have Table with 10000 elements in it and I plot a graphic using ListPlot.
enter image description here



I need find minimum in graph.
enter image description here



Here is a .txt file with data.



I tried to use



Min[Data[[All,1]]]


the result is 0., cuz it search the first point of x but I don't need the first one.










share|improve this question











$endgroup$











  • $begingroup$
    Data[[Ordering[Data[[All,2]],1][[1]],1]]?
    $endgroup$
    – Henrik Schumacher
    Feb 10 at 15:56













2












2








2





$begingroup$


I have Table with 10000 elements in it and I plot a graphic using ListPlot.
enter image description here



I need find minimum in graph.
enter image description here



Here is a .txt file with data.



I tried to use



Min[Data[[All,1]]]


the result is 0., cuz it search the first point of x but I don't need the first one.










share|improve this question











$endgroup$




I have Table with 10000 elements in it and I plot a graphic using ListPlot.
enter image description here



I need find minimum in graph.
enter image description here



Here is a .txt file with data.



I tried to use



Min[Data[[All,1]]]


the result is 0., cuz it search the first point of x but I don't need the first one.







plotting list-manipulation mathematical-optimization peak-detection






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share|improve this question




share|improve this question








edited Feb 11 at 3:56









Community

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1










asked Feb 10 at 15:53









JohnJohn

32016




32016











  • $begingroup$
    Data[[Ordering[Data[[All,2]],1][[1]],1]]?
    $endgroup$
    – Henrik Schumacher
    Feb 10 at 15:56
















  • $begingroup$
    Data[[Ordering[Data[[All,2]],1][[1]],1]]?
    $endgroup$
    – Henrik Schumacher
    Feb 10 at 15:56















$begingroup$
Data[[Ordering[Data[[All,2]],1][[1]],1]]?
$endgroup$
– Henrik Schumacher
Feb 10 at 15:56




$begingroup$
Data[[Ordering[Data[[All,2]],1][[1]],1]]?
$endgroup$
– Henrik Schumacher
Feb 10 at 15:56










3 Answers
3






active

oldest

votes


















4












$begingroup$

This is how I would go about it:



minIndex = data[[All, 2]] // MinDetect // PositionIndex;


Now all the positions of the minima were collected in an Association and given the key 1. We now want to split the list of minima positions into sublists that are "connected", e.g. where each position is separated by a distance of 1. From these sublists we only need the first and the last positions (the "corners"):



minPositions = minIndex[1] // RightComposition[
Split[#, #1 == #2 - 1 &] & (* distance could be made more "soft" of course *)
, Part[#, All, 1, -1] &
, Flatten
]



1, 276, 1167, 2844




We now Extract the data points for these positions dropping the first, which you do not need as you said:



minPoints = Extract[data, List /@ minPositions] // Drop[#, 1] &



0.275, 0., 1.166, 0., 2.843, 0.




Finally:



ListPlot[ data, Epilog -> Red, PointSize -> Large, Point@minPoints , ImageSize -> Large ]


ListPlot of data with minima






share|improve this answer









$endgroup$




















    4












    $begingroup$

    lp = ListLinePlot[data, 
    MeshFunctions -> #2 &,
    MeshStyle -> Directive[Red, PointSize[Large]],
    Mesh -> Min[data[[All, 2]]],
    AxesOrigin -> 0, -.01]


    enter image description here



    To extract the data elements:



    Cases[Normal @ lp, Point[x_] :> x, All]



    0., 0., 0.275, 0., 1.166, 0., 2.843, 0.




    If you want to remove the first point:



    lp /. Point[x_] :> Point[Rest @ x]


    enter image description here






    share|improve this answer











    $endgroup$




















      3












      $begingroup$

      This should do:



      MinimalBy[Data, Last]


      If you have runs of minimal elements and only want the first and last one: assuming that the grid spacing is 0.001 and inserting 10% of tolerance,



      Split[MinimalBy[Data, Last], #2[[1]]-#1[[1]] <= 0.0011 &][[All, 1,-1]]


      maybe combined with Flatten to make into a single list of points.






      share|improve this answer











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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        This is how I would go about it:



        minIndex = data[[All, 2]] // MinDetect // PositionIndex;


        Now all the positions of the minima were collected in an Association and given the key 1. We now want to split the list of minima positions into sublists that are "connected", e.g. where each position is separated by a distance of 1. From these sublists we only need the first and the last positions (the "corners"):



        minPositions = minIndex[1] // RightComposition[
        Split[#, #1 == #2 - 1 &] & (* distance could be made more "soft" of course *)
        , Part[#, All, 1, -1] &
        , Flatten
        ]



        1, 276, 1167, 2844




        We now Extract the data points for these positions dropping the first, which you do not need as you said:



        minPoints = Extract[data, List /@ minPositions] // Drop[#, 1] &



        0.275, 0., 1.166, 0., 2.843, 0.




        Finally:



        ListPlot[ data, Epilog -> Red, PointSize -> Large, Point@minPoints , ImageSize -> Large ]


        ListPlot of data with minima






        share|improve this answer









        $endgroup$

















          4












          $begingroup$

          This is how I would go about it:



          minIndex = data[[All, 2]] // MinDetect // PositionIndex;


          Now all the positions of the minima were collected in an Association and given the key 1. We now want to split the list of minima positions into sublists that are "connected", e.g. where each position is separated by a distance of 1. From these sublists we only need the first and the last positions (the "corners"):



          minPositions = minIndex[1] // RightComposition[
          Split[#, #1 == #2 - 1 &] & (* distance could be made more "soft" of course *)
          , Part[#, All, 1, -1] &
          , Flatten
          ]



          1, 276, 1167, 2844




          We now Extract the data points for these positions dropping the first, which you do not need as you said:



          minPoints = Extract[data, List /@ minPositions] // Drop[#, 1] &



          0.275, 0., 1.166, 0., 2.843, 0.




          Finally:



          ListPlot[ data, Epilog -> Red, PointSize -> Large, Point@minPoints , ImageSize -> Large ]


          ListPlot of data with minima






          share|improve this answer









          $endgroup$















            4












            4








            4





            $begingroup$

            This is how I would go about it:



            minIndex = data[[All, 2]] // MinDetect // PositionIndex;


            Now all the positions of the minima were collected in an Association and given the key 1. We now want to split the list of minima positions into sublists that are "connected", e.g. where each position is separated by a distance of 1. From these sublists we only need the first and the last positions (the "corners"):



            minPositions = minIndex[1] // RightComposition[
            Split[#, #1 == #2 - 1 &] & (* distance could be made more "soft" of course *)
            , Part[#, All, 1, -1] &
            , Flatten
            ]



            1, 276, 1167, 2844




            We now Extract the data points for these positions dropping the first, which you do not need as you said:



            minPoints = Extract[data, List /@ minPositions] // Drop[#, 1] &



            0.275, 0., 1.166, 0., 2.843, 0.




            Finally:



            ListPlot[ data, Epilog -> Red, PointSize -> Large, Point@minPoints , ImageSize -> Large ]


            ListPlot of data with minima






            share|improve this answer









            $endgroup$



            This is how I would go about it:



            minIndex = data[[All, 2]] // MinDetect // PositionIndex;


            Now all the positions of the minima were collected in an Association and given the key 1. We now want to split the list of minima positions into sublists that are "connected", e.g. where each position is separated by a distance of 1. From these sublists we only need the first and the last positions (the "corners"):



            minPositions = minIndex[1] // RightComposition[
            Split[#, #1 == #2 - 1 &] & (* distance could be made more "soft" of course *)
            , Part[#, All, 1, -1] &
            , Flatten
            ]



            1, 276, 1167, 2844




            We now Extract the data points for these positions dropping the first, which you do not need as you said:



            minPoints = Extract[data, List /@ minPositions] // Drop[#, 1] &



            0.275, 0., 1.166, 0., 2.843, 0.




            Finally:



            ListPlot[ data, Epilog -> Red, PointSize -> Large, Point@minPoints , ImageSize -> Large ]


            ListPlot of data with minima







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Feb 10 at 16:53









            gwrgwr

            8,52322761




            8,52322761





















                4












                $begingroup$

                lp = ListLinePlot[data, 
                MeshFunctions -> #2 &,
                MeshStyle -> Directive[Red, PointSize[Large]],
                Mesh -> Min[data[[All, 2]]],
                AxesOrigin -> 0, -.01]


                enter image description here



                To extract the data elements:



                Cases[Normal @ lp, Point[x_] :> x, All]



                0., 0., 0.275, 0., 1.166, 0., 2.843, 0.




                If you want to remove the first point:



                lp /. Point[x_] :> Point[Rest @ x]


                enter image description here






                share|improve this answer











                $endgroup$

















                  4












                  $begingroup$

                  lp = ListLinePlot[data, 
                  MeshFunctions -> #2 &,
                  MeshStyle -> Directive[Red, PointSize[Large]],
                  Mesh -> Min[data[[All, 2]]],
                  AxesOrigin -> 0, -.01]


                  enter image description here



                  To extract the data elements:



                  Cases[Normal @ lp, Point[x_] :> x, All]



                  0., 0., 0.275, 0., 1.166, 0., 2.843, 0.




                  If you want to remove the first point:



                  lp /. Point[x_] :> Point[Rest @ x]


                  enter image description here






                  share|improve this answer











                  $endgroup$















                    4












                    4








                    4





                    $begingroup$

                    lp = ListLinePlot[data, 
                    MeshFunctions -> #2 &,
                    MeshStyle -> Directive[Red, PointSize[Large]],
                    Mesh -> Min[data[[All, 2]]],
                    AxesOrigin -> 0, -.01]


                    enter image description here



                    To extract the data elements:



                    Cases[Normal @ lp, Point[x_] :> x, All]



                    0., 0., 0.275, 0., 1.166, 0., 2.843, 0.




                    If you want to remove the first point:



                    lp /. Point[x_] :> Point[Rest @ x]


                    enter image description here






                    share|improve this answer











                    $endgroup$



                    lp = ListLinePlot[data, 
                    MeshFunctions -> #2 &,
                    MeshStyle -> Directive[Red, PointSize[Large]],
                    Mesh -> Min[data[[All, 2]]],
                    AxesOrigin -> 0, -.01]


                    enter image description here



                    To extract the data elements:



                    Cases[Normal @ lp, Point[x_] :> x, All]



                    0., 0., 0.275, 0., 1.166, 0., 2.843, 0.




                    If you want to remove the first point:



                    lp /. Point[x_] :> Point[Rest @ x]


                    enter image description here







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Feb 10 at 18:57

























                    answered Feb 10 at 17:32









                    kglrkglr

                    187k10203422




                    187k10203422





















                        3












                        $begingroup$

                        This should do:



                        MinimalBy[Data, Last]


                        If you have runs of minimal elements and only want the first and last one: assuming that the grid spacing is 0.001 and inserting 10% of tolerance,



                        Split[MinimalBy[Data, Last], #2[[1]]-#1[[1]] <= 0.0011 &][[All, 1,-1]]


                        maybe combined with Flatten to make into a single list of points.






                        share|improve this answer











                        $endgroup$

















                          3












                          $begingroup$

                          This should do:



                          MinimalBy[Data, Last]


                          If you have runs of minimal elements and only want the first and last one: assuming that the grid spacing is 0.001 and inserting 10% of tolerance,



                          Split[MinimalBy[Data, Last], #2[[1]]-#1[[1]] <= 0.0011 &][[All, 1,-1]]


                          maybe combined with Flatten to make into a single list of points.






                          share|improve this answer











                          $endgroup$















                            3












                            3








                            3





                            $begingroup$

                            This should do:



                            MinimalBy[Data, Last]


                            If you have runs of minimal elements and only want the first and last one: assuming that the grid spacing is 0.001 and inserting 10% of tolerance,



                            Split[MinimalBy[Data, Last], #2[[1]]-#1[[1]] <= 0.0011 &][[All, 1,-1]]


                            maybe combined with Flatten to make into a single list of points.






                            share|improve this answer











                            $endgroup$



                            This should do:



                            MinimalBy[Data, Last]


                            If you have runs of minimal elements and only want the first and last one: assuming that the grid spacing is 0.001 and inserting 10% of tolerance,



                            Split[MinimalBy[Data, Last], #2[[1]]-#1[[1]] <= 0.0011 &][[All, 1,-1]]


                            maybe combined with Flatten to make into a single list of points.







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Feb 10 at 18:44

























                            answered Feb 10 at 17:30









                            RomanRoman

                            2,614717




                            2,614717



























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