Finding minimum from ListPlot












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.






plotting list-manipulation mathematical-optimization peak-detection







share|improve this question











$endgroup$












  • $begingroup$
    Data[[Ordering[Data[[All,2]],1][[1]],1]]?
    $endgroup$
    – Henrik Schumacher
    12 hours ago
















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.






plotting list-manipulation mathematical-optimization peak-detection







share|improve this question











$endgroup$












  • $begingroup$
    Data[[Ordering[Data[[All,2]],1][[1]],1]]?
    $endgroup$
    – Henrik Schumacher
    12 hours ago














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.






plotting list-manipulation mathematical-optimization peak-detection







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




plotting list-manipulation mathematical-optimization peak-detection






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 14 mins ago









Community♦

1




1










asked 12 hours ago









JohnJohn

32016




32016












  • $begingroup$
    Data[[Ordering[Data[[All,2]],1][[1]],1]]?
    $endgroup$
    – Henrik Schumacher
    12 hours ago


















  • $begingroup$
    Data[[Ordering[Data[[All,2]],1][[1]],1]]?
    $endgroup$
    – Henrik Schumacher
    12 hours ago
















$begingroup$
Data[[Ordering[Data[[All,2]],1][[1]],1]]?
$endgroup$
– Henrik Schumacher
12 hours ago




$begingroup$
Data[[Ordering[Data[[All,2]],1][[1]],1]]?
$endgroup$
– Henrik Schumacher
12 hours ago










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











      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "387"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: false,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: null,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f191251%2ffinding-minimum-from-listplot%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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 11 hours ago









            gwrgwr

            8,42322761




            8,42322761























                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 9 hours ago

























                    answered 10 hours ago









                    kglrkglr

                    183k10201416




                    183k10201416























                        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 9 hours ago

























                            answered 10 hours ago









                            RomanRoman

                            1,012511




                            1,012511






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematica Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f191251%2ffinding-minimum-from-listplot%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                CARDNET

                                Image
                                CARDNET は日本カードネットワークが運営するクレジット決済ネットワーク。日本カードネットワークの略称としてもCARDNETが使用されている。 事件・事故 2017年4月15日午前11時8分からクレジットカードの決済がしづらくなる障害が発生した。障害は同日午後5時18分に復旧した。原因は、複数あるCARDNETセンターの拠点のうち1つで、2重化してあるL3スイッチの片方が故障し、1系統にトラフィックが集中したことによる輻輳が発生したため。対処としては、故障したL3スイッチ交換とシステム再起動が行われた [1] 。 出典 ^ 金子寛人 ( 2017年4月17日 ). “CARDNETのクレジット決済に6時間強障害、原因はL3スイッチ故障”. 日経コンピュータ. 2018年10月18日 閲覧。 この項目は、コンピュータに関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（PJ:コンピュータ/P:コンピュータ）。 この項目は、経済に関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（ポータル 経済学、プロジェクト 経済）。 This page is only for reference, If you need detailed information, please check here
                                Read more

                                Boot-repair Failure: Unable to locate package grub-common:i386

                                Image
                                0 I am on a system dual-booting Windows and Debian (usually using the latter), and a recent Windows update appears to have wreaked havoc on the Debian boot system. I managed to boot Windows again, but not Linux. Following online suggestions, I created a Live USB of Ubuntu, and booted from that. Then I have installed and run the boot-repair utility. The utility instructs me to type several lines including sudo chroot "/mnt/boot-sav/sda6" apt-get purge -y grub*-common grub-common:i386 To which I receive the error E: Unable to locate package grub-common:i386 I am given to understand the issue may have to do with having UEFI vs BIOS (I believe I have UEFI), but do not understand this well. In any case I have subsequently run the boot-repair BootInfo Summary utility which created the following file h
                                Read more

                                濃尾地震

                                Image
                                濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸
                                Read more