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How can I make a scatter plot from a matrix with many y-axis values for each x-axis value?


How can I plot more than one value in the same date?Make a scatter plot from two listsHow to plot first and second row from matrix to a graphHow to map the second highest value in each row of a matrixHow to plot multiple graphs with same x-axis values on the same graph?How to overlay ListPlot on a ContourPlot with correct range?How do I plot several y-values to 1 x-value?How can I make Plot or ListPlot scale the plot automatically for a given domain?Plot with two scales for X axisHow do I fill a table with values from my function?













2












$begingroup$


Consider matrix



A = 
0, 2.54343, 3.566, 4.2323, 2.3434,
0.1, 6.432, 4.6465, 5.656, 1.34,
0.2, 0.3423, 1.6342, 3.323, 2.04;


The x-axis values are 0, 0.1, 0.2. There are four y-axis values for each x-axis value.



How do I plot this?



Naive use of



ListPlot[A]


fails.










share|improve this question











$endgroup$
















    2












    $begingroup$


    Consider matrix



    A = 
    0, 2.54343, 3.566, 4.2323, 2.3434,
    0.1, 6.432, 4.6465, 5.656, 1.34,
    0.2, 0.3423, 1.6342, 3.323, 2.04;


    The x-axis values are 0, 0.1, 0.2. There are four y-axis values for each x-axis value.



    How do I plot this?



    Naive use of



    ListPlot[A]


    fails.










    share|improve this question











    $endgroup$














      2












      2








      2


      1



      $begingroup$


      Consider matrix



      A = 
      0, 2.54343, 3.566, 4.2323, 2.3434,
      0.1, 6.432, 4.6465, 5.656, 1.34,
      0.2, 0.3423, 1.6342, 3.323, 2.04;


      The x-axis values are 0, 0.1, 0.2. There are four y-axis values for each x-axis value.



      How do I plot this?



      Naive use of



      ListPlot[A]


      fails.










      share|improve this question











      $endgroup$




      Consider matrix



      A = 
      0, 2.54343, 3.566, 4.2323, 2.3434,
      0.1, 6.432, 4.6465, 5.656, 1.34,
      0.2, 0.3423, 1.6342, 3.323, 2.04;


      The x-axis values are 0, 0.1, 0.2. There are four y-axis values for each x-axis value.



      How do I plot this?



      Naive use of



      ListPlot[A]


      fails.







      plotting list-manipulation matrix






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 hours ago









      m_goldberg

      90.6k873203




      90.6k873203










      asked 8 hours ago









      Nigel1Nigel1

      32616




      32616




















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Transpose the list A , remove its first list and use the option DataRange:



          ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
          PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]


          enter image description here



          Alternatively, create new input list with explicit x values for each point:



          d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
          ListPlot[d1, Frame -> True, Axes->False,
          PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]



          same picture




          You can also use TemporalData using 0, .1, .2 as the common time stamps:



          d2 = TemporalData[Rest/@ A, A[[All,1]]];;
          ListPlot[d2, Frame -> True, Axes -> False,
          BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]



          same picture







          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            I'm guessing that the OP has A[[All,1]] as the x-values with A[[All,2,3,4,5]] as the y-values.
            $endgroup$
            – JimB
            7 hours ago






          • 1




            $begingroup$
            @JimB, thank you. Fixed the errors now.
            $endgroup$
            – kglr
            7 hours ago











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          3












          $begingroup$

          Transpose the list A , remove its first list and use the option DataRange:



          ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
          PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]


          enter image description here



          Alternatively, create new input list with explicit x values for each point:



          d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
          ListPlot[d1, Frame -> True, Axes->False,
          PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]



          same picture




          You can also use TemporalData using 0, .1, .2 as the common time stamps:



          d2 = TemporalData[Rest/@ A, A[[All,1]]];;
          ListPlot[d2, Frame -> True, Axes -> False,
          BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]



          same picture







          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            I'm guessing that the OP has A[[All,1]] as the x-values with A[[All,2,3,4,5]] as the y-values.
            $endgroup$
            – JimB
            7 hours ago






          • 1




            $begingroup$
            @JimB, thank you. Fixed the errors now.
            $endgroup$
            – kglr
            7 hours ago















          3












          $begingroup$

          Transpose the list A , remove its first list and use the option DataRange:



          ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
          PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]


          enter image description here



          Alternatively, create new input list with explicit x values for each point:



          d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
          ListPlot[d1, Frame -> True, Axes->False,
          PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]



          same picture




          You can also use TemporalData using 0, .1, .2 as the common time stamps:



          d2 = TemporalData[Rest/@ A, A[[All,1]]];;
          ListPlot[d2, Frame -> True, Axes -> False,
          BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]



          same picture







          share|improve this answer











          $endgroup$








          • 1




            $begingroup$
            I'm guessing that the OP has A[[All,1]] as the x-values with A[[All,2,3,4,5]] as the y-values.
            $endgroup$
            – JimB
            7 hours ago






          • 1




            $begingroup$
            @JimB, thank you. Fixed the errors now.
            $endgroup$
            – kglr
            7 hours ago













          3












          3








          3





          $begingroup$

          Transpose the list A , remove its first list and use the option DataRange:



          ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
          PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]


          enter image description here



          Alternatively, create new input list with explicit x values for each point:



          d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
          ListPlot[d1, Frame -> True, Axes->False,
          PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]



          same picture




          You can also use TemporalData using 0, .1, .2 as the common time stamps:



          d2 = TemporalData[Rest/@ A, A[[All,1]]];;
          ListPlot[d2, Frame -> True, Axes -> False,
          BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]



          same picture







          share|improve this answer











          $endgroup$



          Transpose the list A , remove its first list and use the option DataRange:



          ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
          PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]


          enter image description here



          Alternatively, create new input list with explicit x values for each point:



          d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
          ListPlot[d1, Frame -> True, Axes->False,
          PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]



          same picture




          You can also use TemporalData using 0, .1, .2 as the common time stamps:



          d2 = TemporalData[Rest/@ A, A[[All,1]]];;
          ListPlot[d2, Frame -> True, Axes -> False,
          BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]



          same picture








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          kglrkglr

          199k10224452




          199k10224452







          • 1




            $begingroup$
            I'm guessing that the OP has A[[All,1]] as the x-values with A[[All,2,3,4,5]] as the y-values.
            $endgroup$
            – JimB
            7 hours ago






          • 1




            $begingroup$
            @JimB, thank you. Fixed the errors now.
            $endgroup$
            – kglr
            7 hours ago












          • 1




            $begingroup$
            I'm guessing that the OP has A[[All,1]] as the x-values with A[[All,2,3,4,5]] as the y-values.
            $endgroup$
            – JimB
            7 hours ago






          • 1




            $begingroup$
            @JimB, thank you. Fixed the errors now.
            $endgroup$
            – kglr
            7 hours ago







          1




          1




          $begingroup$
          I'm guessing that the OP has A[[All,1]] as the x-values with A[[All,2,3,4,5]] as the y-values.
          $endgroup$
          – JimB
          7 hours ago




          $begingroup$
          I'm guessing that the OP has A[[All,1]] as the x-values with A[[All,2,3,4,5]] as the y-values.
          $endgroup$
          – JimB
          7 hours ago




          1




          1




          $begingroup$
          @JimB, thank you. Fixed the errors now.
          $endgroup$
          – kglr
          7 hours ago




          $begingroup$
          @JimB, thank you. Fixed the errors now.
          $endgroup$
          – kglr
          7 hours ago

















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