ListPlot from Table of single input x and many y(x) output












3















Suppose have data for several x and several y(x), for example



mydata = {{11, 1.1, 2.3, 2.4}, {12, 1.3, 1.4, 1.9}, {13, 1.8, 1.7, 1.2}}

mydata // MatrixForm


So



for x=11, have y(x)= 1.1, 2.3 and 2.4



for x=12, have y(x)= 1.3, 1.4 and 1.9



for x=13, have y(x)= 1.8, 1.7 and 1.2



How to do list plot for many y(x) as function of x?



Use of 



ListPlot[mydata]


is no working.






plotting list-manipulation table







share|improve this question




















  • 1





    Related: (21884), (47038)

    – Mr.Wizard♦
    13 hours ago
















3















Suppose have data for several x and several y(x), for example



mydata = {{11, 1.1, 2.3, 2.4}, {12, 1.3, 1.4, 1.9}, {13, 1.8, 1.7, 1.2}}

mydata // MatrixForm


So



for x=11, have y(x)= 1.1, 2.3 and 2.4



for x=12, have y(x)= 1.3, 1.4 and 1.9



for x=13, have y(x)= 1.8, 1.7 and 1.2



How to do list plot for many y(x) as function of x?



Use of 



ListPlot[mydata]


is no working.






plotting list-manipulation table







share|improve this question




















  • 1





    Related: (21884), (47038)

    – Mr.Wizard♦
    13 hours ago














3












3








3


1






Suppose have data for several x and several y(x), for example



mydata = {{11, 1.1, 2.3, 2.4}, {12, 1.3, 1.4, 1.9}, {13, 1.8, 1.7, 1.2}}

mydata // MatrixForm


So



for x=11, have y(x)= 1.1, 2.3 and 2.4



for x=12, have y(x)= 1.3, 1.4 and 1.9



for x=13, have y(x)= 1.8, 1.7 and 1.2



How to do list plot for many y(x) as function of x?



Use of 



ListPlot[mydata]


is no working.






plotting list-manipulation table







share|improve this question
















Suppose have data for several x and several y(x), for example



mydata = {{11, 1.1, 2.3, 2.4}, {12, 1.3, 1.4, 1.9}, {13, 1.8, 1.7, 1.2}}

mydata // MatrixForm


So



for x=11, have y(x)= 1.1, 2.3 and 2.4



for x=12, have y(x)= 1.3, 1.4 and 1.9



for x=13, have y(x)= 1.8, 1.7 and 1.2



How to do list plot for many y(x) as function of x?



Use of 



ListPlot[mydata]


is no working.






plotting list-manipulation table




plotting list-manipulation table






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 11 hours ago







Nigel1

















asked 14 hours ago









Nigel1Nigel1

1756




1756








  • 1





    Related: (21884), (47038)

    – Mr.Wizard♦
    13 hours ago














  • 1





    Related: (21884), (47038)

    – Mr.Wizard♦
    13 hours ago








1




1





Related: (21884), (47038)

– Mr.Wizard♦
13 hours ago





Related: (21884), (47038)

– Mr.Wizard♦
13 hours ago










3 Answers
3






active

oldest

votes


















4














mydata2 = Thread[{First@#, Rest@#}] & /@ mydata;

ListPlot[mydata2]


enter image description here






share|improve this answer































    4














    mydata2 = Map[(mydata[[All, {1, #}]]) &, Range[2, Last@Dimensions@mydata]];
    
ListPlot[mydata2, Joined -> True]


    enter image description here






    share|improve this answer



















    • 1





      Notice how your image is very similar to Mr.Wizard's image. Still, you image somehow appears to be of lower visual quality; it looks 'fuzzy' or something. Why is that? Well, this is because you saved it as a JPG image. JPG works well with photographic images, but introduces artefacts in illustrations, diagrams, screenshots, etc. For this kind of images, PNG is the ideal choice! :)

      – Andreas Rejbrand
      13 hours ago





















    2














    Not applicable to all data, but in this case DataRange appears useful:



    ListLinePlot[mydata[Transpose][[2 ;;]], DataRange -> {11, 13}]


    enter image description here






    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














      mydata2 = Thread[{First@#, Rest@#}] & /@ mydata;
      
ListPlot[mydata2]


      enter image description here






      share|improve this answer




























        4














        mydata2 = Thread[{First@#, Rest@#}] & /@ mydata;
        
ListPlot[mydata2]


        enter image description here






        share|improve this answer


























          4












          4








          4







          mydata2 = Thread[{First@#, Rest@#}] & /@ mydata;
          
ListPlot[mydata2]


          enter image description here






          share|improve this answer













          mydata2 = Thread[{First@#, Rest@#}] & /@ mydata;
          
ListPlot[mydata2]


          enter image description here







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 14 hours ago









          kglrkglr

          178k9198409




          178k9198409























              4














              mydata2 = Map[(mydata[[All, {1, #}]]) &, Range[2, Last@Dimensions@mydata]];
              
ListPlot[mydata2, Joined -> True]


              enter image description here






              share|improve this answer



















              • 1





                Notice how your image is very similar to Mr.Wizard's image. Still, you image somehow appears to be of lower visual quality; it looks 'fuzzy' or something. Why is that? Well, this is because you saved it as a JPG image. JPG works well with photographic images, but introduces artefacts in illustrations, diagrams, screenshots, etc. For this kind of images, PNG is the ideal choice! :)

                – Andreas Rejbrand
                13 hours ago


















              4














              mydata2 = Map[(mydata[[All, {1, #}]]) &, Range[2, Last@Dimensions@mydata]];
              
ListPlot[mydata2, Joined -> True]


              enter image description here






              share|improve this answer



















              • 1





                Notice how your image is very similar to Mr.Wizard's image. Still, you image somehow appears to be of lower visual quality; it looks 'fuzzy' or something. Why is that? Well, this is because you saved it as a JPG image. JPG works well with photographic images, but introduces artefacts in illustrations, diagrams, screenshots, etc. For this kind of images, PNG is the ideal choice! :)

                – Andreas Rejbrand
                13 hours ago
















              4












              4








              4







              mydata2 = Map[(mydata[[All, {1, #}]]) &, Range[2, Last@Dimensions@mydata]];
              
ListPlot[mydata2, Joined -> True]


              enter image description here






              share|improve this answer













              mydata2 = Map[(mydata[[All, {1, #}]]) &, Range[2, Last@Dimensions@mydata]];
              
ListPlot[mydata2, Joined -> True]


              enter image description here







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 14 hours ago









              RMMARMMA

              1,39311229




              1,39311229








              • 1





                Notice how your image is very similar to Mr.Wizard's image. Still, you image somehow appears to be of lower visual quality; it looks 'fuzzy' or something. Why is that? Well, this is because you saved it as a JPG image. JPG works well with photographic images, but introduces artefacts in illustrations, diagrams, screenshots, etc. For this kind of images, PNG is the ideal choice! :)

                – Andreas Rejbrand
                13 hours ago
















              • 1





                Notice how your image is very similar to Mr.Wizard's image. Still, you image somehow appears to be of lower visual quality; it looks 'fuzzy' or something. Why is that? Well, this is because you saved it as a JPG image. JPG works well with photographic images, but introduces artefacts in illustrations, diagrams, screenshots, etc. For this kind of images, PNG is the ideal choice! :)

                – Andreas Rejbrand
                13 hours ago










              1




              1





              Notice how your image is very similar to Mr.Wizard's image. Still, you image somehow appears to be of lower visual quality; it looks 'fuzzy' or something. Why is that? Well, this is because you saved it as a JPG image. JPG works well with photographic images, but introduces artefacts in illustrations, diagrams, screenshots, etc. For this kind of images, PNG is the ideal choice! :)

              – Andreas Rejbrand
              13 hours ago







              Notice how your image is very similar to Mr.Wizard's image. Still, you image somehow appears to be of lower visual quality; it looks 'fuzzy' or something. Why is that? Well, this is because you saved it as a JPG image. JPG works well with photographic images, but introduces artefacts in illustrations, diagrams, screenshots, etc. For this kind of images, PNG is the ideal choice! :)

              – Andreas Rejbrand
              13 hours ago













              2














              Not applicable to all data, but in this case DataRange appears useful:



              ListLinePlot[mydata[Transpose][[2 ;;]], DataRange -> {11, 13}]


              enter image description here






              share|improve this answer




























                2














                Not applicable to all data, but in this case DataRange appears useful:



                ListLinePlot[mydata[Transpose][[2 ;;]], DataRange -> {11, 13}]


                enter image description here






                share|improve this answer


























                  2












                  2








                  2







                  Not applicable to all data, but in this case DataRange appears useful:



                  ListLinePlot[mydata[Transpose][[2 ;;]], DataRange -> {11, 13}]


                  enter image description here






                  share|improve this answer













                  Not applicable to all data, but in this case DataRange appears useful:



                  ListLinePlot[mydata[Transpose][[2 ;;]], DataRange -> {11, 13}]


                  enter image description here







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 13 hours ago









                  Mr.Wizard♦Mr.Wizard

                  230k294741040




                  230k294741040






























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