Padding lists for accurate plottingA question about transforming one List into two Lists with additional...

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Padding lists for accurate plotting


A question about transforming one List into two Lists with additional requirementsEfficiently extracting an array subset given a separate arrayValues (or positions) of array row elements within a specified number of positions from target valueImport a column of data, make a matrix from it and export it WITHOUT curly bracesHow to map the second highest value in each row of a matrixMultiple curves plot from excelPlotting confidence region for empirical interpolated curveOpposite of Part in matrices?Trouble with exporting data with rows and columns switchedLooking for a better way use multiple pure functions to condense repetitive code













1












$begingroup$


I have the following data which is in the form of irregular/non rectangular arrays



list1 = {{1}, {2}, {3}, {4}, {5, 6, 7}, {8, 9, 10}, {11}, {12}}


To transpose it for plotting, I have to use (because of the irregular shape)



list2 = Flatten[list1, {{2}, {1}}]


This is now a $3times1$ column.
I want to plot this data, So I use the ListLinePlot as



ListLinePlot[list2, DataRange -> {1, 3}, Frame -> True] 


The three rows are plotted as three curves, but the problem is that the upper two curves which correspond to the second and third row of list2 also start from 1 on the x-axis.? Shouldn't they start from 2 instead of 1? I thought I could use PadLeft or PadRight with empty entries {} to the left or right of the last two (2 element) rows of list2 (to make them 6 element rows, like the first row of list2) to force the two curves to start from 2, but I failed. Could someone tell any workaround?










share|improve this question









$endgroup$








  • 1




    $begingroup$
    Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}] do what you want?
    $endgroup$
    – J. M. is slightly pensive
    52 mins ago










  • $begingroup$
    @J.M.isslightlypensive Thanks. But it gives zeros on right and left which actually do not do the trick, however, if they are somehow empty, would do the trick.
    $endgroup$
    – AtoZ
    32 mins ago
















1












$begingroup$


I have the following data which is in the form of irregular/non rectangular arrays



list1 = {{1}, {2}, {3}, {4}, {5, 6, 7}, {8, 9, 10}, {11}, {12}}


To transpose it for plotting, I have to use (because of the irregular shape)



list2 = Flatten[list1, {{2}, {1}}]


This is now a $3times1$ column.
I want to plot this data, So I use the ListLinePlot as



ListLinePlot[list2, DataRange -> {1, 3}, Frame -> True] 


The three rows are plotted as three curves, but the problem is that the upper two curves which correspond to the second and third row of list2 also start from 1 on the x-axis.? Shouldn't they start from 2 instead of 1? I thought I could use PadLeft or PadRight with empty entries {} to the left or right of the last two (2 element) rows of list2 (to make them 6 element rows, like the first row of list2) to force the two curves to start from 2, but I failed. Could someone tell any workaround?










share|improve this question









$endgroup$








  • 1




    $begingroup$
    Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}] do what you want?
    $endgroup$
    – J. M. is slightly pensive
    52 mins ago










  • $begingroup$
    @J.M.isslightlypensive Thanks. But it gives zeros on right and left which actually do not do the trick, however, if they are somehow empty, would do the trick.
    $endgroup$
    – AtoZ
    32 mins ago














1












1








1





$begingroup$


I have the following data which is in the form of irregular/non rectangular arrays



list1 = {{1}, {2}, {3}, {4}, {5, 6, 7}, {8, 9, 10}, {11}, {12}}


To transpose it for plotting, I have to use (because of the irregular shape)



list2 = Flatten[list1, {{2}, {1}}]


This is now a $3times1$ column.
I want to plot this data, So I use the ListLinePlot as



ListLinePlot[list2, DataRange -> {1, 3}, Frame -> True] 


The three rows are plotted as three curves, but the problem is that the upper two curves which correspond to the second and third row of list2 also start from 1 on the x-axis.? Shouldn't they start from 2 instead of 1? I thought I could use PadLeft or PadRight with empty entries {} to the left or right of the last two (2 element) rows of list2 (to make them 6 element rows, like the first row of list2) to force the two curves to start from 2, but I failed. Could someone tell any workaround?










share|improve this question









$endgroup$




I have the following data which is in the form of irregular/non rectangular arrays



list1 = {{1}, {2}, {3}, {4}, {5, 6, 7}, {8, 9, 10}, {11}, {12}}


To transpose it for plotting, I have to use (because of the irregular shape)



list2 = Flatten[list1, {{2}, {1}}]


This is now a $3times1$ column.
I want to plot this data, So I use the ListLinePlot as



ListLinePlot[list2, DataRange -> {1, 3}, Frame -> True] 


The three rows are plotted as three curves, but the problem is that the upper two curves which correspond to the second and third row of list2 also start from 1 on the x-axis.? Shouldn't they start from 2 instead of 1? I thought I could use PadLeft or PadRight with empty entries {} to the left or right of the last two (2 element) rows of list2 (to make them 6 element rows, like the first row of list2) to force the two curves to start from 2, but I failed. Could someone tell any workaround?







plotting list-manipulation






share|improve this question













share|improve this question











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share|improve this question










asked 58 mins ago









AtoZAtoZ

1436




1436








  • 1




    $begingroup$
    Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}] do what you want?
    $endgroup$
    – J. M. is slightly pensive
    52 mins ago










  • $begingroup$
    @J.M.isslightlypensive Thanks. But it gives zeros on right and left which actually do not do the trick, however, if they are somehow empty, would do the trick.
    $endgroup$
    – AtoZ
    32 mins ago














  • 1




    $begingroup$
    Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}] do what you want?
    $endgroup$
    – J. M. is slightly pensive
    52 mins ago










  • $begingroup$
    @J.M.isslightlypensive Thanks. But it gives zeros on right and left which actually do not do the trick, however, if they are somehow empty, would do the trick.
    $endgroup$
    – AtoZ
    32 mins ago








1




1




$begingroup$
Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}] do what you want?
$endgroup$
– J. M. is slightly pensive
52 mins ago




$begingroup$
Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}] do what you want?
$endgroup$
– J. M. is slightly pensive
52 mins ago












$begingroup$
@J.M.isslightlypensive Thanks. But it gives zeros on right and left which actually do not do the trick, however, if they are somehow empty, would do the trick.
$endgroup$
– AtoZ
32 mins ago




$begingroup$
@J.M.isslightlypensive Thanks. But it gives zeros on right and left which actually do not do the trick, however, if they are somehow empty, would do the trick.
$endgroup$
– AtoZ
32 mins ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

ListLinePlot[Transpose[PadRight[list1, Automatic, Null]], 
DataRange -> {1, 3}]


enter image description here






share|improve this answer









$endgroup$













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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    ListLinePlot[Transpose[PadRight[list1, Automatic, Null]], 
    DataRange -> {1, 3}]


    enter image description here






    share|improve this answer









    $endgroup$


















      3












      $begingroup$

      ListLinePlot[Transpose[PadRight[list1, Automatic, Null]], 
      DataRange -> {1, 3}]


      enter image description here






      share|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        ListLinePlot[Transpose[PadRight[list1, Automatic, Null]], 
        DataRange -> {1, 3}]


        enter image description here






        share|improve this answer









        $endgroup$



        ListLinePlot[Transpose[PadRight[list1, Automatic, Null]], 
        DataRange -> {1, 3}]


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 47 mins ago









        kglrkglr

        188k10204422




        188k10204422






























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