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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 $3\times1$ 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?

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    $\begingroup$ Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}] do what you want? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Mar 14, 2019 at 6:00
  • $\begingroup$ @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
    Commented Mar 14, 2019 at 6:20

1 Answer 1

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

enter image description here

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  • $\begingroup$ Thanks. It works perfectly.. $\endgroup$
    – AtoZ
    Commented Mar 15, 2019 at 3:24

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