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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













3












$begingroup$


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
    Mar 14 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
    Mar 14 at 6:20















3












$begingroup$


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
    Mar 14 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
    Mar 14 at 6:20













3












3








3





$begingroup$


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











share|improve this question




share|improve this question










asked Mar 14 at 5:54









AtoZAtoZ

1556




1556







  • 1




    $begingroup$
    Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> 1, 3] do what you want?
    $endgroup$
    – J. M. is slightly pensive
    Mar 14 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
    Mar 14 at 6:20












  • 1




    $begingroup$
    Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> 1, 3] do what you want?
    $endgroup$
    – J. M. is slightly pensive
    Mar 14 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
    Mar 14 at 6:20







1




1




$begingroup$
Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> 1, 3] do what you want?
$endgroup$
– J. M. is slightly pensive
Mar 14 at 6:00




$begingroup$
Does ListLinePlot[Transpose[PadRight[list1]], DataRange -> 1, 3] do what you want?
$endgroup$
– J. M. is slightly pensive
Mar 14 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
Mar 14 at 6:20




$begingroup$
@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
Mar 14 at 6:20










1 Answer
1






active

oldest

votes


















5












$begingroup$

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


enter image description here






share|improve this answer









$endgroup$












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










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

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


enter image description here






share|improve this answer









$endgroup$












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















5












$begingroup$

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


enter image description here






share|improve this answer









$endgroup$












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













5












5








5





$begingroup$

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 Mar 14 at 6:05









kglrkglr

189k10206424




189k10206424











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
















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















$begingroup$
Thanks. It works perfectly..
$endgroup$
– AtoZ
Mar 15 at 3:24




$begingroup$
Thanks. It works perfectly..
$endgroup$
– AtoZ
Mar 15 at 3:24

















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