Padding lists for accurate plotting
$begingroup$
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?
plotting list-manipulation
asked yesterday
AtoZAtoZ
1556
$endgroup$
add a comment |
$begingroup$
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?
plotting list-manipulation
asked yesterday
AtoZAtoZ
1556
$endgroup$
1
$begingroup$
DoesListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}]do what you want?
$endgroup$
– J. M. is slightly pensive♦
yesterday
$begingroup$
@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
23 hours ago
add a comment |
$begingroup$
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?
plotting list-manipulation
asked yesterday
AtoZAtoZ
1556
$endgroup$
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
plotting list-manipulation
asked yesterday
AtoZAtoZ
1556
asked yesterday
AtoZAtoZ
1556
asked yesterday
AtoZAtoZ
1556
asked yesterday
AtoZAtoZ
1556
asked yesterday
AtoZAtoZ
1556
1556
1
$begingroup$
DoesListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}]do what you want?
$endgroup$
– J. M. is slightly pensive♦
yesterday
$begingroup$
@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
23 hours ago
add a comment |
1
$begingroup$
DoesListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}]do what you want?
$endgroup$
– J. M. is slightly pensive♦
yesterday
$begingroup$
@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
23 hours ago
1
1
$begingroup$
Does
ListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}] do what you want?$endgroup$
– J. M. is slightly pensive♦
yesterday
$begingroup$
Does
ListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}] do what you want?$endgroup$
– J. M. is slightly pensive♦
yesterday
$begingroup$
@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
23 hours 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
23 hours ago
add a comment |
1 Answer
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$begingroup$
ListLinePlot[Transpose[PadRight[list1, Automatic, Null]],
DataRange -> {1, 3}]
answered yesterday
kglrkglr
189k10205422
$endgroup$
$begingroup$
Thanks. It works perfectly..
$endgroup$
– AtoZ
2 hours ago
add a comment |
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$begingroup$
ListLinePlot[Transpose[PadRight[list1, Automatic, Null]],
DataRange -> {1, 3}]
answered yesterday
kglrkglr
189k10205422
$endgroup$
$begingroup$
Thanks. It works perfectly..
$endgroup$
– AtoZ
2 hours ago
add a comment |
$begingroup$
ListLinePlot[Transpose[PadRight[list1, Automatic, Null]],
DataRange -> {1, 3}]
answered yesterday
kglrkglr
189k10205422
$endgroup$
$begingroup$
Thanks. It works perfectly..
$endgroup$
– AtoZ
2 hours ago
add a comment |
$begingroup$
ListLinePlot[Transpose[PadRight[list1, Automatic, Null]],
DataRange -> {1, 3}]
answered yesterday
kglrkglr
189k10205422
$endgroup$
ListLinePlot[Transpose[PadRight[list1, Automatic, Null]],
DataRange -> {1, 3}]
answered yesterday
kglrkglr
189k10205422
answered yesterday
kglrkglr
189k10205422
answered yesterday
kglrkglr
189k10205422
answered yesterday
kglrkglr
189k10205422
189k10205422
$begingroup$
Thanks. It works perfectly..
$endgroup$
– AtoZ
2 hours ago
add a comment |
$begingroup$
Thanks. It works perfectly..
$endgroup$
– AtoZ
2 hours ago
$begingroup$
Thanks. It works perfectly..
$endgroup$
– AtoZ
2 hours ago
$begingroup$
Thanks. It works perfectly..
$endgroup$
– AtoZ
2 hours ago
add a comment |
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$begingroup$
Does
ListLinePlot[Transpose[PadRight[list1]], DataRange -> {1, 3}]do what you want?$endgroup$
– J. M. is slightly pensive♦
yesterday
$begingroup$
@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
23 hours ago