Padding lists for accurate plottingA 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
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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
$begingroup$
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 1 hour ago
AtoZAtoZ
1436
$endgroup$
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 1 hour ago
AtoZAtoZ
1436
$endgroup$
1
$begingroup$
DoesListLinePlot[Transpose[PadRight[list1]], DataRange -> 1, 3]do what you want?
$endgroup$
– J. M. is slightly pensive♦
53 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
33 mins 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 1 hour ago
AtoZAtoZ
1436
$endgroup$
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 1 hour ago
AtoZAtoZ
1436
asked 1 hour ago
AtoZAtoZ
1436
asked 1 hour ago
AtoZAtoZ
1436
asked 1 hour ago
AtoZAtoZ
1436
asked 1 hour ago
AtoZAtoZ
1436
1436
1
$begingroup$
DoesListLinePlot[Transpose[PadRight[list1]], DataRange -> 1, 3]do what you want?
$endgroup$
– J. M. is slightly pensive♦
53 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
33 mins ago
add a comment |
1
$begingroup$
DoesListLinePlot[Transpose[PadRight[list1]], DataRange -> 1, 3]do what you want?
$endgroup$
– J. M. is slightly pensive♦
53 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
33 mins ago
1
1
$begingroup$
Does
ListLinePlot[Transpose[PadRight[list1]], DataRange -> 1, 3] do what you want?$endgroup$
– J. M. is slightly pensive♦
53 mins ago
$begingroup$
Does
ListLinePlot[Transpose[PadRight[list1]], DataRange -> 1, 3] do what you want?$endgroup$
– J. M. is slightly pensive♦
53 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
33 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
33 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
ListLinePlot[Transpose[PadRight[list1, Automatic, Null]],
DataRange -> 1, 3]
answered 49 mins ago
kglrkglr
188k10204422
$endgroup$
add a comment |
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$begingroup$
ListLinePlot[Transpose[PadRight[list1, Automatic, Null]],
DataRange -> 1, 3]
answered 49 mins ago
kglrkglr
188k10204422
$endgroup$
add a comment |
$begingroup$
ListLinePlot[Transpose[PadRight[list1, Automatic, Null]],
DataRange -> 1, 3]
answered 49 mins ago
kglrkglr
188k10204422
$endgroup$
add a comment |
$begingroup$
ListLinePlot[Transpose[PadRight[list1, Automatic, Null]],
DataRange -> 1, 3]
answered 49 mins ago
kglrkglr
188k10204422
$endgroup$
ListLinePlot[Transpose[PadRight[list1, Automatic, Null]],
DataRange -> 1, 3]
answered 49 mins ago
kglrkglr
188k10204422
answered 49 mins ago
kglrkglr
188k10204422
answered 49 mins ago
kglrkglr
188k10204422
answered 49 mins ago
kglrkglr
188k10204422
188k10204422
add a comment |
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♦
53 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
33 mins ago