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How can I make a scatter plot from a matrix with many y-axis values for each x-axis value?
How can I plot more than one value in the same date?Make a scatter plot from two listsHow to plot first and second row from matrix to a graphHow to map the second highest value in each row of a matrixHow to plot multiple graphs with same x-axis values on the same graph?How to overlay ListPlot on a ContourPlot with correct range?How do I plot several y-values to 1 x-value?How can I make Plot or ListPlot scale the plot automatically for a given domain?Plot with two scales for X axisHow do I fill a table with values from my function?
$begingroup$
Consider matrix
A =
0, 2.54343, 3.566, 4.2323, 2.3434,
0.1, 6.432, 4.6465, 5.656, 1.34,
0.2, 0.3423, 1.6342, 3.323, 2.04;
The x-axis values are 0, 0.1, 0.2. There are four y-axis values for each x-axis value.
How do I plot this?
Naive use of
ListPlot[A]
fails.
plotting list-manipulation matrix
$endgroup$
add a comment |
$begingroup$
Consider matrix
A =
0, 2.54343, 3.566, 4.2323, 2.3434,
0.1, 6.432, 4.6465, 5.656, 1.34,
0.2, 0.3423, 1.6342, 3.323, 2.04;
The x-axis values are 0, 0.1, 0.2. There are four y-axis values for each x-axis value.
How do I plot this?
Naive use of
ListPlot[A]
fails.
plotting list-manipulation matrix
$endgroup$
add a comment |
$begingroup$
Consider matrix
A =
0, 2.54343, 3.566, 4.2323, 2.3434,
0.1, 6.432, 4.6465, 5.656, 1.34,
0.2, 0.3423, 1.6342, 3.323, 2.04;
The x-axis values are 0, 0.1, 0.2. There are four y-axis values for each x-axis value.
How do I plot this?
Naive use of
ListPlot[A]
fails.
plotting list-manipulation matrix
$endgroup$
Consider matrix
A =
0, 2.54343, 3.566, 4.2323, 2.3434,
0.1, 6.432, 4.6465, 5.656, 1.34,
0.2, 0.3423, 1.6342, 3.323, 2.04;
The x-axis values are 0, 0.1, 0.2. There are four y-axis values for each x-axis value.
How do I plot this?
Naive use of
ListPlot[A]
fails.
plotting list-manipulation matrix
plotting list-manipulation matrix
edited 2 hours ago


m_goldberg
90.6k873203
90.6k873203
asked 8 hours ago
Nigel1Nigel1
32616
32616
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Transpose
the list A
, remove its first list and use the option DataRange
:
ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]
Alternatively, create new input list with explicit x values for each point:
d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
ListPlot[d1, Frame -> True, Axes->False,
PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]
same picture
You can also use TemporalData
using 0, .1, .2
as the common time stamps:
d2 = TemporalData[Rest/@ A, A[[All,1]]];;
ListPlot[d2, Frame -> True, Axes -> False,
BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]
same picture
$endgroup$
1
$begingroup$
I'm guessing that the OP hasA[[All,1]]
as the x-values withA[[All,2,3,4,5]]
as the y-values.
$endgroup$
– JimB
7 hours ago
1
$begingroup$
@JimB, thank you. Fixed the errors now.
$endgroup$
– kglr
7 hours ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Transpose
the list A
, remove its first list and use the option DataRange
:
ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]
Alternatively, create new input list with explicit x values for each point:
d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
ListPlot[d1, Frame -> True, Axes->False,
PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]
same picture
You can also use TemporalData
using 0, .1, .2
as the common time stamps:
d2 = TemporalData[Rest/@ A, A[[All,1]]];;
ListPlot[d2, Frame -> True, Axes -> False,
BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]
same picture
$endgroup$
1
$begingroup$
I'm guessing that the OP hasA[[All,1]]
as the x-values withA[[All,2,3,4,5]]
as the y-values.
$endgroup$
– JimB
7 hours ago
1
$begingroup$
@JimB, thank you. Fixed the errors now.
$endgroup$
– kglr
7 hours ago
add a comment |
$begingroup$
Transpose
the list A
, remove its first list and use the option DataRange
:
ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]
Alternatively, create new input list with explicit x values for each point:
d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
ListPlot[d1, Frame -> True, Axes->False,
PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]
same picture
You can also use TemporalData
using 0, .1, .2
as the common time stamps:
d2 = TemporalData[Rest/@ A, A[[All,1]]];;
ListPlot[d2, Frame -> True, Axes -> False,
BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]
same picture
$endgroup$
1
$begingroup$
I'm guessing that the OP hasA[[All,1]]
as the x-values withA[[All,2,3,4,5]]
as the y-values.
$endgroup$
– JimB
7 hours ago
1
$begingroup$
@JimB, thank you. Fixed the errors now.
$endgroup$
– kglr
7 hours ago
add a comment |
$begingroup$
Transpose
the list A
, remove its first list and use the option DataRange
:
ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]
Alternatively, create new input list with explicit x values for each point:
d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
ListPlot[d1, Frame -> True, Axes->False,
PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]
same picture
You can also use TemporalData
using 0, .1, .2
as the common time stamps:
d2 = TemporalData[Rest/@ A, A[[All,1]]];;
ListPlot[d2, Frame -> True, Axes -> False,
BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]
same picture
$endgroup$
Transpose
the list A
, remove its first list and use the option DataRange
:
ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]
Alternatively, create new input list with explicit x values for each point:
d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
ListPlot[d1, Frame -> True, Axes->False,
PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]
same picture
You can also use TemporalData
using 0, .1, .2
as the common time stamps:
d2 = TemporalData[Rest/@ A, A[[All,1]]];;
ListPlot[d2, Frame -> True, Axes -> False,
BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]
same picture
edited 7 hours ago
answered 7 hours ago
kglrkglr
199k10224452
199k10224452
1
$begingroup$
I'm guessing that the OP hasA[[All,1]]
as the x-values withA[[All,2,3,4,5]]
as the y-values.
$endgroup$
– JimB
7 hours ago
1
$begingroup$
@JimB, thank you. Fixed the errors now.
$endgroup$
– kglr
7 hours ago
add a comment |
1
$begingroup$
I'm guessing that the OP hasA[[All,1]]
as the x-values withA[[All,2,3,4,5]]
as the y-values.
$endgroup$
– JimB
7 hours ago
1
$begingroup$
@JimB, thank you. Fixed the errors now.
$endgroup$
– kglr
7 hours ago
1
1
$begingroup$
I'm guessing that the OP has
A[[All,1]]
as the x-values with A[[All,2,3,4,5]]
as the y-values.$endgroup$
– JimB
7 hours ago
$begingroup$
I'm guessing that the OP has
A[[All,1]]
as the x-values with A[[All,2,3,4,5]]
as the y-values.$endgroup$
– JimB
7 hours ago
1
1
$begingroup$
@JimB, thank you. Fixed the errors now.
$endgroup$
– kglr
7 hours ago
$begingroup$
@JimB, thank you. Fixed the errors now.
$endgroup$
– kglr
7 hours ago
add a comment |
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