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How can I make a scatter plot from a matrix with many y-axis values for each x-axis value?

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3

1

$begingroup$

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

share|improve this question

edited Jun 17 at 0:32

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

asked Jun 16 at 18:30

Nigel1Nigel1

33111 silver badge66 bronze badges

$endgroup$

add a comment | 

3

1

$begingroup$

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

share|improve this question

edited Jun 17 at 0:32

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

asked Jun 16 at 18:30

Nigel1Nigel1

33111 silver badge66 bronze badges

$endgroup$

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

share|improve this question

edited Jun 17 at 0:32

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

asked Jun 16 at 18:30

Nigel1Nigel1

33111 silver badge66 bronze badges

$endgroup$

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;

ListPlot[A]

share|improve this question

edited Jun 17 at 0:32

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

asked Jun 16 at 18:30

Nigel1Nigel1

33111 silver badge66 bronze badges

share|improve this question

edited Jun 17 at 0:32

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

asked Jun 16 at 18:30

Nigel1Nigel1

33111 silver badge66 bronze badges

edited Jun 17 at 0:32

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

edited Jun 17 at 0:32

m_goldberg

90.6k88 gold badges7575 silver badges203203 bronze badges

asked Jun 16 at 18:30

Nigel1Nigel1

33111 silver badge66 bronze badges

asked Jun 16 at 18:30

Nigel1Nigel1

33111 silver badge66 bronze badges

1 Answer
1

active

oldest

votes

4

$begingroup$

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

share|improve this answer

edited Jun 16 at 19:24

answered Jun 16 at 18:38

kglrkglr

200k1010 gold badges230230 silver badges456456 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Jun 16 at 19:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JimB, thank you. Fixed the errors now.
  $endgroup$
  – kglr
  Jun 16 at 19:25
  

  
  
  
  

add a comment | 

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4

$begingroup$

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

share|improve this answer

edited Jun 16 at 19:24

answered Jun 16 at 18:38

kglrkglr

200k1010 gold badges230230 silver badges456456 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Jun 16 at 19:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JimB, thank you. Fixed the errors now.
  $endgroup$
  – kglr
  Jun 16 at 19:25
  

  
  
  
  

add a comment | 

4

$begingroup$

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

share|improve this answer

edited Jun 16 at 19:24

answered Jun 16 at 18:38

kglrkglr

200k1010 gold badges230230 silver badges456456 bronze badges

$endgroup$

• 
  
  
  

  
  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
  Jun 16 at 19:09
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JimB, thank you. Fixed the errors now.
  $endgroup$
  – kglr
  Jun 16 at 19:25
  

  
  
  
  

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

share|improve this answer

edited Jun 16 at 19:24

answered Jun 16 at 18:38

kglrkglr

200k1010 gold badges230230 silver badges456456 bronze badges

$endgroup$

ListPlot[Rest[Transpose@A], DataRange -> 0, .2, Frame -> True, Axes -> False,
 PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]

d1 = Thread[A[[All, 1]], #]& /@ Rest[Transpose@A];;
ListPlot[d1, Frame -> True, Axes->False, 
 PlotRangePadding -> Scaled[.1], BaseStyle -> PointSize[Large]]

d2 = TemporalData[Rest/@ A, A[[All,1]]];;
ListPlot[d2, Frame -> True, Axes -> False, 
 BaseStyle -> PointSize[Large], PlotRangePadding -> Scaled[.1]]

share|improve this answer

edited Jun 16 at 19:24

answered Jun 16 at 18:38

kglrkglr

200k1010 gold badges230230 silver badges456456 bronze badges

answered Jun 16 at 18:38

kglrkglr

200k1010 gold badges230230 silver badges456456 bronze badges

answered Jun 16 at 18:38

kglrkglr

200k1010 gold badges230230 silver badges456456 bronze badges