Plot list with structure of each item x, y, z

up vote
6
down vote

favorite

Given the data below:

x, y, z = 
 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 1.0054247, 
 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 1.0398293, 
 2.1, 0.28706229,1.0712175, 2.6, 0.3643249, 1.1155575

How can I plot column x with column z and z with y?

edited 21 mins ago

m_goldberg

81.9k869190

asked 12 hours ago

Gallagher

745

add a commentÂ |Â

up vote
6
down vote

favorite

Given the data below:

x, y, z = 
 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 1.0054247, 
 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 1.0398293, 
 2.1, 0.28706229,1.0712175, 2.6, 0.3643249, 1.1155575

How can I plot column x with column z and z with y?

edited 21 mins ago

m_goldberg

81.9k869190

asked 12 hours ago

Gallagher

745

add a commentÂ |Â

up vote
6
down vote

favorite

Given the data below:

x, y, z = 
 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 1.0054247, 
 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 1.0398293, 
 2.1, 0.28706229,1.0712175, 2.6, 0.3643249, 1.1155575

How can I plot column x with column z and z with y?

edited 21 mins ago

m_goldberg

81.9k869190

asked 12 hours ago

Gallagher

745

Given the data below:

x, y, z = 
 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 1.0054247, 
 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 1.0398293, 
 2.1, 0.28706229,1.0712175, 2.6, 0.3643249, 1.1155575

How can I plot column x with column z and z with y?

plotting data

edited 21 mins ago

m_goldberg

81.9k869190

asked 12 hours ago

Gallagher

745

edited 21 mins ago

m_goldberg

81.9k869190

asked 12 hours ago

Gallagher

745

edited 21 mins ago

m_goldberg

81.9k869190

edited 21 mins ago

m_goldberg

81.9k869190

edited 21 mins ago

m_goldberg

81.9k869190

asked 12 hours ago

Gallagher

745

asked 12 hours ago

Gallagher

745

asked 12 hours ago

Gallagher

745

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
7
down vote

This is also a good place to apply patterns to re-arrange the elements of the data.

list = 0.1, 0.013070604, 1.00015, 0.6, 0.078698955,1.0054247, 
 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577,1.0398293,
 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249,1.1155575;

ListPlot[list /. x_, y_, z_ -> x, z, list /. x_, y_, z_ -> z, y]

answered 12 hours ago

bill s

50.9k373144

add a commentÂ |Â

up vote
6
down vote

How about this

data = Flatten /@ 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 
 1.0054247, 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 
 1.0398293, 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249, 
 1.1155575;

ListPlot[data[[All, 3, 1]], data[[All, 2 ;;]]]

answered 12 hours ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

1,828721

add a commentÂ |Â

up vote
2
down vote

Here is another way:

a = 
 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 
 1.0054247, 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 
 1.0398293, 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249, 
 1.1155575
 ; 
x, y, z = MapAt[Transpose, Transpose[a], 2];
ListPlot[Transpose[x, z], Transpose[z, y]]

enter image description here

answered 10 hours ago

Henrik Schumacher

38.4k253114

add a commentÂ |Â

up vote
2
down vote

Yet another way:

data2 = Flatten /@ data;
yz, xz = data2[[All, 2, 3]], data2[[All, 1, 3]] ;

ListPlot[yz , xz, 
 PlotStyle -> Red, Blue,
 PlotLegends -> "y Vs z", "x Vs z" ,
 PlotRange -> .8, 1.2,
 AxesOrigin -> 0, .8,
 AxesLabel -> Column[Style[#, 16] & /@ "x", "y"], Style[ "z", 16]]Ã‚Â

enter image description here

ListPlot[-#[[1]], #[[2]] & /@ yz , xz, 
 Ã‚Â PlotStyle -> Red, Blue,
 Ã‚Â PlotLegends -> "y Vs z", "x Vs z" ,
 Ã‚Â PlotRange -> -3, 3, .8, 1.2,
 Ã‚Â AxesOrigin -> 0, .8,
 Ã‚Â AxesLabel -> None, Style[ "z", 16], 
 Ã‚Â Epilog -> Text[Style["y", 16], Offset[-15, 0, Scaled@0, 0] ], 
 Text[Style["x", 16], Offset[15, 0, Scaled@1, 0]],
 Ã‚Â PlotRangeClipping -> False, 
 Ã‚Â ImagePadding -> Scaled[.05], 
 Ã‚Â ImageSize -> 600, 
 Ticks -> Join[Charting`FindTicks[3, 0, -3, 0][-3, 0], 
 Charting`FindTicks[0, 1, 0, 1][0, 3] ], Automatic]

enter image description here

edited 2 hours ago

answered 3 hours ago

kglr

160k8184384

add a commentÂ |Â

up vote
1
down vote

Another way

data=0.1,0.013070604,1.00015,0.6,0.078698955,1.0054247,
 1.1,0.14552025,1.0184426,1.6,0.21458577,1.0398293,
 2.1,0.28706229,1.0712175,2.6,0.3643249,1.1155575
x,y,z=#[[;;,1]],#[[;;,2,1]],#[[;;,2,2]]&@data
ListPlot[MapThread[List,#]&/@z,x,y,z]

enter image description here

answered 10 hours ago

That Gravity Guy

54637

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
7
down vote

This is also a good place to apply patterns to re-arrange the elements of the data.

list = 0.1, 0.013070604, 1.00015, 0.6, 0.078698955,1.0054247, 
 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577,1.0398293,
 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249,1.1155575;

ListPlot[list /. x_, y_, z_ -> x, z, list /. x_, y_, z_ -> z, y]

answered 12 hours ago

bill s

50.9k373144

add a commentÂ |Â

up vote
7
down vote

This is also a good place to apply patterns to re-arrange the elements of the data.

list = 0.1, 0.013070604, 1.00015, 0.6, 0.078698955,1.0054247, 
 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577,1.0398293,
 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249,1.1155575;

ListPlot[list /. x_, y_, z_ -> x, z, list /. x_, y_, z_ -> z, y]

answered 12 hours ago

bill s

50.9k373144

add a commentÂ |Â

up vote
7
down vote

This is also a good place to apply patterns to re-arrange the elements of the data.

list = 0.1, 0.013070604, 1.00015, 0.6, 0.078698955,1.0054247, 
 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577,1.0398293,
 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249,1.1155575;

ListPlot[list /. x_, y_, z_ -> x, z, list /. x_, y_, z_ -> z, y]

answered 12 hours ago

bill s

50.9k373144

This is also a good place to apply patterns to re-arrange the elements of the data.

list = 0.1, 0.013070604, 1.00015, 0.6, 0.078698955,1.0054247, 
 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577,1.0398293,
 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249,1.1155575;

ListPlot[list /. x_, y_, z_ -> x, z, list /. x_, y_, z_ -> z, y]

answered 12 hours ago

bill s

50.9k373144

answered 12 hours ago

bill s

50.9k373144

answered 12 hours ago

bill s

50.9k373144

answered 12 hours ago

bill s

50.9k373144

add a commentÂ |Â

up vote
6
down vote

How about this

data = Flatten /@ 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 
 1.0054247, 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 
 1.0398293, 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249, 
 1.1155575;

ListPlot[data[[All, 3, 1]], data[[All, 2 ;;]]]

answered 12 hours ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

1,828721

add a commentÂ |Â

up vote
6
down vote

How about this

data = Flatten /@ 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 
 1.0054247, 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 
 1.0398293, 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249, 
 1.1155575;

ListPlot[data[[All, 3, 1]], data[[All, 2 ;;]]]

answered 12 hours ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

1,828721

add a commentÂ |Â

up vote
6
down vote

How about this

data = Flatten /@ 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 
 1.0054247, 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 
 1.0398293, 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249, 
 1.1155575;

ListPlot[data[[All, 3, 1]], data[[All, 2 ;;]]]

answered 12 hours ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

1,828721

How about this

data = Flatten /@ 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 
 1.0054247, 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 
 1.0398293, 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249, 
 1.1155575;

ListPlot[data[[All, 3, 1]], data[[All, 2 ;;]]]

answered 12 hours ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

1,828721

answered 12 hours ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

1,828721

answered 12 hours ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

1,828721

answered 12 hours ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

1,828721

add a commentÂ |Â

up vote
2
down vote

Here is another way:

a = 
 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 
 1.0054247, 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 
 1.0398293, 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249, 
 1.1155575
 ; 
x, y, z = MapAt[Transpose, Transpose[a], 2];
ListPlot[Transpose[x, z], Transpose[z, y]]

enter image description here

answered 10 hours ago

Henrik Schumacher

38.4k253114

add a commentÂ |Â

up vote
2
down vote

Here is another way:

a = 
 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 
 1.0054247, 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 
 1.0398293, 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249, 
 1.1155575
 ; 
x, y, z = MapAt[Transpose, Transpose[a], 2];
ListPlot[Transpose[x, z], Transpose[z, y]]

enter image description here

answered 10 hours ago

Henrik Schumacher

38.4k253114

add a commentÂ |Â

up vote
2
down vote

Here is another way:

a = 
 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 
 1.0054247, 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 
 1.0398293, 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249, 
 1.1155575
 ; 
x, y, z = MapAt[Transpose, Transpose[a], 2];
ListPlot[Transpose[x, z], Transpose[z, y]]

enter image description here

answered 10 hours ago

Henrik Schumacher

38.4k253114

Here is another way:

a = 
 0.1, 0.013070604, 1.00015, 0.6, 0.078698955, 
 1.0054247, 1.1, 0.14552025, 1.0184426, 1.6, 0.21458577, 
 1.0398293, 2.1, 0.28706229, 1.0712175, 2.6, 0.3643249, 
 1.1155575
 ; 
x, y, z = MapAt[Transpose, Transpose[a], 2];
ListPlot[Transpose[x, z], Transpose[z, y]]

enter image description here

answered 10 hours ago

Henrik Schumacher

38.4k253114

answered 10 hours ago

Henrik Schumacher

38.4k253114

answered 10 hours ago

Henrik Schumacher

38.4k253114

answered 10 hours ago

Henrik Schumacher

38.4k253114

add a commentÂ |Â

up vote
2
down vote

Yet another way:

data2 = Flatten /@ data;
yz, xz = data2[[All, 2, 3]], data2[[All, 1, 3]] ;

ListPlot[yz , xz, 
 PlotStyle -> Red, Blue,
 PlotLegends -> "y Vs z", "x Vs z" ,
 PlotRange -> .8, 1.2,
 AxesOrigin -> 0, .8,
 AxesLabel -> Column[Style[#, 16] & /@ "x", "y"], Style[ "z", 16]]Ã‚Â

enter image description here

ListPlot[-#[[1]], #[[2]] & /@ yz , xz, 
 Ã‚Â PlotStyle -> Red, Blue,
 Ã‚Â PlotLegends -> "y Vs z", "x Vs z" ,
 Ã‚Â PlotRange -> -3, 3, .8, 1.2,
 Ã‚Â AxesOrigin -> 0, .8,
 Ã‚Â AxesLabel -> None, Style[ "z", 16], 
 Ã‚Â Epilog -> Text[Style["y", 16], Offset[-15, 0, Scaled@0, 0] ], 
 Text[Style["x", 16], Offset[15, 0, Scaled@1, 0]],
 Ã‚Â PlotRangeClipping -> False, 
 Ã‚Â ImagePadding -> Scaled[.05], 
 Ã‚Â ImageSize -> 600, 
 Ticks -> Join[Charting`FindTicks[3, 0, -3, 0][-3, 0], 
 Charting`FindTicks[0, 1, 0, 1][0, 3] ], Automatic]

enter image description here

edited 2 hours ago

answered 3 hours ago

kglr

160k8184384

add a commentÂ |Â

up vote
2
down vote

Yet another way:

data2 = Flatten /@ data;
yz, xz = data2[[All, 2, 3]], data2[[All, 1, 3]] ;

ListPlot[yz , xz, 
 PlotStyle -> Red, Blue,
 PlotLegends -> "y Vs z", "x Vs z" ,
 PlotRange -> .8, 1.2,
 AxesOrigin -> 0, .8,
 AxesLabel -> Column[Style[#, 16] & /@ "x", "y"], Style[ "z", 16]]Ã‚Â

enter image description here

ListPlot[-#[[1]], #[[2]] & /@ yz , xz, 
 Ã‚Â PlotStyle -> Red, Blue,
 Ã‚Â PlotLegends -> "y Vs z", "x Vs z" ,
 Ã‚Â PlotRange -> -3, 3, .8, 1.2,
 Ã‚Â AxesOrigin -> 0, .8,
 Ã‚Â AxesLabel -> None, Style[ "z", 16], 
 Ã‚Â Epilog -> Text[Style["y", 16], Offset[-15, 0, Scaled@0, 0] ], 
 Text[Style["x", 16], Offset[15, 0, Scaled@1, 0]],
 Ã‚Â PlotRangeClipping -> False, 
 Ã‚Â ImagePadding -> Scaled[.05], 
 Ã‚Â ImageSize -> 600, 
 Ticks -> Join[Charting`FindTicks[3, 0, -3, 0][-3, 0], 
 Charting`FindTicks[0, 1, 0, 1][0, 3] ], Automatic]

enter image description here

edited 2 hours ago

answered 3 hours ago

kglr

160k8184384

add a commentÂ |Â

up vote
2
down vote

Yet another way:

data2 = Flatten /@ data;
yz, xz = data2[[All, 2, 3]], data2[[All, 1, 3]] ;

ListPlot[yz , xz, 
 PlotStyle -> Red, Blue,
 PlotLegends -> "y Vs z", "x Vs z" ,
 PlotRange -> .8, 1.2,
 AxesOrigin -> 0, .8,
 AxesLabel -> Column[Style[#, 16] & /@ "x", "y"], Style[ "z", 16]]Ã‚Â

enter image description here

ListPlot[-#[[1]], #[[2]] & /@ yz , xz, 
 Ã‚Â PlotStyle -> Red, Blue,
 Ã‚Â PlotLegends -> "y Vs z", "x Vs z" ,
 Ã‚Â PlotRange -> -3, 3, .8, 1.2,
 Ã‚Â AxesOrigin -> 0, .8,
 Ã‚Â AxesLabel -> None, Style[ "z", 16], 
 Ã‚Â Epilog -> Text[Style["y", 16], Offset[-15, 0, Scaled@0, 0] ], 
 Text[Style["x", 16], Offset[15, 0, Scaled@1, 0]],
 Ã‚Â PlotRangeClipping -> False, 
 Ã‚Â ImagePadding -> Scaled[.05], 
 Ã‚Â ImageSize -> 600, 
 Ticks -> Join[Charting`FindTicks[3, 0, -3, 0][-3, 0], 
 Charting`FindTicks[0, 1, 0, 1][0, 3] ], Automatic]

enter image description here

edited 2 hours ago

answered 3 hours ago

kglr

160k8184384

Yet another way:

data2 = Flatten /@ data;
yz, xz = data2[[All, 2, 3]], data2[[All, 1, 3]] ;

ListPlot[yz , xz, 
 PlotStyle -> Red, Blue,
 PlotLegends -> "y Vs z", "x Vs z" ,
 PlotRange -> .8, 1.2,
 AxesOrigin -> 0, .8,
 AxesLabel -> Column[Style[#, 16] & /@ "x", "y"], Style[ "z", 16]]Ã‚Â

enter image description here

ListPlot[-#[[1]], #[[2]] & /@ yz , xz, 
 Ã‚Â PlotStyle -> Red, Blue,
 Ã‚Â PlotLegends -> "y Vs z", "x Vs z" ,
 Ã‚Â PlotRange -> -3, 3, .8, 1.2,
 Ã‚Â AxesOrigin -> 0, .8,
 Ã‚Â AxesLabel -> None, Style[ "z", 16], 
 Ã‚Â Epilog -> Text[Style["y", 16], Offset[-15, 0, Scaled@0, 0] ], 
 Text[Style["x", 16], Offset[15, 0, Scaled@1, 0]],
 Ã‚Â PlotRangeClipping -> False, 
 Ã‚Â ImagePadding -> Scaled[.05], 
 Ã‚Â ImageSize -> 600, 
 Ticks -> Join[Charting`FindTicks[3, 0, -3, 0][-3, 0], 
 Charting`FindTicks[0, 1, 0, 1][0, 3] ], Automatic]

enter image description here

edited 2 hours ago

answered 3 hours ago

kglr

160k8184384

edited 2 hours ago

answered 3 hours ago

kglr

160k8184384

answered 3 hours ago

kglr

160k8184384

answered 3 hours ago

kglr

160k8184384

add a commentÂ |Â

up vote
1
down vote

Another way

data=0.1,0.013070604,1.00015,0.6,0.078698955,1.0054247,
 1.1,0.14552025,1.0184426,1.6,0.21458577,1.0398293,
 2.1,0.28706229,1.0712175,2.6,0.3643249,1.1155575
x,y,z=#[[;;,1]],#[[;;,2,1]],#[[;;,2,2]]&@data
ListPlot[MapThread[List,#]&/@z,x,y,z]

enter image description here

answered 10 hours ago

That Gravity Guy

54637

add a commentÂ |Â

up vote
1
down vote

Another way

data=0.1,0.013070604,1.00015,0.6,0.078698955,1.0054247,
 1.1,0.14552025,1.0184426,1.6,0.21458577,1.0398293,
 2.1,0.28706229,1.0712175,2.6,0.3643249,1.1155575
x,y,z=#[[;;,1]],#[[;;,2,1]],#[[;;,2,2]]&@data
ListPlot[MapThread[List,#]&/@z,x,y,z]

enter image description here

answered 10 hours ago

That Gravity Guy

54637

add a commentÂ |Â

up vote
1
down vote

Another way

data=0.1,0.013070604,1.00015,0.6,0.078698955,1.0054247,
 1.1,0.14552025,1.0184426,1.6,0.21458577,1.0398293,
 2.1,0.28706229,1.0712175,2.6,0.3643249,1.1155575
x,y,z=#[[;;,1]],#[[;;,2,1]],#[[;;,2,2]]&@data
ListPlot[MapThread[List,#]&/@z,x,y,z]

enter image description here

answered 10 hours ago

That Gravity Guy

54637

Another way

data=0.1,0.013070604,1.00015,0.6,0.078698955,1.0054247,
 1.1,0.14552025,1.0184426,1.6,0.21458577,1.0398293,
 2.1,0.28706229,1.0712175,2.6,0.3643249,1.1155575
x,y,z=#[[;;,1]],#[[;;,2,1]],#[[;;,2,2]]&@data
ListPlot[MapThread[List,#]&/@z,x,y,z]

enter image description here

answered 10 hours ago

That Gravity Guy

54637

answered 10 hours ago

That Gravity Guy

54637

answered 10 hours ago

That Gravity Guy

54637

answered 10 hours ago

That Gravity Guy

54637

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