Why Z has to be 2-dimensional for 3d plotting in matplotlib









up vote
1
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I am trying to plot 3d Surface plots using code from this site using matplotlib:



X,Y and Z are obtained as below:



from math import pi
from numpy import cos, meshgrid
alpha = 0.7
phi_ext = 2 * pi * 0.5

def flux_qubit_potential(phi_m, phi_p):
return 2 + alpha - 2 * cos(phi_p)*cos(phi_m) - alpha * cos(phi_ext - 2*phi_p)

phi_m = linspace(0, 2*pi, 100)
phi_p = linspace(0, 2*pi, 100)
X,Y = meshgrid(phi_p, phi_m)
Z = flux_qubit_potential(X, Y).T


And 3d plotting is done with following code:



from mpl_toolkits.mplot3d.axes3d import Axes3D

fig = plt.figure(figsize=(14,6))

# `ax` is a 3D-aware axis instance, because of the projection='3d' keyword argument to add_subplot
ax = fig.add_subplot(1, 2, 1, projection='3d')

p = ax.plot_surface(X, Y, Z, rstride=4, cstride=4, linewidth=0)

# surface_plot with color grading and color bar
ax = fig.add_subplot(1, 2, 2, projection='3d')
p = ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.coolwarm, linewidth=0, antialiased=False)
cb = fig.colorbar(p, shrink=0.5)


However, if I replace X,Y and Z by my x,y,z 3d data (sample give below), there is an error that Z has to be 2 dimensional. How can I plot with usual x,y,z values, as following:



 x y z
0 12 0 0.1
1 13 1 0.8
2 14 3 1.0
3 16 4 1.2
4 18 4 0.7









share|improve this question























  • What do you mean by 'my x,y,z 3d data'? There is no reference to any such thing anywhere in your code.
    – Thierry Lathuille
    Nov 11 at 8:24










  • I have added a sample of my data above.
    – rnso
    Nov 11 at 9:03










  • Concerning the "why?", Why does pyplot.contour() require Z to be a 2D array?. I would be very much inclined to close this as duplicate of Simplest way to plot 3d surface given 3d points unless the question makes clear why it isn't.
    – ImportanceOfBeingErnest
    Nov 11 at 12:08














up vote
1
down vote

favorite












I am trying to plot 3d Surface plots using code from this site using matplotlib:



X,Y and Z are obtained as below:



from math import pi
from numpy import cos, meshgrid
alpha = 0.7
phi_ext = 2 * pi * 0.5

def flux_qubit_potential(phi_m, phi_p):
return 2 + alpha - 2 * cos(phi_p)*cos(phi_m) - alpha * cos(phi_ext - 2*phi_p)

phi_m = linspace(0, 2*pi, 100)
phi_p = linspace(0, 2*pi, 100)
X,Y = meshgrid(phi_p, phi_m)
Z = flux_qubit_potential(X, Y).T


And 3d plotting is done with following code:



from mpl_toolkits.mplot3d.axes3d import Axes3D

fig = plt.figure(figsize=(14,6))

# `ax` is a 3D-aware axis instance, because of the projection='3d' keyword argument to add_subplot
ax = fig.add_subplot(1, 2, 1, projection='3d')

p = ax.plot_surface(X, Y, Z, rstride=4, cstride=4, linewidth=0)

# surface_plot with color grading and color bar
ax = fig.add_subplot(1, 2, 2, projection='3d')
p = ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.coolwarm, linewidth=0, antialiased=False)
cb = fig.colorbar(p, shrink=0.5)


However, if I replace X,Y and Z by my x,y,z 3d data (sample give below), there is an error that Z has to be 2 dimensional. How can I plot with usual x,y,z values, as following:



 x y z
0 12 0 0.1
1 13 1 0.8
2 14 3 1.0
3 16 4 1.2
4 18 4 0.7









share|improve this question























  • What do you mean by 'my x,y,z 3d data'? There is no reference to any such thing anywhere in your code.
    – Thierry Lathuille
    Nov 11 at 8:24










  • I have added a sample of my data above.
    – rnso
    Nov 11 at 9:03










  • Concerning the "why?", Why does pyplot.contour() require Z to be a 2D array?. I would be very much inclined to close this as duplicate of Simplest way to plot 3d surface given 3d points unless the question makes clear why it isn't.
    – ImportanceOfBeingErnest
    Nov 11 at 12:08












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I am trying to plot 3d Surface plots using code from this site using matplotlib:



X,Y and Z are obtained as below:



from math import pi
from numpy import cos, meshgrid
alpha = 0.7
phi_ext = 2 * pi * 0.5

def flux_qubit_potential(phi_m, phi_p):
return 2 + alpha - 2 * cos(phi_p)*cos(phi_m) - alpha * cos(phi_ext - 2*phi_p)

phi_m = linspace(0, 2*pi, 100)
phi_p = linspace(0, 2*pi, 100)
X,Y = meshgrid(phi_p, phi_m)
Z = flux_qubit_potential(X, Y).T


And 3d plotting is done with following code:



from mpl_toolkits.mplot3d.axes3d import Axes3D

fig = plt.figure(figsize=(14,6))

# `ax` is a 3D-aware axis instance, because of the projection='3d' keyword argument to add_subplot
ax = fig.add_subplot(1, 2, 1, projection='3d')

p = ax.plot_surface(X, Y, Z, rstride=4, cstride=4, linewidth=0)

# surface_plot with color grading and color bar
ax = fig.add_subplot(1, 2, 2, projection='3d')
p = ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.coolwarm, linewidth=0, antialiased=False)
cb = fig.colorbar(p, shrink=0.5)


However, if I replace X,Y and Z by my x,y,z 3d data (sample give below), there is an error that Z has to be 2 dimensional. How can I plot with usual x,y,z values, as following:



 x y z
0 12 0 0.1
1 13 1 0.8
2 14 3 1.0
3 16 4 1.2
4 18 4 0.7









share|improve this question















I am trying to plot 3d Surface plots using code from this site using matplotlib:



X,Y and Z are obtained as below:



from math import pi
from numpy import cos, meshgrid
alpha = 0.7
phi_ext = 2 * pi * 0.5

def flux_qubit_potential(phi_m, phi_p):
return 2 + alpha - 2 * cos(phi_p)*cos(phi_m) - alpha * cos(phi_ext - 2*phi_p)

phi_m = linspace(0, 2*pi, 100)
phi_p = linspace(0, 2*pi, 100)
X,Y = meshgrid(phi_p, phi_m)
Z = flux_qubit_potential(X, Y).T


And 3d plotting is done with following code:



from mpl_toolkits.mplot3d.axes3d import Axes3D

fig = plt.figure(figsize=(14,6))

# `ax` is a 3D-aware axis instance, because of the projection='3d' keyword argument to add_subplot
ax = fig.add_subplot(1, 2, 1, projection='3d')

p = ax.plot_surface(X, Y, Z, rstride=4, cstride=4, linewidth=0)

# surface_plot with color grading and color bar
ax = fig.add_subplot(1, 2, 2, projection='3d')
p = ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.coolwarm, linewidth=0, antialiased=False)
cb = fig.colorbar(p, shrink=0.5)


However, if I replace X,Y and Z by my x,y,z 3d data (sample give below), there is an error that Z has to be 2 dimensional. How can I plot with usual x,y,z values, as following:



 x y z
0 12 0 0.1
1 13 1 0.8
2 14 3 1.0
3 16 4 1.2
4 18 4 0.7






python matplotlib 3d






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 11 at 11:15

























asked Nov 11 at 8:03









rnso

11.4k134092




11.4k134092











  • What do you mean by 'my x,y,z 3d data'? There is no reference to any such thing anywhere in your code.
    – Thierry Lathuille
    Nov 11 at 8:24










  • I have added a sample of my data above.
    – rnso
    Nov 11 at 9:03










  • Concerning the "why?", Why does pyplot.contour() require Z to be a 2D array?. I would be very much inclined to close this as duplicate of Simplest way to plot 3d surface given 3d points unless the question makes clear why it isn't.
    – ImportanceOfBeingErnest
    Nov 11 at 12:08
















  • What do you mean by 'my x,y,z 3d data'? There is no reference to any such thing anywhere in your code.
    – Thierry Lathuille
    Nov 11 at 8:24










  • I have added a sample of my data above.
    – rnso
    Nov 11 at 9:03










  • Concerning the "why?", Why does pyplot.contour() require Z to be a 2D array?. I would be very much inclined to close this as duplicate of Simplest way to plot 3d surface given 3d points unless the question makes clear why it isn't.
    – ImportanceOfBeingErnest
    Nov 11 at 12:08















What do you mean by 'my x,y,z 3d data'? There is no reference to any such thing anywhere in your code.
– Thierry Lathuille
Nov 11 at 8:24




What do you mean by 'my x,y,z 3d data'? There is no reference to any such thing anywhere in your code.
– Thierry Lathuille
Nov 11 at 8:24












I have added a sample of my data above.
– rnso
Nov 11 at 9:03




I have added a sample of my data above.
– rnso
Nov 11 at 9:03












Concerning the "why?", Why does pyplot.contour() require Z to be a 2D array?. I would be very much inclined to close this as duplicate of Simplest way to plot 3d surface given 3d points unless the question makes clear why it isn't.
– ImportanceOfBeingErnest
Nov 11 at 12:08




Concerning the "why?", Why does pyplot.contour() require Z to be a 2D array?. I would be very much inclined to close this as duplicate of Simplest way to plot 3d surface given 3d points unless the question makes clear why it isn't.
– ImportanceOfBeingErnest
Nov 11 at 12:08












2 Answers
2






active

oldest

votes

















up vote
1
down vote













This is because, in my understanding, to draw a surface you need to form a polygon mesh. To draw a 3d surface, you need to have small squares, for example, on the xy-plane and then have 1 corresponding z value for all the x-y points. The smaller the area of the square means finer mesh-grid and better resolution(smooth-looking surface.) Now if you have an arbitrary set of xyz points, how matplotlib can determine which surface to draw. That is why a mesh is required. You can of course plot 3d scatter or line plots with your data.






share|improve this answer




















  • Is there any way I can plot x,y,z values as a mesh, wireframe or surface?
    – rnso
    Nov 11 at 11:19










  • Difficult because you can then, in theory, have 2 data points and could plot a surface which will have the same problem when you try to interpolate. This is essentially an interpolation in 3d. So you will then have a non-smooth surface with not a nice triangulation(or any tessellation) . Anyway you can do that stackoverflow.com/questions/12423601/… ( the 2nd answer)
    – anotherone
    Nov 11 at 11:27










  • check this for the doc: matplotlib.org/api/_as_gen/…
    – anotherone
    Nov 11 at 11:29

















up vote
1
down vote













In the documentation you will find that x, y and z need to a 2D array. For the coordinates x and y you will need to use numpy.meshgrid as you show in the first piece of code. This creates a 2D array for each coordinate where x and y are constant along the other direction and vary on its own direction.



With respect to z, this also needs to be a 2D array since Axes3D.surface_plot maps each element of the 2D array z with the 2D grid defined by x and y.



Hence, when you use your own x, y and z make sure that you use numpy.meshgrid for x and y and, then, define z = f(x,y) (e.g. the function flux_qubit_potential you show).



Edit:



After OP's comment, is clear that the desired output is a plot where the function g is g = f(x,y,z). This would mean that g is a 3D array in the end. To do this in terms of iso-surfaces have a look at these answers.






share|improve this answer






















  • By this approach, I will not be using my z values. How do I incorporate my z values?
    – rnso
    Nov 11 at 11:13










  • So is z a coordinate? The code you show is to plot z = f(x,y). Not a function, say g to which is g = f(x,y,z).
    – b-fg
    Nov 11 at 11:16










  • No, in my data, z is value along z-axis (just as x and y are values for corresponding axes).
    – rnso
    Nov 11 at 11:17











  • Ok, then you need another function, not Axes3D.surface_plot. Give me a sec and I will point you on the right direction.
    – b-fg
    Nov 11 at 11:18










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













This is because, in my understanding, to draw a surface you need to form a polygon mesh. To draw a 3d surface, you need to have small squares, for example, on the xy-plane and then have 1 corresponding z value for all the x-y points. The smaller the area of the square means finer mesh-grid and better resolution(smooth-looking surface.) Now if you have an arbitrary set of xyz points, how matplotlib can determine which surface to draw. That is why a mesh is required. You can of course plot 3d scatter or line plots with your data.






share|improve this answer




















  • Is there any way I can plot x,y,z values as a mesh, wireframe or surface?
    – rnso
    Nov 11 at 11:19










  • Difficult because you can then, in theory, have 2 data points and could plot a surface which will have the same problem when you try to interpolate. This is essentially an interpolation in 3d. So you will then have a non-smooth surface with not a nice triangulation(or any tessellation) . Anyway you can do that stackoverflow.com/questions/12423601/… ( the 2nd answer)
    – anotherone
    Nov 11 at 11:27










  • check this for the doc: matplotlib.org/api/_as_gen/…
    – anotherone
    Nov 11 at 11:29














up vote
1
down vote













This is because, in my understanding, to draw a surface you need to form a polygon mesh. To draw a 3d surface, you need to have small squares, for example, on the xy-plane and then have 1 corresponding z value for all the x-y points. The smaller the area of the square means finer mesh-grid and better resolution(smooth-looking surface.) Now if you have an arbitrary set of xyz points, how matplotlib can determine which surface to draw. That is why a mesh is required. You can of course plot 3d scatter or line plots with your data.






share|improve this answer




















  • Is there any way I can plot x,y,z values as a mesh, wireframe or surface?
    – rnso
    Nov 11 at 11:19










  • Difficult because you can then, in theory, have 2 data points and could plot a surface which will have the same problem when you try to interpolate. This is essentially an interpolation in 3d. So you will then have a non-smooth surface with not a nice triangulation(or any tessellation) . Anyway you can do that stackoverflow.com/questions/12423601/… ( the 2nd answer)
    – anotherone
    Nov 11 at 11:27










  • check this for the doc: matplotlib.org/api/_as_gen/…
    – anotherone
    Nov 11 at 11:29












up vote
1
down vote










up vote
1
down vote









This is because, in my understanding, to draw a surface you need to form a polygon mesh. To draw a 3d surface, you need to have small squares, for example, on the xy-plane and then have 1 corresponding z value for all the x-y points. The smaller the area of the square means finer mesh-grid and better resolution(smooth-looking surface.) Now if you have an arbitrary set of xyz points, how matplotlib can determine which surface to draw. That is why a mesh is required. You can of course plot 3d scatter or line plots with your data.






share|improve this answer












This is because, in my understanding, to draw a surface you need to form a polygon mesh. To draw a 3d surface, you need to have small squares, for example, on the xy-plane and then have 1 corresponding z value for all the x-y points. The smaller the area of the square means finer mesh-grid and better resolution(smooth-looking surface.) Now if you have an arbitrary set of xyz points, how matplotlib can determine which surface to draw. That is why a mesh is required. You can of course plot 3d scatter or line plots with your data.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 11 at 11:12









anotherone

397317




397317











  • Is there any way I can plot x,y,z values as a mesh, wireframe or surface?
    – rnso
    Nov 11 at 11:19










  • Difficult because you can then, in theory, have 2 data points and could plot a surface which will have the same problem when you try to interpolate. This is essentially an interpolation in 3d. So you will then have a non-smooth surface with not a nice triangulation(or any tessellation) . Anyway you can do that stackoverflow.com/questions/12423601/… ( the 2nd answer)
    – anotherone
    Nov 11 at 11:27










  • check this for the doc: matplotlib.org/api/_as_gen/…
    – anotherone
    Nov 11 at 11:29
















  • Is there any way I can plot x,y,z values as a mesh, wireframe or surface?
    – rnso
    Nov 11 at 11:19










  • Difficult because you can then, in theory, have 2 data points and could plot a surface which will have the same problem when you try to interpolate. This is essentially an interpolation in 3d. So you will then have a non-smooth surface with not a nice triangulation(or any tessellation) . Anyway you can do that stackoverflow.com/questions/12423601/… ( the 2nd answer)
    – anotherone
    Nov 11 at 11:27










  • check this for the doc: matplotlib.org/api/_as_gen/…
    – anotherone
    Nov 11 at 11:29















Is there any way I can plot x,y,z values as a mesh, wireframe or surface?
– rnso
Nov 11 at 11:19




Is there any way I can plot x,y,z values as a mesh, wireframe or surface?
– rnso
Nov 11 at 11:19












Difficult because you can then, in theory, have 2 data points and could plot a surface which will have the same problem when you try to interpolate. This is essentially an interpolation in 3d. So you will then have a non-smooth surface with not a nice triangulation(or any tessellation) . Anyway you can do that stackoverflow.com/questions/12423601/… ( the 2nd answer)
– anotherone
Nov 11 at 11:27




Difficult because you can then, in theory, have 2 data points and could plot a surface which will have the same problem when you try to interpolate. This is essentially an interpolation in 3d. So you will then have a non-smooth surface with not a nice triangulation(or any tessellation) . Anyway you can do that stackoverflow.com/questions/12423601/… ( the 2nd answer)
– anotherone
Nov 11 at 11:27












check this for the doc: matplotlib.org/api/_as_gen/…
– anotherone
Nov 11 at 11:29




check this for the doc: matplotlib.org/api/_as_gen/…
– anotherone
Nov 11 at 11:29












up vote
1
down vote













In the documentation you will find that x, y and z need to a 2D array. For the coordinates x and y you will need to use numpy.meshgrid as you show in the first piece of code. This creates a 2D array for each coordinate where x and y are constant along the other direction and vary on its own direction.



With respect to z, this also needs to be a 2D array since Axes3D.surface_plot maps each element of the 2D array z with the 2D grid defined by x and y.



Hence, when you use your own x, y and z make sure that you use numpy.meshgrid for x and y and, then, define z = f(x,y) (e.g. the function flux_qubit_potential you show).



Edit:



After OP's comment, is clear that the desired output is a plot where the function g is g = f(x,y,z). This would mean that g is a 3D array in the end. To do this in terms of iso-surfaces have a look at these answers.






share|improve this answer






















  • By this approach, I will not be using my z values. How do I incorporate my z values?
    – rnso
    Nov 11 at 11:13










  • So is z a coordinate? The code you show is to plot z = f(x,y). Not a function, say g to which is g = f(x,y,z).
    – b-fg
    Nov 11 at 11:16










  • No, in my data, z is value along z-axis (just as x and y are values for corresponding axes).
    – rnso
    Nov 11 at 11:17











  • Ok, then you need another function, not Axes3D.surface_plot. Give me a sec and I will point you on the right direction.
    – b-fg
    Nov 11 at 11:18














up vote
1
down vote













In the documentation you will find that x, y and z need to a 2D array. For the coordinates x and y you will need to use numpy.meshgrid as you show in the first piece of code. This creates a 2D array for each coordinate where x and y are constant along the other direction and vary on its own direction.



With respect to z, this also needs to be a 2D array since Axes3D.surface_plot maps each element of the 2D array z with the 2D grid defined by x and y.



Hence, when you use your own x, y and z make sure that you use numpy.meshgrid for x and y and, then, define z = f(x,y) (e.g. the function flux_qubit_potential you show).



Edit:



After OP's comment, is clear that the desired output is a plot where the function g is g = f(x,y,z). This would mean that g is a 3D array in the end. To do this in terms of iso-surfaces have a look at these answers.






share|improve this answer






















  • By this approach, I will not be using my z values. How do I incorporate my z values?
    – rnso
    Nov 11 at 11:13










  • So is z a coordinate? The code you show is to plot z = f(x,y). Not a function, say g to which is g = f(x,y,z).
    – b-fg
    Nov 11 at 11:16










  • No, in my data, z is value along z-axis (just as x and y are values for corresponding axes).
    – rnso
    Nov 11 at 11:17











  • Ok, then you need another function, not Axes3D.surface_plot. Give me a sec and I will point you on the right direction.
    – b-fg
    Nov 11 at 11:18












up vote
1
down vote










up vote
1
down vote









In the documentation you will find that x, y and z need to a 2D array. For the coordinates x and y you will need to use numpy.meshgrid as you show in the first piece of code. This creates a 2D array for each coordinate where x and y are constant along the other direction and vary on its own direction.



With respect to z, this also needs to be a 2D array since Axes3D.surface_plot maps each element of the 2D array z with the 2D grid defined by x and y.



Hence, when you use your own x, y and z make sure that you use numpy.meshgrid for x and y and, then, define z = f(x,y) (e.g. the function flux_qubit_potential you show).



Edit:



After OP's comment, is clear that the desired output is a plot where the function g is g = f(x,y,z). This would mean that g is a 3D array in the end. To do this in terms of iso-surfaces have a look at these answers.






share|improve this answer














In the documentation you will find that x, y and z need to a 2D array. For the coordinates x and y you will need to use numpy.meshgrid as you show in the first piece of code. This creates a 2D array for each coordinate where x and y are constant along the other direction and vary on its own direction.



With respect to z, this also needs to be a 2D array since Axes3D.surface_plot maps each element of the 2D array z with the 2D grid defined by x and y.



Hence, when you use your own x, y and z make sure that you use numpy.meshgrid for x and y and, then, define z = f(x,y) (e.g. the function flux_qubit_potential you show).



Edit:



After OP's comment, is clear that the desired output is a plot where the function g is g = f(x,y,z). This would mean that g is a 3D array in the end. To do this in terms of iso-surfaces have a look at these answers.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 11 at 11:51

























answered Nov 11 at 11:11









b-fg

1,47811422




1,47811422











  • By this approach, I will not be using my z values. How do I incorporate my z values?
    – rnso
    Nov 11 at 11:13










  • So is z a coordinate? The code you show is to plot z = f(x,y). Not a function, say g to which is g = f(x,y,z).
    – b-fg
    Nov 11 at 11:16










  • No, in my data, z is value along z-axis (just as x and y are values for corresponding axes).
    – rnso
    Nov 11 at 11:17











  • Ok, then you need another function, not Axes3D.surface_plot. Give me a sec and I will point you on the right direction.
    – b-fg
    Nov 11 at 11:18
















  • By this approach, I will not be using my z values. How do I incorporate my z values?
    – rnso
    Nov 11 at 11:13










  • So is z a coordinate? The code you show is to plot z = f(x,y). Not a function, say g to which is g = f(x,y,z).
    – b-fg
    Nov 11 at 11:16










  • No, in my data, z is value along z-axis (just as x and y are values for corresponding axes).
    – rnso
    Nov 11 at 11:17











  • Ok, then you need another function, not Axes3D.surface_plot. Give me a sec and I will point you on the right direction.
    – b-fg
    Nov 11 at 11:18















By this approach, I will not be using my z values. How do I incorporate my z values?
– rnso
Nov 11 at 11:13




By this approach, I will not be using my z values. How do I incorporate my z values?
– rnso
Nov 11 at 11:13












So is z a coordinate? The code you show is to plot z = f(x,y). Not a function, say g to which is g = f(x,y,z).
– b-fg
Nov 11 at 11:16




So is z a coordinate? The code you show is to plot z = f(x,y). Not a function, say g to which is g = f(x,y,z).
– b-fg
Nov 11 at 11:16












No, in my data, z is value along z-axis (just as x and y are values for corresponding axes).
– rnso
Nov 11 at 11:17





No, in my data, z is value along z-axis (just as x and y are values for corresponding axes).
– rnso
Nov 11 at 11:17













Ok, then you need another function, not Axes3D.surface_plot. Give me a sec and I will point you on the right direction.
– b-fg
Nov 11 at 11:18




Ok, then you need another function, not Axes3D.surface_plot. Give me a sec and I will point you on the right direction.
– b-fg
Nov 11 at 11:18

















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