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How can I rotate a graph that contains an imshow in matplotlib?

I have some microscopic scan data in a rectangular grid that was scanned by an X/Y scanner. The object I'm scanning contains markers in a nice, rectangular, orthogonal pattern on its surface.
Due to practicalities, we can only align the object to the X/Y scanner to within a few degrees when fixing it to the X/Y scanner table. We compensate for this by some calibration measurement, and then performing scans that are nicely aligned with the object we're trying to scan (by moving the X and Y axes simultaneously).
This works well, but now I want to display this data in a coordinate system that corresponds to the X/Y scanner's axes. So the scanned, rectangular dataset will be rotated a bit in that coordinate system. (See the second image below for what I mean).
For presentation purposes however, I would prefer to have the scanned image nicely horizontal, and the axes rotated. I am trying to accomplish this in Matplotlib but I'm failing miserably.
I can get a rotated Axes (see the third image), but I am unable to figure out what kind of transforms I should do to get the ax3.imshow() to show its data at its intended position.
Here's where I am so far:
The program to generate the test data and the image I have so far is shown below.
I'd appreciate any help I can get. I'm afraid to say the documentation of matplotlib leaves a lot to be desired, so I hope somebody who knows matplotlib inside out may chime in.
#! /usr/bin/env python3
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.transforms import Affine2D
import mpl_toolkits.axisartist.floating_axes as floating_axes
import random
angle_deg = 5.0 # degrees CCW.
# Test data
NX = 5
NY = 3
x = np.zeros(shape=(NY, NX))
y = np.zeros(shape=(NY, NX))
c = np.zeros(shape=(NY, NX))
angle_rad = np.deg2rad(angle_deg)
for iy in range(NY):
for ix in range(NX):
icx = 0.5 * (NX - 1)
icy = 0.5 * (NY - 1)
# To centered coordinates, in physical units
xcp = (ix - icx) * 0.1
ycp = (iy - icy) * 0.1
# Rotate
xcpr = xcp * np.cos(angle_rad) - ycp * np.sin(angle_rad)
ycpr = xcp * np.sin(angle_rad) + ycp * np.cos(angle_rad)
# Rotate, and translated to the physical center
x[iy, ix] = xcpr + 40.0
y[iy, ix] = ycpr + 30.0
c[iy, ix] = 100 + ix + iy
fig = plt.figure()
fig.set_size_inches(15, 9)
# Add axis #1.
ax1 = fig.add_subplot(131)
ax1.set_title("imshow()")
ax1.imshow(c, origin='lower')
ax1.grid()
# Add axis #2.
ax2 = fig.add_subplot(132)
ax2.set_title("pcolormesh()")
ax2.pcolormesh(x, y, c, shading='nearest')
ax2.axis('equal')
ax2.grid()
# Add axis #3.
min_x = x.min()
max_x = x.max()
min_y = y.min()
max_y = y.max()
cx = 0.5 * (min_x + max_x)
cy = 0.5 * (min_y + max_y)
transform = Affine2D().rotate_deg(-angle_deg)
grid_helper = floating_axes.GridHelperCurveLinear(transform, extremes=(min_x, max_x, min_y, max_y))
ax3 = floating_axes.FloatingSubplot(fig, 133, grid_helper=grid_helper)
ax3.grid()
ax3.set_title("rotate the entire graph.\nHow do I do get the data to show up here,\nhorizontally like in the leftmost image!?\n")
ax3.imshow(c, origin='lower')
fig.add_subplot(ax3)
plt.show()

Using a linear equation to create a radially interpolated circle

Given a linear equation, I want to use the slope to create a circle of values around a given point, defined by the slope of the linear equation if possible
Im currently a bit far away - can only make the radial plot but do not know how to connect this with an input equation. My first thought would be to change the opacity using import matplotlib.animation as animation and looping matplotlib's alpha argument to become gradually more and more opaque. However the alpha doesnt seem to change opacity.
Code:
# lenth of radius
distance = 200
# create radius
radialVals = np.linspace(0,distance)
# 2 pi radians = full circle
azm = np.linspace(0, 2 * np.pi)
r, th = np.meshgrid(radialVals, azm)
z = (r ** 2.0) / 4.0
# creates circle
plt.subplot(projection="polar")
# add color gradient
plt.pcolormesh(th, r, z)
plt.plot(azm, r,alpha=1, ls='', drawstyle = 'steps')
#gridlines
# plt.grid()
plt.show()

Here is one way to solve it, the idea is to create a mesh, calculate the colors with a function then use imshow to visualize the mesh.
from matplotlib import pyplot as plt
import numpy as np
def create_mesh(slope,center,radius,t_x,t_y,ax,xlim,ylim):
"""
slope: the slope of the linear function
center: the center of the circle
raadius: the radius of the circle
t_x: the number of grids in x direction
t_y: the number of grids in y direction
ax: the canvas
xlim,ylim: the lims of the ax
"""
def cart2pol(x,y):
rho = np.sqrt(x**2 + y**2)
phi = np.arctan2(y,x)
return rho,phi
def linear_func(slope):
# initialize a patch and grids
patch = np.empty((t_x,t_y))
patch[:,:] = np.nan
x = np.linspace(xlim[0],xlim[1],t_x)
y = np.linspace(ylim[0],ylim[1],t_y)
x_grid,y_grid = np.meshgrid(x, y)
# centered grid
xc = np.linspace(xlim[0]-center[0],xlim[1]-center[0],t_x)
yc = np.linspace(ylim[0]-center[1],ylim[1]-center[1],t_y)
xc_grid,yc_grid = np.meshgrid(xc, yc)
rho,phi = cart2pol(xc_grid,yc_grid)
linear_values = slope * rho
# threshold controls the size of the gaussian
circle_mask = (x_grid-center[0])**2 + (y_grid-center[1])**2 < radius
patch[circle_mask] = linear_values[circle_mask]
return patch
# modify the patch
patch = linear_func(slope)
extent = xlim[0],xlim[1],ylim[0],ylim[1]
ax.imshow(patch,alpha=.6,interpolation='bilinear',extent=extent,
cmap=plt.cm.YlGn,vmin=v_min,vmax=v_max)
fig,ax = plt.subplots(nrows=1,ncols=2,figsize=(12,6))
slopes = [40,30]
centroids = [[2,2],[4,3]]
radii = [1,4]
for item in ax:item.set_xlim(0,8);item.set_ylim(0,8)
v_max,v_min = max(slopes),0
create_mesh(slopes[0],centroids[0],radii[0],t_x=300,t_y=300,ax=ax[0],xlim=(0,8),ylim=(0,8))
create_mesh(slopes[1],centroids[1],radii[1],t_x=300,t_y=300,ax=ax[1],xlim=(0,8),ylim=(0,8))
plt.show()
The output of this code is
As you can see, the color gradient of the figure on the left is not as sharp as the figure on the right because of the different slopes ([40,30]).
Also note that, these two lines of code
v_max,v_min = max(slopes),0
ax.imshow(patch,alpha=.6,interpolation='bilinear',extent=extent,
cmap=plt.cm.YlGn,vmin=v_min,vmax=v_max)
are added in order to let the two subplots share the same colormap.

How to change the length of axes for 3D plots in matplotlib [duplicate]

I have this so far:
x,y,z = data.nonzero()
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(x, y, z, zdir='z', c= 'red')
plt.savefig("plot.png")
Which creates:
What I'd like to do is stretch this out to make the Z axis 9 times taller and keep X and Y the same. I'd like to keep the same coordinates though.
So far I tried this guy:
fig = plt.figure(figsize=(4.,35.))
But that just stretches out the plot.png image.

The code example below provides a way to scale each axis relative to the others. However, to do so you need to modify the Axes3D.get_proj function. Below is an example based on the example provided by matplot lib: http://matplotlib.org/1.4.0/mpl_toolkits/mplot3d/tutorial.html#line-plots
(There is a shorter version at the end of this answer)
from mpl_toolkits.mplot3d.axes3d import Axes3D
from mpl_toolkits.mplot3d import proj3d
import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt
#Make sure these are floating point values:
scale_x = 1.0
scale_y = 2.0
scale_z = 3.0
#Axes are scaled down to fit in scene
max_scale=max(scale_x, scale_y, scale_z)
scale_x=scale_x/max_scale
scale_y=scale_y/max_scale
scale_z=scale_z/max_scale
#Create scaling matrix
scale = np.array([[scale_x,0,0,0],
[0,scale_y,0,0],
[0,0,scale_z,0],
[0,0,0,1]])
print scale
def get_proj_scale(self):
"""
Create the projection matrix from the current viewing position.
elev stores the elevation angle in the z plane
azim stores the azimuth angle in the x,y plane
dist is the distance of the eye viewing point from the object
point.
"""
relev, razim = np.pi * self.elev/180, np.pi * self.azim/180
xmin, xmax = self.get_xlim3d()
ymin, ymax = self.get_ylim3d()
zmin, zmax = self.get_zlim3d()
# transform to uniform world coordinates 0-1.0,0-1.0,0-1.0
worldM = proj3d.world_transformation(
xmin, xmax,
ymin, ymax,
zmin, zmax)
# look into the middle of the new coordinates
R = np.array([0.5, 0.5, 0.5])
xp = R[0] + np.cos(razim) * np.cos(relev) * self.dist
yp = R[1] + np.sin(razim) * np.cos(relev) * self.dist
zp = R[2] + np.sin(relev) * self.dist
E = np.array((xp, yp, zp))
self.eye = E
self.vvec = R - E
self.vvec = self.vvec / proj3d.mod(self.vvec)
if abs(relev) > np.pi/2:
# upside down
V = np.array((0, 0, -1))
else:
V = np.array((0, 0, 1))
zfront, zback = -self.dist, self.dist
viewM = proj3d.view_transformation(E, R, V)
perspM = proj3d.persp_transformation(zfront, zback)
M0 = np.dot(viewM, worldM)
M = np.dot(perspM, M0)
return np.dot(M, scale);
Axes3D.get_proj=get_proj_scale
"""
You need to include all the code above.
From here on you should be able to plot as usual.
"""
mpl.rcParams['legend.fontsize'] = 10
fig = plt.figure(figsize=(5,5))
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
ax.plot(x, y, z, label='parametric curve')
ax.legend()
plt.show()
Standard output:
Scaled by (1, 2, 3):
Scaled by (1, 1, 3):
The reason I particularly like this method,
Swap z and x, scale by (3, 1, 1):
Below is a shorter version of the code.
from mpl_toolkits.mplot3d.axes3d import Axes3D
from mpl_toolkits.mplot3d import proj3d
import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt
mpl.rcParams['legend.fontsize'] = 10
fig = plt.figure(figsize=(5,5))
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
"""
Scaling is done from here...
"""
x_scale=1
y_scale=1
z_scale=2
scale=np.diag([x_scale, y_scale, z_scale, 1.0])
scale=scale*(1.0/scale.max())
scale[3,3]=1.0
def short_proj():
return np.dot(Axes3D.get_proj(ax), scale)
ax.get_proj=short_proj
"""
to here
"""
ax.plot(z, y, x, label='parametric curve')
ax.legend()
plt.show()

Please note that the answer below simplifies the patch, but uses the same underlying principle as the answer by #ChristianSarofeen.
Solution
As already indicated in other answers, it is not a feature that is currently implemented in matplotlib. However, since what you are requesting is simply a 3D transformation that can be applied to the existing projection matrix used by matplotlib, and thanks to the wonderful features of Python, this problem can be solved with a simple oneliner:
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([scale_x, scale_y, scale_z, 1]))
where scale_x, scale_y and scale_z are values from 0 to 1 that will re-scale your plot along each of the axes accordingly. ax is simply the 3D axes which can be obtained with ax = fig.gca(projection='3d')
Explanation
To explain, the function get_proj of Axes3D generates the projection matrix from the current viewing position. Multiplying it by a scaling matrix:
scale_x, 0, 0
0, scale_y, 0
0, 0, scale_z
0, 0, 1
includes the scaling into the projection used by the renderer. So, what we are doing here is substituting the original get_proj function with an expression taking the result of the original get_proj and multiplying it by the scaling matrix.
Example
To illustrate the result with the standard parametric function example:
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z ** 2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
# OUR ONE LINER ADDED HERE:
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([0.5, 0.5, 1, 1]))
ax.plot(x, y, z)
plt.show()
for values 0.5, 0.5, 1, we get:
while for values 0.2, 1.0, 0.2, we get:

In my case I wanted to stretch z-axis 2 times for better point visibility
from mpl_toolkits import mplot3d
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
# plt.rcParams["figure.figsize"] = (10,200)
# plt.rcParams["figure.autolayout"] = True
ax = plt.axes(projection='3d')
ax.set_box_aspect(aspect = (1,1,2))
ax.plot(dataX,dataY,dataZ)

I looks like by default, mplot3d will leave quite a bit of room at the top and bottom of a very tall plot. But, you can trick it into filling that space using fig.subplots_adjust, and extending the top and bottom out of the normal plotting area (i.e. top > 1 and bottom < 0). Some trial and error here is probably needed for your particular plot.
I've created some random arrays for x, y, and z with limits similar to your plot, and have found the parameters below (bottom=-0.15, top = 1.2) seem to work ok.
You might also want to change ax.view_init to set a nice viewing angle.
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d
from numpy import random
# Make some random data with similar limits to the OP's example
x,y,z=random.rand(3,100)
z*=250
y*=800
y+=900
x*=350
x+=1200
fig=plt.figure(figsize=(4,35))
# Set the bottom and top outside the actual figure limits,
# to stretch the 3D axis
fig.subplots_adjust(bottom=-0.15,top=1.2)
ax = fig.add_subplot(111, projection='3d')
# Change the viewing angle to an agreeable one
ax.view_init(2,None)
ax.scatter(x, y, z, zdir='z', c= 'red')
plt.savefig("plot.png")

Sounds like you're trying to adjust the scale of the plot. I don't think there's a way to stretch a linear scale to user specifications, but you can use set_yscale(), set_xscale(), set_zscale() to alter the scales with respect to each other.
Intuitively, set_yscale(log), set_xscale(log), set_zscale(linear) might solve your problems.
A likely better option: specify a stretch, set them all to symlog with the same log base and then specify the Z-axis's symlog scale with the linscalex/linscaley kwargs to your specifications.
More here:
http://matplotlib.org/mpl_toolkits/mplot3d/api.html

I found this while searching on a similar problem. After experimenting a bit, perhaps I can share some of my prelim findings here..matplotlib library is VAST!! (am a newcomer). Note that quite akin to this question, all i wanted was to 'visually' stretch the chart without distorting it.
Background story (only key code snippets are shown to avoid unnecessary clutter for those who know the library, and if you want a run-able code please drop a comment):
I have three 1-d ndarrays representing the X,Y and Z data points respectively. Clearly I can't use plot_surface (as it requires 2d ndarrays for each dim) so I went for the extremely useful plot_trisurf:
fig = plt.figure()
ax = Axes3D(fig)
3d_surf_obj = ax.plot_trisurf(X, Y, Z_defl, cmap=cm.jet,linewidth=0,antialiased=True)
You can think of the plot like a floating barge deforming in waves...As you can see, the axes stretch make it pretty deceiving visually (note that x is supposed to be at x6 times longer than y and >>>>> z). While the plot points are correct, I wanted something more visually 'stretched' at the very least. Was looking for A QUICK FIX, if I may. Long story cut short, I found a bit of success with...'figure.figsize' general setting (see snippet below).
matplotlib.rcParams.update({'font.serif': 'Times New Roman',
'font.size': 10.0,
'axes.labelsize': 'Medium',
'axes.labelweight': 'normal',
'axes.linewidth': 0.8,
###########################################
# THIS IS THE IMPORTANT ONE FOR STRETCHING
# default is [6,4] but...i changed it to
'figure.figsize':[15,5] # THIS ONE #
})
For [15,5] I got something like...
Pretty neat!!
So I started to push it.... and got up to [20,6] before deciding to settle there..
If you want to try for visually stretching the vertical axis, try with ratios like... [7,10], which in this case gives me ...
Not too shabby !
Should do it for visual prowess.

Multiply all your z values by 9,
ax.scatter(x, y, 9*z, zdir='z', c= 'red')
And then give the z-axis custom plot labels and spacing.
ax.ZTick = [0,-9*50, -9*100, -9*150, -9*200];
ax.ZTickLabel = {'0','-50','-100','-150','-200'};

matplotlib (mplot3d) - how to increase the size of an axis (stretch) in a 3D Plot?

I have this so far:
x,y,z = data.nonzero()
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(x, y, z, zdir='z', c= 'red')
plt.savefig("plot.png")
Which creates:
What I'd like to do is stretch this out to make the Z axis 9 times taller and keep X and Y the same. I'd like to keep the same coordinates though.
So far I tried this guy:
fig = plt.figure(figsize=(4.,35.))
But that just stretches out the plot.png image.

The code example below provides a way to scale each axis relative to the others. However, to do so you need to modify the Axes3D.get_proj function. Below is an example based on the example provided by matplot lib: http://matplotlib.org/1.4.0/mpl_toolkits/mplot3d/tutorial.html#line-plots
(There is a shorter version at the end of this answer)
from mpl_toolkits.mplot3d.axes3d import Axes3D
from mpl_toolkits.mplot3d import proj3d
import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt
#Make sure these are floating point values:
scale_x = 1.0
scale_y = 2.0
scale_z = 3.0
#Axes are scaled down to fit in scene
max_scale=max(scale_x, scale_y, scale_z)
scale_x=scale_x/max_scale
scale_y=scale_y/max_scale
scale_z=scale_z/max_scale
#Create scaling matrix
scale = np.array([[scale_x,0,0,0],
[0,scale_y,0,0],
[0,0,scale_z,0],
[0,0,0,1]])
print scale
def get_proj_scale(self):
"""
Create the projection matrix from the current viewing position.
elev stores the elevation angle in the z plane
azim stores the azimuth angle in the x,y plane
dist is the distance of the eye viewing point from the object
point.
"""
relev, razim = np.pi * self.elev/180, np.pi * self.azim/180
xmin, xmax = self.get_xlim3d()
ymin, ymax = self.get_ylim3d()
zmin, zmax = self.get_zlim3d()
# transform to uniform world coordinates 0-1.0,0-1.0,0-1.0
worldM = proj3d.world_transformation(
xmin, xmax,
ymin, ymax,
zmin, zmax)
# look into the middle of the new coordinates
R = np.array([0.5, 0.5, 0.5])
xp = R[0] + np.cos(razim) * np.cos(relev) * self.dist
yp = R[1] + np.sin(razim) * np.cos(relev) * self.dist
zp = R[2] + np.sin(relev) * self.dist
E = np.array((xp, yp, zp))
self.eye = E
self.vvec = R - E
self.vvec = self.vvec / proj3d.mod(self.vvec)
if abs(relev) > np.pi/2:
# upside down
V = np.array((0, 0, -1))
else:
V = np.array((0, 0, 1))
zfront, zback = -self.dist, self.dist
viewM = proj3d.view_transformation(E, R, V)
perspM = proj3d.persp_transformation(zfront, zback)
M0 = np.dot(viewM, worldM)
M = np.dot(perspM, M0)
return np.dot(M, scale);
Axes3D.get_proj=get_proj_scale
"""
You need to include all the code above.
From here on you should be able to plot as usual.
"""
mpl.rcParams['legend.fontsize'] = 10
fig = plt.figure(figsize=(5,5))
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
ax.plot(x, y, z, label='parametric curve')
ax.legend()
plt.show()
Standard output:
Scaled by (1, 2, 3):
Scaled by (1, 1, 3):
The reason I particularly like this method,
Swap z and x, scale by (3, 1, 1):
Below is a shorter version of the code.
from mpl_toolkits.mplot3d.axes3d import Axes3D
from mpl_toolkits.mplot3d import proj3d
import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt
mpl.rcParams['legend.fontsize'] = 10
fig = plt.figure(figsize=(5,5))
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
"""
Scaling is done from here...
"""
x_scale=1
y_scale=1
z_scale=2
scale=np.diag([x_scale, y_scale, z_scale, 1.0])
scale=scale*(1.0/scale.max())
scale[3,3]=1.0
def short_proj():
return np.dot(Axes3D.get_proj(ax), scale)
ax.get_proj=short_proj
"""
to here
"""
ax.plot(z, y, x, label='parametric curve')
ax.legend()
plt.show()

Please note that the answer below simplifies the patch, but uses the same underlying principle as the answer by #ChristianSarofeen.
Solution
As already indicated in other answers, it is not a feature that is currently implemented in matplotlib. However, since what you are requesting is simply a 3D transformation that can be applied to the existing projection matrix used by matplotlib, and thanks to the wonderful features of Python, this problem can be solved with a simple oneliner:
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([scale_x, scale_y, scale_z, 1]))
where scale_x, scale_y and scale_z are values from 0 to 1 that will re-scale your plot along each of the axes accordingly. ax is simply the 3D axes which can be obtained with ax = fig.gca(projection='3d')
Explanation
To explain, the function get_proj of Axes3D generates the projection matrix from the current viewing position. Multiplying it by a scaling matrix:
scale_x, 0, 0
0, scale_y, 0
0, 0, scale_z
0, 0, 1
includes the scaling into the projection used by the renderer. So, what we are doing here is substituting the original get_proj function with an expression taking the result of the original get_proj and multiplying it by the scaling matrix.
Example
To illustrate the result with the standard parametric function example:
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z ** 2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
# OUR ONE LINER ADDED HERE:
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([0.5, 0.5, 1, 1]))
ax.plot(x, y, z)
plt.show()
for values 0.5, 0.5, 1, we get:
while for values 0.2, 1.0, 0.2, we get:

In my case I wanted to stretch z-axis 2 times for better point visibility
from mpl_toolkits import mplot3d
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
# plt.rcParams["figure.figsize"] = (10,200)
# plt.rcParams["figure.autolayout"] = True
ax = plt.axes(projection='3d')
ax.set_box_aspect(aspect = (1,1,2))
ax.plot(dataX,dataY,dataZ)

I looks like by default, mplot3d will leave quite a bit of room at the top and bottom of a very tall plot. But, you can trick it into filling that space using fig.subplots_adjust, and extending the top and bottom out of the normal plotting area (i.e. top > 1 and bottom < 0). Some trial and error here is probably needed for your particular plot.
I've created some random arrays for x, y, and z with limits similar to your plot, and have found the parameters below (bottom=-0.15, top = 1.2) seem to work ok.
You might also want to change ax.view_init to set a nice viewing angle.
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d
from numpy import random
# Make some random data with similar limits to the OP's example
x,y,z=random.rand(3,100)
z*=250
y*=800
y+=900
x*=350
x+=1200
fig=plt.figure(figsize=(4,35))
# Set the bottom and top outside the actual figure limits,
# to stretch the 3D axis
fig.subplots_adjust(bottom=-0.15,top=1.2)
ax = fig.add_subplot(111, projection='3d')
# Change the viewing angle to an agreeable one
ax.view_init(2,None)
ax.scatter(x, y, z, zdir='z', c= 'red')
plt.savefig("plot.png")

Sounds like you're trying to adjust the scale of the plot. I don't think there's a way to stretch a linear scale to user specifications, but you can use set_yscale(), set_xscale(), set_zscale() to alter the scales with respect to each other.
Intuitively, set_yscale(log), set_xscale(log), set_zscale(linear) might solve your problems.
A likely better option: specify a stretch, set them all to symlog with the same log base and then specify the Z-axis's symlog scale with the linscalex/linscaley kwargs to your specifications.
More here:
http://matplotlib.org/mpl_toolkits/mplot3d/api.html

I found this while searching on a similar problem. After experimenting a bit, perhaps I can share some of my prelim findings here..matplotlib library is VAST!! (am a newcomer). Note that quite akin to this question, all i wanted was to 'visually' stretch the chart without distorting it.
Background story (only key code snippets are shown to avoid unnecessary clutter for those who know the library, and if you want a run-able code please drop a comment):
I have three 1-d ndarrays representing the X,Y and Z data points respectively. Clearly I can't use plot_surface (as it requires 2d ndarrays for each dim) so I went for the extremely useful plot_trisurf:
fig = plt.figure()
ax = Axes3D(fig)
3d_surf_obj = ax.plot_trisurf(X, Y, Z_defl, cmap=cm.jet,linewidth=0,antialiased=True)
You can think of the plot like a floating barge deforming in waves...As you can see, the axes stretch make it pretty deceiving visually (note that x is supposed to be at x6 times longer than y and >>>>> z). While the plot points are correct, I wanted something more visually 'stretched' at the very least. Was looking for A QUICK FIX, if I may. Long story cut short, I found a bit of success with...'figure.figsize' general setting (see snippet below).
matplotlib.rcParams.update({'font.serif': 'Times New Roman',
'font.size': 10.0,
'axes.labelsize': 'Medium',
'axes.labelweight': 'normal',
'axes.linewidth': 0.8,
###########################################
# THIS IS THE IMPORTANT ONE FOR STRETCHING
# default is [6,4] but...i changed it to
'figure.figsize':[15,5] # THIS ONE #
})
For [15,5] I got something like...
Pretty neat!!
So I started to push it.... and got up to [20,6] before deciding to settle there..
If you want to try for visually stretching the vertical axis, try with ratios like... [7,10], which in this case gives me ...
Not too shabby !
Should do it for visual prowess.

Multiply all your z values by 9,
ax.scatter(x, y, 9*z, zdir='z', c= 'red')
And then give the z-axis custom plot labels and spacing.
ax.ZTick = [0,-9*50, -9*100, -9*150, -9*200];
ax.ZTickLabel = {'0','-50','-100','-150','-200'};

Setting aspect ratio of 3D plot

I am trying to plot a 3D image of the seafloor from the data of a sonar run over a 500m by 40m portion of the seafloor. I am using matplotlib/mplot3d with Axes3D and I want to be able to change the aspect ratio of the axes so that the x & y axis are to scale. An example script with generated data rather than the real data is:
import matplotlib.pyplot as plt
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
# Create figure.
fig = plt.figure()
ax = fig.gca(projection = '3d')
# Generate example data.
R, Y = np.meshgrid(np.arange(0, 500, 0.5), np.arange(0, 40, 0.5))
z = 0.1 * np.abs(np.sin(R/40) * np.sin(Y/6))
# Plot the data.
surf = ax.plot_surface(R, Y, z, cmap=cm.jet, linewidth=0)
fig.colorbar(surf)
# Set viewpoint.
ax.azim = -160
ax.elev = 30
# Label axes.
ax.set_xlabel('Along track (m)')
ax.set_ylabel('Range (m)')
ax.set_zlabel('Height (m)')
# Save image.
fig.savefig('data.png')
And the output image from this script:
Now I would like to change it so that 1 metre in the along-track (x) axis is the same as 1 metre in the range (y) axis (or maybe a different ratio depending on the relative sizes involved). I would also like to set the ratio of the z-axis, again not neccessarily to 1:1 due to the relative sizes in the data, but so the axis is smaller than the current plot.
I have tried building and using this branch of matplotlib, following the example script in this message from the mailing list, but adding the ax.pbaspect = [1.0, 1.0, 0.25] line to my script (having uninstalled the 'standard' version of matplotlib to ensure the custom version was being used) didn't make any difference in the generated image.
Edit: So the desired output would be something like the following (crudely edited with Inkscape) image. In this case I haven't set a 1:1 ratio on the x/y axes because that looks ridiculously thin, but I have spread it out so it isn't square as on the original output.

Add following code before savefig:
ax.auto_scale_xyz([0, 500], [0, 500], [0, 0.15])
If you want no square axis:
edit the get_proj function inside site-packages\mpl_toolkits\mplot3d\axes3d.py:
xmin, xmax = np.divide(self.get_xlim3d(), self.pbaspect[0])
ymin, ymax = np.divide(self.get_ylim3d(), self.pbaspect[1])
zmin, zmax = np.divide(self.get_zlim3d(), self.pbaspect[2])
then add one line to set pbaspect:
ax = fig.gca(projection = '3d')
ax.pbaspect = [2.0, 0.6, 0.25]

The answer to this question works perfectly for me. And you do not need to set up any ratio, it does everything automatically.

An issue has been opened over at github: https://github.com/matplotlib/matplotlib/issues/8593
The above solutions don't seem to work any more. One has to edit the get_proj function inside site-packages\mpl_toolkits\mplot3d\axes3d.py in the following way now:
try:
self.localPbAspect = self.pbaspect
except AttributeError:
self.localPbAspect = [1,1,1]
xmin, xmax = ( lim / self.localPbAspect[0] for lim in self.get_xlim3d() )
ymin, ymax = ( lim / self.localPbAspect[1] for lim in self.get_ylim3d() )
zmin, zmax = ( lim / self.localPbAspect[2] for lim in self.get_zlim3d() )

That how i solved the wasted space problem:
try:
self.localPbAspect=self.pbaspect
zoom_out = (self.localPbAspect[0]+self.localPbAspect[1]+self.localPbAspect[2])
except AttributeError:
self.localPbAspect=[1,1,1]
zoom_out = 0
xmin, xmax = self.get_xlim3d() / self.localPbAspect[0]
ymin, ymax = self.get_ylim3d() / self.localPbAspect[1]
zmin, zmax = self.get_zlim3d() / self.localPbAspect[2]
# transform to uniform world coordinates 0-1.0,0-1.0,0-1.0
worldM = proj3d.world_transformation(xmin, xmax,
ymin, ymax,
zmin, zmax)
# look into the middle of the new coordinates
R = np.array([0.5*self.localPbAspect[0], 0.5*self.localPbAspect[1], 0.5*self.localPbAspect[2]])
xp = R[0] + np.cos(razim) * np.cos(relev) * (self.dist+zoom_out)
yp = R[1] + np.sin(razim) * np.cos(relev) * (self.dist+zoom_out)
zp = R[2] + np.sin(relev) * (self.dist+zoom_out)
E = np.array((xp, yp, zp))