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Plotting stochastic processes in Python
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Basically, I want to plot a scatter plot between two variables with varying percentile, I've plotted the scatter plot with the following toy code but I'm unable to plot it for different percentile (quantile).
quantiles = [1,10,25,50,50,75,90,99]
grays = ["#DCDCDC", "#A9A9A9", "#2F4F4F","#A9A9A9", "#DCDCDC"]
alpha = 0.3
data = df[['area_log','mr_ecdf']]
y = data['mr_ecdf']
x = data['area_log']
idx = np.argsort(x)
x = np.array(x)[idx]
y = np.array(y)[idx]
for i in range(len(quantiles)//2):
plt.fill_between(x, y, y, color='black', alpha = alpha, label=f"{quantiles[i]}")
lower_lim = np.percentile(y, quantiles[i])
upper_lim = np.percentile(y, 100-quantiles[i])
data = data[data['mr_ecdf'] >= lower_lim]
data = data[data['mr_ecdf'] <= upper_lim]
y = data['mr_ecdf']
x = data['area_log']
idx = np.argsort(x)
x = np.array(x)[idx]
y = np.array(y)[idx]
data = df[['area_log','mr_ecdf']]
y = data['mr_ecdf']
x = data['area_log']
plt.scatter(x, y,s=1, color = 'r', label = 'data')
plt.legend()
# axes.set_ylim([0,1])
enter image description here
data link : here
I want plot something like this (First- (1,1)):
As was mentioned by #Mr. T, one way to do that is to calculate the CIs yourself and then plot them using plt.fill_between. The data you show pose a problem since there is not enough points and variance so you'll never get what is on your pictures (and the separation in my figure is also not clear so I have put another example below to show how it works). If you have data for that, post it, I will update. Anyway, you should check the post I mentioned in the comment and some way of doing it follows:
import numpy as np
import matplotlib.pyplot as plt
x = np.array([5,7,8,7,2,17,2,9,4,11,12,9,6])
y = np.array([99,86,87,88,111,86,103,87,94,78,77,85,86])
idx = np.argsort(x)
x = np.array(x)[idx]
y = np.array(y)[idx]
# Create a list of quantiles to calculate
quantiles = [0.05, 0.25, 0.75, 0.95]
grays = ["#DCDCDC", "#A9A9A9", "#2F4F4F","#A9A9A9", "#DCDCDC"]
alpha = 0.3
plt.fill_between(x, y-np.percentile(y, 0.5), y+np.percentile(y, 0.5), color=grays[2], alpha = alpha, label="0.50")
# if the percentiles are symmetrical and we want labels on both sides
for i in range(len(quantiles)//2):
plt.fill_between(x, y, y+np.percentile(y, quantiles[i]), color=grays[i], alpha = alpha, label=f"{quantiles[i]}")
plt.fill_between(x, y-np.percentile(y, quantiles[-(i+1)]),y, color=grays[-(i+1)], alpha = alpha, label=f"{quantiles[-(i+1)]}")
plt.scatter(x, y, color = 'r', label = 'data')
plt.legend()
EDIT:
Some explanation. I am not sure what is not correct in my code, but I would be happy if you can tell me -- there is always a way for improvement (Thanks to #Mr T. again for the catch). Nevertheless, the fill between function does the following:
Fill the area between two horizontal curves.
The curves are defined by the points (x, y1) and (x, y2)
So you specify by the y1 and y2 where you want to have the graph filled with a colour. Let me bring another example:
X = np.linspace(120, 50, 71)
Y = X + 20*np.random.randn(71)
plt.fill_between(X, Y-np.percentile(Y, 95),Y+np.percentile(Y, 95), color="k", alpha = alpha)
plt.fill_between(X, Y-np.percentile(Y, 80),Y+np.percentile(Y, 80), color="r", alpha = alpha)
plt.fill_between(X, Y-np.percentile(Y, 60),Y, color="b", alpha = alpha)
plt.scatter(X, Y, color = 'r', label = 'data')
I generated some random data to see what is happening. The line plt.fill_between(X, Y-np.percentile(Y, 60),Y, color="b", alpha = alpha) is plotting the fill only from the 60th percentile below Y up to Y. The other two lines are covering the space always from both sides of Y (hence the +-). You can see that the percentiles overlap, of course they do, they must -- a 90 percentile includes the 60 as well. So you see only the differences between them. You could plot the data in the opposite order (or change z-factor) but then all would be covered by the highest percentile. I hope this clarifies the answer. Also, your question is perfectly fine, sorry if my answer feels not neutral. Just if you had also the data for the graphs and not only the picture, my/others answer could be more tailored :).
A good way to show the concentration of the data points in a plot is using a scatter plot with non-unit transparency. As a result, the areas with more concentration would appear darker.
# this is synthetic example
N = 10000 # a very very large number
x = np.random.normal(0, 1, N)
y = np.random.normal(0, 1, N)
plt.scatter(x, y, marker='.', alpha=0.1) # an area full of dots, darker wherever the number of dots is more
which gives something like this:
Imagine the case we want to emphasize on the outliers. So the situation is almost reversed: A plot in which the less-concentrated areas are bolder. (There might be a trick to apply for my simple example, but imagine a general case where a distribution of points are not known prior, or it's difficult to define a rule for transparency/weight on color.)
I was thinking if there's anything handy same as alpha that is designed for this job specifically. Although other ideas for emphasizing on outliers are also welcomed.
UPDATE: This is what happens when more then one data point is scattered on the same area:
I'm looking for something like the picture below, the more data point, the less transparent the marker.
To answer the question: You can calculate the density of points, normalize it and encode it in the alpha channel of a colormap.
import numpy as np
from scipy import stats
import matplotlib.pyplot as plt
from matplotlib.colors import LinearSegmentedColormap
# this is synthetic example
N = 10000 # a very very large number
x = np.random.normal(0, 1, N)
y = np.random.normal(0, 1, N)
fig, (ax,ax2) = plt.subplots(ncols=2, figsize=(8,5))
ax.scatter(x, y, marker='.', alpha=0.1)
values = np.vstack([x,y])
kernel = stats.gaussian_kde(values)
weights = kernel(values)
weights = weights/weights.max()
cols = plt.cm.Blues([0.8, 0.5])
cols[:,3] = [1., 0.005]
cmap = LinearSegmentedColormap.from_list("", cols)
ax2.scatter(x, y, c=weights, s = 1, marker='.', cmap=cmap)
plt.show()
Left is the original image, right is the image where higher density points have a lower alpha.
Note, however, that this is undesireable, because high density transparent points are undistinguishable from low density. I.e. in the right image it really looks as though you have a hole in the middle of your distribution.
Clearly, a solution with a colormap which does not contain the color of the background is a lot less confusing to the reader.
import numpy as np
from scipy import stats
import matplotlib.pyplot as plt
# this is synthetic example
N = 10000 # a very very large number
x = np.random.normal(0, 1, N)
y = np.random.normal(0, 1, N)
fig, ax = plt.subplots(figsize=(5,5))
values = np.vstack([x,y])
kernel = stats.gaussian_kde(values)
weights = kernel(values)
weights = weights/weights.max()
ax.scatter(x, y, c = weights, s=9, edgecolor="none", marker='.', cmap="magma")
plt.show()
Here, low density points are still emphazised by darker color, but at the same time it's clear to the viewer that the highest density lies in the middle.
As far as I know, there is no "direct" solution to this quite interesting problem. As a workaround, I propose this solution:
N = 10000 # a very very large number
x = np.random.normal(0, 1, N)
y = np.random.normal(0, 1, N)
fig = plt.figure() # create figure directly to be able to extract the bg color
ax = fig.gca()
ax.scatter(x, y, marker='.') # plot all markers without alpha
bgcolor = ax.get_facecolor() # extract current background color
# plot with alpha, "overwriting" dense points
ax.scatter(x, y, marker='.', color=bgcolor, alpha=0.2)
This will plot all points without transparency and then plot all points again with some transparency, "overwriting" those points with the highest density the most. Setting the alpha value to other higher values will put more emphasis to outliers and vice versa.
Of course the color of the second scatter plot needs to be adjusted to your background color. In my example this is done by extracting the background color and setting it as the new scatter plot's color.
This solution is independent of the kind of distribution. It only depends on the density of the points. However it produces twice the amount of points, thus may take slightly longer to render.
Reproducing the edit in the question, my solution is showing exactly the desired behavior. The leftmost point is a single point and is the darkest, the rightmost is consisting of three points and is the lightest color.
x = [0, 1, 1, 2, 2, 2]
y = [0, 0, 0, 0, 0, 0]
fig = plt.figure() # create figure directly to be able to extract the bg color
ax = fig.gca()
ax.scatter(x, y, marker='.', s=10000) # plot all markers without alpha
bgcolor = ax.get_facecolor() # extract current background color
# plot with alpha, "overwriting" dense points
ax.scatter(x, y, marker='.', color=bgcolor, alpha=0.2, s=10000)
Assuming that the distributions are centered around a specific point (e.g. (0,0) in this case), I would use this:
import numpy as np
import matplotlib.pyplot as plt
N = 500
# 0 mean, 0.2 std
x = np.random.normal(0,0.2,N)
y = np.random.normal(0,0.2,N)
# calculate the distance to (0, 0).
color = np.sqrt((x-0)**2 + (y-0)**2)
plt.scatter(x , y, c=color, cmap='plasma', alpha=0.7)
plt.show()
Results:
I don't know if it helps you, because it's not exactly you asked for, but you can simply color points, which values are bigger than some threshold. For example:
import matplotlib.pyplot as plt
num = 100
threshold = 80
x = np.linspace(0, 100, num=num)
y = np.random.normal(size=num)*45
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
ax.scatter(x[np.abs(y) < threshold], y[np.abs(y) < threshold], color="#00FFAA")
ax.scatter(x[np.abs(y) >= threshold], y[np.abs(y) >= threshold], color="#AA00FF")
plt.show()
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'};
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'};
I am trying to make a discrete colorbar for a scatterplot in matplotlib
I have my x, y data and for each point an integer tag value which I want to be represented with a unique colour, e.g.
plt.scatter(x, y, c=tag)
typically tag will be an integer ranging from 0-20, but the exact range may change
so far I have just used the default settings, e.g.
plt.colorbar()
which gives a continuous range of colours. Ideally i would like a set of n discrete colours (n=20 in this example). Even better would be to get a tag value of 0 to produce a gray colour and 1-20 be colourful.
I have found some 'cookbook' scripts but they are very complicated and I cannot think they are the right way to solve a seemingly simple problem
You can create a custom discrete colorbar quite easily by using a BoundaryNorm as normalizer for your scatter. The quirky bit (in my method) is making 0 showup as grey.
For images i often use the cmap.set_bad() and convert my data to a numpy masked array. That would be much easier to make 0 grey, but i couldnt get this to work with the scatter or the custom cmap.
As an alternative you can make your own cmap from scratch, or read-out an existing one and override just some specific entries.
import numpy as np
import matplotlib as mpl
import matplotlib.pylab as plt
fig, ax = plt.subplots(1, 1, figsize=(6, 6)) # setup the plot
x = np.random.rand(20) # define the data
y = np.random.rand(20) # define the data
tag = np.random.randint(0, 20, 20)
tag[10:12] = 0 # make sure there are some 0 values to show up as grey
cmap = plt.cm.jet # define the colormap
# extract all colors from the .jet map
cmaplist = [cmap(i) for i in range(cmap.N)]
# force the first color entry to be grey
cmaplist[0] = (.5, .5, .5, 1.0)
# create the new map
cmap = mpl.colors.LinearSegmentedColormap.from_list(
'Custom cmap', cmaplist, cmap.N)
# define the bins and normalize
bounds = np.linspace(0, 20, 21)
norm = mpl.colors.BoundaryNorm(bounds, cmap.N)
# make the scatter
scat = ax.scatter(x, y, c=tag, s=np.random.randint(100, 500, 20),
cmap=cmap, norm=norm)
# create a second axes for the colorbar
ax2 = fig.add_axes([0.95, 0.1, 0.03, 0.8])
cb = plt.colorbar.ColorbarBase(ax2, cmap=cmap, norm=norm,
spacing='proportional', ticks=bounds, boundaries=bounds, format='%1i')
ax.set_title('Well defined discrete colors')
ax2.set_ylabel('Very custom cbar [-]', size=12)
I personally think that with 20 different colors its a bit hard to read the specific value, but thats up to you of course.
You could follow this example below or the newly added example in the documentation
#!/usr/bin/env python
"""
Use a pcolor or imshow with a custom colormap to make a contour plot.
Since this example was initially written, a proper contour routine was
added to matplotlib - see contour_demo.py and
http://matplotlib.sf.net/matplotlib.pylab.html#-contour.
"""
from pylab import *
delta = 0.01
x = arange(-3.0, 3.0, delta)
y = arange(-3.0, 3.0, delta)
X,Y = meshgrid(x, y)
Z1 = bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z2 = bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
Z = Z2 - Z1 # difference of Gaussians
cmap = cm.get_cmap('PiYG', 11) # 11 discrete colors
im = imshow(Z, cmap=cmap, interpolation='bilinear',
vmax=abs(Z).max(), vmin=-abs(Z).max())
axis('off')
colorbar()
show()
which produces the following image:
The above answers are good, except they don't have proper tick placement on the colorbar. I like having the ticks in the middle of the color so that the number -> color mapping is more clear. You can solve this problem by changing the limits of the matshow call:
import matplotlib.pyplot as plt
import numpy as np
def discrete_matshow(data):
# get discrete colormap
cmap = plt.get_cmap('RdBu', np.max(data) - np.min(data) + 1)
# set limits .5 outside true range
mat = plt.matshow(data, cmap=cmap, vmin=np.min(data) - 0.5,
vmax=np.max(data) + 0.5)
# tell the colorbar to tick at integers
cax = plt.colorbar(mat, ticks=np.arange(np.min(data), np.max(data) + 1))
# generate data
a = np.random.randint(1, 9, size=(10, 10))
discrete_matshow(a)
To set a values above or below the range of the colormap, you'll want to use the set_over and set_under methods of the colormap. If you want to flag a particular value, mask it (i.e. create a masked array), and use the set_bad method. (Have a look at the documentation for the base colormap class: http://matplotlib.org/api/colors_api.html#matplotlib.colors.Colormap )
It sounds like you want something like this:
import matplotlib.pyplot as plt
import numpy as np
# Generate some data
x, y, z = np.random.random((3, 30))
z = z * 20 + 0.1
# Set some values in z to 0...
z[:5] = 0
cmap = plt.get_cmap('jet', 20)
cmap.set_under('gray')
fig, ax = plt.subplots()
cax = ax.scatter(x, y, c=z, s=100, cmap=cmap, vmin=0.1, vmax=z.max())
fig.colorbar(cax, extend='min')
plt.show()
This topic is well covered already but I wanted to add something more specific : I wanted to be sure that a certain value would be mapped to that color (not to any color).
It is not complicated but as it took me some time, it might help others not lossing as much time as I did :)
import matplotlib
from matplotlib.colors import ListedColormap
# Let's design a dummy land use field
A = np.reshape([7,2,13,7,2,2], (2,3))
vals = np.unique(A)
# Let's also design our color mapping: 1s should be plotted in blue, 2s in red, etc...
col_dict={1:"blue",
2:"red",
13:"orange",
7:"green"}
# We create a colormar from our list of colors
cm = ListedColormap([col_dict[x] for x in col_dict.keys()])
# Let's also define the description of each category : 1 (blue) is Sea; 2 (red) is burnt, etc... Order should be respected here ! Or using another dict maybe could help.
labels = np.array(["Sea","City","Sand","Forest"])
len_lab = len(labels)
# prepare normalizer
## Prepare bins for the normalizer
norm_bins = np.sort([*col_dict.keys()]) + 0.5
norm_bins = np.insert(norm_bins, 0, np.min(norm_bins) - 1.0)
print(norm_bins)
## Make normalizer and formatter
norm = matplotlib.colors.BoundaryNorm(norm_bins, len_lab, clip=True)
fmt = matplotlib.ticker.FuncFormatter(lambda x, pos: labels[norm(x)])
# Plot our figure
fig,ax = plt.subplots()
im = ax.imshow(A, cmap=cm, norm=norm)
diff = norm_bins[1:] - norm_bins[:-1]
tickz = norm_bins[:-1] + diff / 2
cb = fig.colorbar(im, format=fmt, ticks=tickz)
fig.savefig("example_landuse.png")
plt.show()
I have been investigating these ideas and here is my five cents worth. It avoids calling BoundaryNorm as well as specifying norm as an argument to scatter and colorbar. However I have found no way of eliminating the rather long-winded call to matplotlib.colors.LinearSegmentedColormap.from_list.
Some background is that matplotlib provides so-called qualitative colormaps, intended to use with discrete data. Set1, e.g., has 9 easily distinguishable colors, and tab20 could be used for 20 colors. With these maps it could be natural to use their first n colors to color scatter plots with n categories, as the following example does. The example also produces a colorbar with n discrete colors approprately labelled.
import matplotlib, numpy as np, matplotlib.pyplot as plt
n = 5
from_list = matplotlib.colors.LinearSegmentedColormap.from_list
cm = from_list(None, plt.cm.Set1(range(0,n)), n)
x = np.arange(99)
y = x % 11
z = x % n
plt.scatter(x, y, c=z, cmap=cm)
plt.clim(-0.5, n-0.5)
cb = plt.colorbar(ticks=range(0,n), label='Group')
cb.ax.tick_params(length=0)
which produces the image below. The n in the call to Set1 specifies
the first n colors of that colormap, and the last n in the call to from_list
specifies to construct a map with n colors (the default being 256). In order to set cm as the default colormap with plt.set_cmap, I found it to be necessary to give it a name and register it, viz:
cm = from_list('Set15', plt.cm.Set1(range(0,n)), n)
plt.cm.register_cmap(None, cm)
plt.set_cmap(cm)
...
plt.scatter(x, y, c=z)
I think you'd want to look at colors.ListedColormap to generate your colormap, or if you just need a static colormap I've been working on an app that might help.