For example, I have 4 control point, and I know it should be a closed contour.
Firstly, I use the scipy.interpolate to interpolate the contour, and the code is:
import numpy as np
import matplotlib.pyplot as plt
from scipy import interpolate
t = np.arange(0, 1.25, 0.25)
x = np.sin(2*np.pi*t)
y = np.cos(2*np.pi*t)
tck,u = interpolate.splprep([x,y], s=0)
unew = np.arange(0, 1.01, 0.01)
out = interpolate.splev(unew, tck)
plt.figure()
plt.plot(x, y, 'x', out[0], out[1])
plt.axis([-1.05, 1.05, -1.05, 1.05])
plt.title('Spline of parametrically-defined curve')
plt.show()
And the result is:
In my thought, the four point should be interpolated into a circle. However, the result is kindly differ from a circle. Then, I find interpolate.CubicSpline may be a solution. The code is:
theta = 2 * np.pi * np.linspace(0, 1, 5)
y = np.c_[np.cos(theta), np.sin(theta)]
cs = interpolate.CubicSpline(theta, y, bc_type='periodic')
# cs = interpolate.CubicSpline(y[:, 0], y[:, 1], bc_type='periodic')
xs = 2 * np.pi * np.linspace(0, 1, 100)
ys = cs(xs)
fig, ax = plt.subplots(figsize=(6.5, 4))
ax.plot(np.cos(xs), np.sin(xs), 'r', label='true')
ax.plot(ys[:, 0], ys[:, 1], label='spline')
ax.axes.set_aspect('equal')
ax.legend(loc='center')
plt.show()
And the result is:
The result is actually what I need. However, in the really application, I don't know the theta, because the real control may be not a circle. In this way, if I only have the control point (n*2), how can we interpolate it into a close contour?
Any suggestion is appreciated!
Related
How to rotate a function by the desired angle (for instance, 30 degrees)?
import matplotlib.pyplot as plt
import numpy as np
from numpy import exp, sin
def g(y):
return exp(-y)*sin(4*y)
y = np.linspace(0, 1.8, 501)
values = g(y)
fig, ax = plt.subplots(figsize=(5,5))
plt.plot(y, values)
plt.show()
Using the cosine and sine of the angle, you can create a rotation matrix. Multiplying each point (y, g(y)) with that matrix create a rotation around 0,0.
Here is some Python/numpy code to illustrate how everything could work together:
import matplotlib.pyplot as plt
import numpy as np
def g(y):
return np.exp(-y) * np.sin(4 * y)
y = np.linspace(0, 1.8, 501)
values = g(y)
theta = np.radians(30)
c, s = np.cos(theta), np.sin(theta)
rot_matrix = np.array(((c, s), (-s, c)))
xy = np.array([y, values]).T # rot_matrix
fig, ax = plt.subplots(figsize=(5, 5))
plt.plot(y, values)
plt.plot(xy[:, 0], xy[:, 1])
plt.axis('equal') # so angles on the screen look like the real angles
plt.show()
PS: To rotate around another point, first subtract the rotation center, do the rotation and then add it again:
center = np.array([0.9, 0])
xy = (np.array([y, values]).T - center) # rot_matrix + center
My code so far, I'm very new to programming and have been trying for a while.
Here I apply the Box-Muller transform to approximate two Gaussian normal distributions starting from a random uniform sampling. Then, I create a histogram for both of them.
Now, I would like to compare the obtained histograms with "the real thing": a standard Bell curve. How to draw such a curve to match the histograms?
import numpy as np
import matplotlib.pyplot as plt
N = 10000
z1 = np.random.uniform(0, 1.0, N)
z2 = np.random.uniform(0, 1.0, N)
R_sq = -2 * np.log(z1)
theta = 2 * np.pi * z2
z1 = np.sqrt(R_sq) * np.cos(theta)
z2 = np.sqrt(R_sq) * np.sin(theta)
fig = plt.figure()
ax = fig.add_subplot(2, 1, 1)
ax.hist(z1, bins=40, range=(-4, 4), color='red')
plt.title("Histgram")
plt.xlabel("z1")
plt.ylabel("frequency")
ax2 = fig.add_subplot(2, 1, 2)
ax2.hist(z2, bins=40, range=(-4, 4), color='blue')
plt.xlabel("z2")
plt.show()
To obtain the 'kernel density estimation', scipy.stats.gaussian_kde calculates a function to fit the data.
To just draw a Gaussian normal curve, there is [scipy.stats.norm]. Subtracting the mean and dividing by the standard deviation, adapts the position to the given data.
Both curves would be drawn such that the area below the curve sums to one. To adjust them to the size of the histogram, these curves need to be scaled by the length of the data times the bin-width. Alternatively, this scaling can stay at 1, and the histogram scaled by adding the parameter hist(..., density=True).
In the demo code the data is mutilated to illustrate the difference between the kde and the Gaussian normal.
import numpy as np
import matplotlib.pyplot as plt
import scipy.stats as stats
x = np.linspace(-4,4,1000)
N = 10000
z1 = np.random.randint(1, 3, N) * np.random.uniform(0, .4, N)
z2 = np.random.uniform(0, 1, N)
R_sq = -2 * np.log(z1)
theta = 2 * np.pi * z2
z1 = np.sqrt(R_sq) * np.cos(theta)
z2 = np.sqrt(R_sq) * np.sin(theta)
fig = plt.figure(figsize=(12,4))
for ind_subplot, zi, col in zip((1, 2), (z1, z2), ('crimson', 'dodgerblue')):
ax = fig.add_subplot(1, 2, ind_subplot)
ax.hist(zi, bins=40, range=(-4, 4), color=col, label='histogram')
ax.set_xlabel("z"+str(ind_subplot))
ax.set_ylabel("frequency")
binwidth = 8 / 40
scale_factor = len(zi) * binwidth
gaussian_kde_zi = stats.gaussian_kde(z1)
ax.plot(x, gaussian_kde_zi(x)*scale_factor, color='springgreen', linewidth=3, label='kde')
std_zi = np.std(zi)
mean_zi = np.mean(zi)
ax.plot(x, stats.norm.pdf((x-mean_zi)/std_zi)*scale_factor, color='black', linewidth=2, label='normal')
ax.legend()
plt.show()
The original values for z1 and z2 very much resemble a normal distribution, and so the black line (the Gaussian normal for the data) and the green line (the KDE) very much resemble each other.
The current code first calculates the real mean and the real standard deviation of the data. As you want to mimic a perfect Gaussian normal, you should compare to the curve with mean zero and standard deviatio one. You'll see they're almost identical on the plot.
I have an array of shape(512,512).
Looks like, (row=x, column=y, density=z=the number of the array)
[[0.012825 0.020408 0.022976 ... 0.015938 0.02165 0.024357]
[0.036332 0.031904 0.025462 ... 0.031095 0.019812 0.024523]
[0.015831 0.027392 0.031939 ... 0.016249 0.01697 0.028686]
...
[0.024545 0.011895 0.022235 ... 0.033226 0.03223 0.030235]]
I had already drawn it into a 2D density plot. My goal is to find the center of the circle and draw a vertical and horizontal cross-section in one figure.
Now, I have the trouble to find the center of the circle and combine two cross-sections in one figure.
Please help.
This is my code:
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
import scipy.ndimage
data = pd.read_csv('D:/BFP.csv', header=None)
# create data
data = np.array(data)
print(data)
#plot data
side = np.linspace(-1.5,1.5,512)
x,y = np.meshgrid(side,side)
z = [[data[i][j] for i in range(len(data[0]))]for j in range(len(data))]
#-- Extract the line...
# Make a line with "num" points...
x0, y0 = 270, 0 # These are in _pixel_ coordinates!!
x1, y1 = 270, 500
num = 512
x_, y_ = np.linspace(x0, x1, num), np.linspace(y0, y1, num)
# Extract the values along the line, using cubic interpolation
zi = scipy.ndimage.map_coordinates(z, np.vstack((x_,y_)))
#-- Plot...
fig, axes = plt.subplots(nrows=2)
axes[0].imshow(z,origin='lower')
axes[0].plot([x0, x1], [y0, y1], 'ro-')
#axes[0].axis('image')
axes[1].plot(zi)
plt.savefig('D:/vertical.png')
plt.show()
image here:
I cannot help you with finding the center of the circle, but you can create a nice visualization of the cross section by creating 3 axes in a grid. Usually, I would use GridSpec for this, but imhsow has a tendency to mess up the relative size of the axes to maintain square pixels. Thankfully, the AxesGrid toolkit can help.
The base of the code is inspired by this matplotlib example.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.axes_grid1 import make_axes_locatable
from scipy.stats import multivariate_normal
import scipy
fig, main_ax = plt.subplots(figsize=(5, 5))
divider = make_axes_locatable(main_ax)
top_ax = divider.append_axes("top", 1.05, pad=0.1, sharex=main_ax)
right_ax = divider.append_axes("right", 1.05, pad=0.1, sharey=main_ax)
# make some labels invisible
top_ax.xaxis.set_tick_params(labelbottom=False)
right_ax.yaxis.set_tick_params(labelleft=False)
main_ax.set_xlabel('dim 1')
main_ax.set_ylabel('dim 2')
top_ax.set_ylabel('Z profile')
right_ax.set_xlabel('Z profile')
x, y = np.mgrid[-1:1:.01, -1:1:.01]
pos = np.empty(x.shape + (2,))
pos[:, :, 0] = x; pos[:, :, 1] = y
rv = multivariate_normal([-0.2, 0.2], [[1, 1.5], [0.25, 0.25]])
z = rv.pdf(pos)
z_max = z.max()
cur_x = 110
cur_y = 40
main_ax.imshow(z, origin='lower')
main_ax.autoscale(enable=False)
right_ax.autoscale(enable=False)
top_ax.autoscale(enable=False)
right_ax.set_xlim(right=z_max)
top_ax.set_ylim(top=z_max)
v_line = main_ax.axvline(cur_x, color='r')
h_line = main_ax.axhline(cur_y, color='g')
v_prof, = right_ax.plot(z[:,int(cur_x)],np.arange(x.shape[1]), 'r-')
h_prof, = top_ax.plot(np.arange(x.shape[0]),z[int(cur_y),:], 'g-')
plt.show()
Just for fun, you can even make it interactive
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.axes_grid1 import make_axes_locatable
from scipy.stats import multivariate_normal
import scipy
fig, main_ax = plt.subplots(figsize=(5, 5))
divider = make_axes_locatable(main_ax)
top_ax = divider.append_axes("top", 1.05, pad=0.1, sharex=main_ax)
right_ax = divider.append_axes("right", 1.05, pad=0.1, sharey=main_ax)
# make some labels invisible
top_ax.xaxis.set_tick_params(labelbottom=False)
right_ax.yaxis.set_tick_params(labelleft=False)
main_ax.set_xlabel('dim 1')
main_ax.set_ylabel('dim 2')
top_ax.set_ylabel('Z profile')
right_ax.set_xlabel('Z profile')
x, y = np.mgrid[-1:1:.01, -1:1:.01]
pos = np.empty(x.shape + (2,))
pos[:, :, 0] = x; pos[:, :, 1] = y
rv = multivariate_normal([-0.2, 0.2], [[1, 1.5], [0.25, 0.25]])
z = rv.pdf(pos)
z_max = z.max()
main_ax.imshow(z, origin='lower')
main_ax.autoscale(enable=False)
right_ax.autoscale(enable=False)
top_ax.autoscale(enable=False)
right_ax.set_xlim(right=z_max)
top_ax.set_ylim(top=z_max)
v_line = main_ax.axvline(np.nan, color='r')
h_line = main_ax.axhline(np.nan, color='g')
v_prof, = right_ax.plot(np.zeros(x.shape[1]),np.arange(x.shape[1]), 'r-')
h_prof, = top_ax.plot(np.arange(x.shape[0]),np.zeros(x.shape[0]), 'g-')
def on_move(event):
if event.inaxes is main_ax:
cur_x = event.xdata
cur_y = event.ydata
v_line.set_xdata([cur_x,cur_x])
h_line.set_ydata([cur_y,cur_y])
v_prof.set_xdata(z[:,int(cur_x)])
h_prof.set_ydata(z[int(cur_y),:])
fig.canvas.draw_idle()
fig.canvas.mpl_connect('motion_notify_event', on_move)
plt.show()
NB: the lag is just due to the convertion in gif, the update is much smoother on my machine
I am trying to remove the edge color in the plot of a cylinder where I have set an alpha and facecolors. However, if I also set the facecolors, I can still see the edge colors. If I remove the alpha = 0.5 statement then the problem is resolved, however I need the alpha to be <1 . Here is an example:
You can still see the blue edgecolors even tough I have set the edgecolor to None.
This is the code where I use plot_surface()
ax.plot_surface(X, Y,Z, edgecolor = "None", facecolors = col1, alpha = 0.5)
Yet the edge colors are still there? However, if I remove the facecolors statement inside plot_surface() then the edge colors are no longer there. Here is the complete code:
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
import random
import numpy as np
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
origin = np.array([0, 0, 0])
#axis and radius
p0 = np.array([0, 0, 0])
p1 = np.array([8, 8, 8])
R = 4
#vector in direction of axis
v = p1 - p0
#find magnitude of vector
mag = norm(v)
#unit vector in direction of axis
v = v / mag
#make some vector not in the same direction as v
not_v = np.array([1, 0, 0])
if (v == not_v).all():
not_v = np.array([0, 1, 0])
#make vector perpendicular to v
n1 = np.cross(v, not_v)
#normalize n1
n1 /= norm(n1)
#make unit vector perpendicular to v and n1
n2 = np.cross(v, n1)
#surface ranges over t from 0 to length of axis and 0 to 2*pi
t = np.linspace(0, mag, 200)
theta = np.linspace(0, 2 * np.pi, 100)
#use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
#generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R * np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
col1 = plt.cm.Blues(np.linspace(0,1,200)) # linear gradient along the t-axis
col1 = np.repeat(col1[np.newaxis,:, :], 100, axis=0) # expand over the theta- axis
ax.plot_surface(X, Y,Z, edgecolor = None, facecolors = col1, alpha = 0.5)
#plot axis
ax.plot(*zip(p0, p1), color = 'red')
ax.set_xlim(0, 10)
ax.set_ylim(0, 10)
ax.set_zlim(0, 10)
plt.axis('off')
ax.axes.get_xaxis().set_visible(False)
ax.axes.get_yaxis().set_visible(False)
plt.show()
Setting linewidth=0 in plot_surface() solves this problem:
ax.plot_surface(X, Y, Z, edgecolor=None, facecolors=col1, alpha=0.5, linewidth=0)
p.s.: I didn't find this worth an answer, but per: Question with no answers, but issue solved in the comments (or extended in chat), I added it as a quick answer so the question can be marked as solved
scikit-learn has a very nice demo that creates an outlier analysis tool. Here is the
import numpy as np
import pylab as pl
import matplotlib.font_manager
from scipy import stats
from sklearn import svm
from sklearn.covariance import EllipticEnvelope
# Example settings
n_samples = 200
outliers_fraction = 0.25
clusters_separation = [0, 1, 2]
# define two outlier detection tools to be compared
classifiers = {
"One-Class SVM": svm.OneClassSVM(nu=0.95 * outliers_fraction + 0.05,
kernel="rbf", gamma=0.1),
"robust covariance estimator": EllipticEnvelope(contamination=.1)}
# Compare given classifiers under given settings
xx, yy = np.meshgrid(np.linspace(-7, 7, 500), np.linspace(-7, 7, 500))
n_inliers = int((1. - outliers_fraction) * n_samples)
n_outliers = int(outliers_fraction * n_samples)
ground_truth = np.ones(n_samples, dtype=int)
ground_truth[-n_outliers:] = 0
# Fit the problem with varying cluster separation
for i, offset in enumerate(clusters_separation):
np.random.seed(42)
# Data generation
X1 = 0.3 * np.random.randn(0.5 * n_inliers, 2) - offset
X2 = 0.3 * np.random.randn(0.5 * n_inliers, 2) + offset
X = np.r_[X1, X2]
# Add outliers
X = np.r_[X, np.random.uniform(low=-6, high=6, size=(n_outliers, 2))]
# Fit the model with the One-Class SVM
pl.figure(figsize=(10, 5))
for i, (clf_name, clf) in enumerate(classifiers.items()):
# fit the data and tag outliers
clf.fit(X)
y_pred = clf.decision_function(X).ravel()
threshold = stats.scoreatpercentile(y_pred,
100 * outliers_fraction)
y_pred = y_pred > threshold
n_errors = (y_pred != ground_truth).sum()
# plot the levels lines and the points
Z = clf.decision_function(np.c_[xx.ravel(), yy.ravel()])
Z = Z.reshape(xx.shape)
subplot = pl.subplot(1, 2, i + 1)
subplot.set_title("Outlier detection")
subplot.contourf(xx, yy, Z, levels=np.linspace(Z.min(), threshold, 7),
cmap=pl.cm.Blues_r)
a = subplot.contour(xx, yy, Z, levels=[threshold],
linewidths=2, colors='red')
subplot.contourf(xx, yy, Z, levels=[threshold, Z.max()],
colors='orange')
b = subplot.scatter(X[:-n_outliers, 0], X[:-n_outliers, 1], c='white')
c = subplot.scatter(X[-n_outliers:, 0], X[-n_outliers:, 1], c='black')
subplot.axis('tight')
subplot.legend(
[a.collections[0], b, c],
['learned decision function', 'true inliers', 'true outliers'],
prop=matplotlib.font_manager.FontProperties(size=11))
subplot.set_xlabel("%d. %s (errors: %d)" % (i + 1, clf_name, n_errors))
subplot.set_xlim((-7, 7))
subplot.set_ylim((-7, 7))
pl.subplots_adjust(0.04, 0.1, 0.96, 0.94, 0.1, 0.26)
pl.show()
And here is what it looks like:
Is that cool or what?
However, I want the plot to be mouse-sensitive. That is, I want to be able to click on dots and find out what they are, with either a tool-tip or with a pop-up window, or something in a scroller. And I'd also like to be able to click-to-zoom, rather than zoom with a bounding box.
Is there any way to do this?
Not to plug my own project to much, but have a look at mpldatacursor. If you'd prefer, it's also quite easy to implement from scratch.
As a quick example:
import matplotlib.pyplot as plt
import numpy as np
from mpldatacursor import datacursor
x1, y1 = np.random.random((2, 5))
x2, y2 = np.random.random((2, 5))
fig, ax = plt.subplots()
ax.plot(x1, y1, 'ro', markersize=12, label='Series A')
ax.plot(x2, y2, 'bo', markersize=12, label='Series B')
ax.legend()
datacursor()
plt.show()
For this to work with the example code you posted, you'd need to change things slightly. As it is, the artist labels are set in the call to legend, instead of when the artist is created. This means that there's no way to retrieve what's displayed in the legend for a particular artist. All you'd need to do is just pass in the labels as a kwarg to scatter instead of as the second argument to legend, and things should work as you were wanting.