I have been trying to get a plot of vector lines going using the matplotlib library and I keep getting something like this:
Not sure what is happening since the code I'm running seems to follow the syntax for how to make a basic quiver plot. I've tried messing with the array type to see if that's the issue but no luck. Some points on the plot just don't seem to be getting any vector data.
import matplotlib.pyplot as plt
import numpy as np
X = np.arange(-2,2,.1)
Y = np.arange(-2,2,.1)
x,y = np.meshgrid(X,Y)
m1 =1
m2 =2
x1 =4/3
x2 =2/3
omega = 3/8
u = -(m1/(abs(x-x1))**3)*(x-x1)-(m2/(abs(x-x2))**3)*(x-x2)+ x*omega
v = -(m1/(abs(y))**3)*(y)-(m2/(abs(y))**3)*(y)+ y*omega
fig, ax = plt.subplots()
ax.quiver(x,y,u,v)
plt.show()
A nice way, I find, to have a look at your data is to normalise the vector field and colour it by intensity. You can always mask glyphs for which the intensity is too low by using a Numpy MaskedArray. Have a look below.
import matplotlib.colors as cl
import matplotlib.pyplot as plt
import matplotlib.ticker as tck
import numpy as np
x, y = np.meshgrid(np.linspace(-2, 2, 41), np.linspace(-2, 2, 41))
m1, m2, x1, x2, omega = (1, 2, 4 / 3, 2 / 3, 3 / 8)
u = -(m1 / abs(x - x1) ** 3 * (x - x1) - m2 / abs(x - x2) ** 3 * (x - x2)
+ x * omega)
v = y * (omega - (m1 + m2) / abs(y) ** 3)
fig, (ax, bx) = plt.subplots(ncols=2, figsize=(20, 10))
ax.quiver(x, y, u, v, antialiased=True, scale=1e4, width=6e-3, headwidth=3,
headlength=4, headaxislength=3.5, pivot='tail',
edgecolors='xkcd:white', linewidths=1)
ax.set_aspect('equal')
w = np.sqrt(u ** 2 + v ** 2)
quiv = bx.quiver(x, y, u / w, v / w, w, antialiased=True, scale=3e1,
width=6e-3, headwidth=3, headlength=4, headaxislength=3.5,
pivot='tail', edgecolors='xkcd:white', linewidths=1,
norm=cl.LogNorm(vmin=1e-1, vmax=1e3))
bx.set_aspect('equal')
fig.colorbar(quiv, cax=fig.add_axes([0.93, 0.1, 0.02, 0.8]),
extend='both', ticks=tck.LogLocator(),
format=tck.LogFormatterSciNotation())
Some of the y values are close to 0 so that you get crazily large v values. I would check the equation because the plot is actually correct (the arrows are infinitely large when y ~= 0).
Related
How can I get from a plot in Python an exact value on y - axis? I have two arrays vertical_data and gradient(temperature_data) and I plotted them as:
plt.plot(gradient(temperature_data),vertical_data)
plt.show()
Plot shown here:
I need the zero value but it is not exactly zero, it's a float.
I did not find a good answer to the question of how to find the roots or zeros of a numpy array, so here is a solution, using simple linear interpolation.
import numpy as np
N = 750
x = .4+np.sort(np.random.rand(N))*3.5
y = (x-4)*np.cos(x*9.)*np.cos(x*6+0.05)+0.1
def find_roots(x,y):
s = np.abs(np.diff(np.sign(y))).astype(bool)
return x[:-1][s] + np.diff(x)[s]/(np.abs(y[1:][s]/y[:-1][s])+1)
z = find_roots(x,y)
import matplotlib.pyplot as plt
plt.plot(x,y)
plt.plot(z, np.zeros(len(z)), marker="o", ls="", ms=4)
plt.show()
Of course you can invert the roles of x and y to get
plt.plot(y,x)
plt.plot(np.zeros(len(z)),z, marker="o", ls="", ms=4)
Because people where asking how to get the intercepts at non-zero values y0, note that one may simply find the zeros of y-y0 then.
y0 = 1.4
z = find_roots(x,y-y0)
# ...
plt.plot(z, np.zeros(len(z))+y0)
People were also asking how to get the intersection between two curves. In that case it's again about finding the roots of the difference between the two, e.g.
x = .4 + np.sort(np.random.rand(N)) * 3.5
y1 = (x - 4) * np.cos(x * 9.) * np.cos(x * 6 + 0.05) + 0.1
y2 = (x - 2) * np.cos(x * 8.) * np.cos(x * 5 + 0.03) + 0.3
z = find_roots(x,y2-y1)
plt.plot(x,y1)
plt.plot(x,y2, color="C2")
plt.plot(z, np.interp(z, x, y1), marker="o", ls="", ms=4, color="C1")
I'm making multiple density plots in 3D with a circle shape, using contourf.
What I want to do is similar to a clip_path for a pcolormesh plot.
How can I clip the plot in 3D in particular for a contourf plot?
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.path import Path
import matplotlib.patches as patches
fig = plt.figure()
ax = fig.gca(projection='3d')
x = np.linspace(0, 1, 100)
X, Y = np.meshgrid(x, x)
levels = np.linspace(-0.1, 0.4, 100) #(z_min,z_max,number of contour),
a=0
b=1
c=2
Z1 = a+.1*np.sin(2*X)*np.sin(4*Y)
Z2 = b+.1*np.sin(3*X)*np.sin(4*Y)
Z3 = c+.1*np.sin(4*X)*np.sin(5*Y)
plt.contourf(X, Y,Z1, levels=a+levels,cmap=plt.get_cmap('rainbow'))
plt.contourf(X, Y,Z2, levels=b+levels,cmap=plt.get_cmap('rainbow'))
plt.contourf(X, Y,Z3, levels=c+levels,cmap=plt.get_cmap('rainbow'))
ax.set_xlim3d(0, 1)
ax.set_ylim3d(0, 1)
ax.set_zlim3d(0, 2)
plt.show()
Are you sure your code generated that output image with the given parameters? I only got some blue squares. Tweaking a bit, it starts looking like yours.
To clip the contours, consider masking the Z arrays. As in:
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
import numpy.ma as ma
fig = plt.figure()
ax = fig.gca(projection='3d')
x = np.linspace(0, 1, 100)
X, Y = np.meshgrid(x, x)
levels = np.linspace(-0.1, 0.4, 100) #(z_min,z_max,number of contour),
a = 0
b = 1
c = 2
Z1 = a + .3 * np.sin(2 * X) * np.sin(4 * Y)
Z2 = b + .3 * np.sin(3 * X) * np.sin(4 * Y)
Z3 = c + .3 * np.sin(4 * X) * np.sin(5 * Y)
mask_cond = (X - .5) ** 2 + (Y - .5) ** 2 > .25
Z1 = ma.masked_where(mask_cond, Z1)
Z2 = ma.masked_where(mask_cond, Z2)
Z3 = ma.masked_where(mask_cond, Z3)
plt.contourf(X, Y, Z1, levels=a + levels, cmap='rainbow')
plt.contourf(X, Y, Z2, levels=b + levels, cmap='rainbow')
plt.contourf(X, Y, Z3, levels=c + levels, cmap='rainbow')
ax.set_xlim3d(0, 1)
ax.set_ylim3d(0, 1)
ax.set_zlim3d(0, 3)
plt.show()
I have been trying to figure an easy way to define u,v,w of mpl_toolkits.mplot3d.Axes3D.quiver such that all the quivers (from all over the 3D scatter plot) are all pointing towards the origin. Many thanks for all help rendered!!
You can do this fairly easily by starting with unit-length quivers:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
fig = plt.figure()
ax = fig.gca(projection='3d')
n = 25
# Generate some random data
x = (np.random.random(n) - 0.5) * 10
y = (np.random.random(n) - 0.5) * 10
z = (np.random.random(n) - 0.5) * 10
r = np.sqrt(np.power(x,2) + np.power(y,2) + np.power(z,2))
# Create unit-length quivers by dividing out the magnitude
u = - x / r
v = - y / r
w = - z / r
ax.quiver(x, y, z, u, v, w)
plt.show()
And then if you want quivers of a certain size or if you have an array of sizes you can simply multiply this in when creating u, v, and w:
sizes = np.random.random(n) * 4
u = -x / r * sizes
v = -y / r * sizes
w = -z / r * sizes
ax.quiver(x, y, z, u, v, w)
plt.show()
This should also support other methods of analytical sizing - provided you calculate the r vectors correctly.
The idea is to plot the following vector field:
I have two main issue with it:
1) I do not know how to make sure that the arrows are not too long (I know I have to use length, but how?).
2) I am told to use Numpyto draw the vector field but again, how?
This is what I have tried:
# The components of the vector field
F_x = y*e**x
F_y = x**2 + e**x
F_z = z**2*e**z
# The grid
xf = np.linspace(-0.15, 2.25, 8)
yf = np.linspace(-0.15, 2.25, 8)
zf = np.linspace(-0.75, 2.50, 8)
X_grid, Y_grid, Z_grid = np.meshgrid(xf, yf, zf)
# The arrows; how to deal with them?
dx = 1
#dy = ...
#dz = ...
# Standardize the arrows; In this way all arrows have the same length.
length = np.sqrt(dx**2 + dy**2 + dz**2)
dx_N = dx/length
dy_N = dy/length
dz_N = dz/length
#how to involve numpy in the process??
# Drawing the figure
fig, ax = plt.subplots(1, 1)
ax.quiver(X_grid, Y_grid, Z_grid, dx_N, dy_N, dz_N, dy, dz, cmap=plt.get_cmap('gnuplot2'))
plt.show()
Thanks
EDIT
Based on the provided link I tried:
from sympy import *
x,y,z = sp.symbols('x y z', real = True)
import matplotlib.pyplot as plt
x, y, z = np.meshgrid(np.arange(0, 2 * np.pi, .2), np.arange(0, 2 * np.pi, .2), np.arange(0, 2 * np.pi, .2))
F_x = y * exp(x)
F_y = x**2 + exp(x)
F_z = z**2 * exp(z)
# Normalize the arrows:
F_x = F_x / np.sqrt(F_x**2 + F_y**2 + F_z**2)
F_y = F_y / np.sqrt(F_x**2 + F_y**2 + F_z**2)
F_z = F_z / np.sqrt(F_x**2 + F_y**2 + F_z**2)
plt.figure()
plt.title('Vector field')
Q = plt.quiver(x, y, z, F_x, F_y, F_z, units='width')
qk = plt.quiverkey(Q, 0.9, 0.9, 2, r'$2 \frac{m}{s}$', labelpos='E',
coordinates='figure')#I don't understand this line
The TypeError: Shape should contain integers only comes up.
The problem is that I don't understand this part of the code:
qk = plt.quiverkey(Q, 0.9, 0.9, 2, r'$2 \frac{m}{s}$', labelpos='E',
coordinates='figure')
I am still stuck on how to plot this vector field
Assume that you want a 3D quiver, you can check out the matplotlib tutorial on quiver3D. And to control the arrow size, check out the Axes3d.quiver library doc, especially the parameters.
A quick snippet:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
x, y, z = np.meshgrid(np.arange(0, 2*np.pi, .5), np.arange(0, 2*np.pi, .5), np.arange(0, 2*np.pi, .5))
F_x = y * np.exp(x)
F_y = x**2 + np.exp(x)
F_z = z**2 * np.exp(z)
fig = plt.figure()
ax = fig.gca(projection='3d')
Q = ax.quiver(x, y, z, F_x, F_y, F_z, length=0.3, normalize=True)
But 3d quiver plot can be very crowded! : )
The quiver() method is a great tool to render vector fields. Since Matplotlib is a two-dimensional plotting library, we need to import the mplot3d toolkit to generate a three-dimensional plot.
Here's a good example:
Dependencies:
Axes3D for 3D rendering
Pyplot to get a MATLAB-like plotting framework
Numpy for numeric-array manipulation
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
x, y, z = np.meshgrid(np.arange(-0.8, 1, 0.25),
np.arange(-0.8, 1, 0.25),
np.arange(-0.8, 1, 0.8))
u = np.sin(np.pi * x) * np.cos(np.pi * y) * np.cos(np.pi * z)
v = -np.cos(np.pi * x) * np.sin(np.pi * y) * np.cos(np.pi * z)
w = (np.sqrt(2.0 / 3.0) * np.cos(np.pi * x) * np.cos(np.pi * y) * np.sin(np.pi * z))
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.quiver(x, y, z, u, v, w,
length=0.15,
color='Purple'
)
ax.view_init(elev=10, azim=30)
ax.dist=8
plt.show()
I'm trying to plot a Chord diagram using Matplotlib. I am aware that already existing libraries, such as Plotly give me that functionality but I would really like to do it in matplotlib.
The code I have so far looks like this:
import itertools
import matplotlib.patches as patches
import matplotlib.pyplot as plt
import numpy as np
%matplotlib inline
fig, ax = plt.subplots()
ax.axhline(0, color='black', linestyle='--')
ax.axvline(0, color='black', linestyle='--')
npoints = 3
# Calculate the xy coords for each point on the circle
s = 2 * np.pi / npoints
verts = np.zeros((npoints, 2))
for i in np.arange(npoints):
angle = s * i
x = npoints * np.cos(angle)
y = npoints * np.sin(angle)
verts[i] = [x, y]
# Plot the arcs
numbers = [i for i in xrange(npoints)]
for i, j in itertools.product(numbers, repeat=2):
if i == j:
continue
x1y1 = x1, y1 = verts[i]
x2y2 = x2, y2 = verts[j]
# Calculate the centre of the Arc
mxmy = mx, my = [(x1 + x2) / 2, (y1 + y2) / 2]
r = np.sqrt((x1 - mx)**2 + (y1 - my)**2)
xy = [mx - r, my - r]
width = 2 * r
height = 2 * r
start_angle = np.arctan2(y1 - my, x1 - mx) * 180 / np.pi
end_angle = np.arctan2(y2 - my, x2 - mx) * 180 / np.pi
arc = patches.Arc(mxmy, width, height, start_angle, end_angle)
ax.add_patch(arc)
# Plot the points
x, y = verts.T
ax.scatter(x, y, marker='o', s=50, c='r')
ax.annotate("1", (x[0], y[0]), xytext=(x[0] + .5, y[0] + .5))
ax.annotate("2", (x[1], y[1]), xytext=(x[1] - 1, y[1] + .5))
ax.annotate("3", (x[2], y[2]), xytext=(x[2] - 1, y[2] - 1))
ax.set_xlim(-npoints - 5, npoints + 6)
ax.set_ylim(-npoints - 5, npoints + 6)
ax.set(aspect=1)
Is anyone able to tell me why my plot looks like this?
I'm expecting something more like the following (image taken from Plotly)
Edit 1
I would like to draw arcs between the following points:
1 and 2
1 and 3
2 and 3
These arcs should ideally be on the inside.
Edit 2
After some further investigation I figured that the end_angle seems to be the root of the problem.
After #f5r5e5d pointing out the Bézier curve used in plotly, I've decided to give this one a go. It looks like this is the way to go in my case, too.
import itertools
import matplotlib.patches as patches
import matplotlib.pyplot as plt
import numpy as np
import sys
%matplotlib inline
fig, ax = plt.subplots()
npoints = 5
# Calculate the xy coords for each point on the circle
s = 2 * np.pi / npoints
verts = np.zeros((npoints, 2))
for i in np.arange(npoints):
angle = s * i
x = npoints * np.cos(angle)
y = npoints * np.sin(angle)
verts[i] = [x, y]
# Plot the Bezier curves
numbers = [i for i in xrange(npoints)]
bezier_path = np.arange(0, 1.01, 0.01)
for a, b in itertools.product(numbers, repeat=2):
if a == b:
continue
x1y1 = x1, y1 = verts[a]
x2y2 = x2, y2 = verts[b]
xbyb = xb, yb = [0, 0]
# Compute and store the Bezier curve points
x = (1 - bezier_path)** 2 * x1 + 2 * (1 - bezier_path) * bezier_path * xb + bezier_path** 2 * x2
y = (1 - bezier_path)** 2 * y1 + 2 * (1 - bezier_path) * bezier_path * yb + bezier_path** 2 * y2
ax.plot(x, y, 'k-')
x, y = verts.T
ax.scatter(x, y, marker='o', s=50, c='r')
ax.set_xlim(-npoints - 5, npoints + 6)
ax.set_ylim(-npoints - 5, npoints + 6)
ax.set(aspect=1)
The code above plots what I wanted it do to. Some modifications on the style and it should be good to go.
Since the underlying problem was "how can I draw a chord diagram in matplotlib", I just want to let you know that there is now a python library to do that: mpl-chord-diagram.
You can just do pip install mpl-chord-diagram.
[disclaimer] I am the current maintainer [/disclaimer]