I'm trying to change a colorbar attached to a scatter plot so that the minimum and maximum of the colorbar are the minimum and maximum of the data, but I want the data to be centred at zero as I'm using a colormap with white at zero. Here is my example
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(0, 1, 61)
y = np.linspace(0, 1, 61)
C = np.linspace(-10, 50, 61)
M = np.abs(C).max() # used for vmin and vmax
fig, ax = plt.subplots(1, 1, figsize=(5,3), dpi=150)
sc=ax.scatter(x, y, c=C, marker='o', edgecolor='k', vmin=-M, vmax=M, cmap=plt.cm.RdBu_r)
cbar=fig.colorbar(sc, ax=ax, label='$R - R_0$ (mm)')
ax.set_xlabel('x')
ax.set_ylabel('y')
As you can see from the attached figure, the colorbar goes down to -M, where as I want the bar to just go down to -10, but if I let vmin=-10 then the colorbar won't be zerod at white. Normally, setting vmin to +/- M when using contourf the colorbar automatically sorts to how I want. This sort of behaviour is what I expect when contourf uses levels=np.linspace(-M,M,61) rather than setting it with vmin and vmax with levels=62. An example showing the default contourf colorbar behaviour I want in my scatter example is shown below
plt.figure(figsize=(6,5), dpi=150)
plt.contourf(x, x, np.reshape(np.linspace(-10, 50, 61*61), (61,61)),
levels=62, vmin=-M, vmax=M, cmap=plt.cm.RdBu_r)
plt.colorbar(label='$R - R_0$ (mm)')
Does anyone have any thoughts? I found this link which I thought might solve the problem, but when executing the cbar.outline.set_ydata line I get this error AttributeError: 'Polygon' object has no attribute 'set_ydata' .
EDIT a little annoyed that someone has closed this question without allowing me to clarify any questions they might have, as none of the proposed solutions are what I'm asking for.
As for Normalize.TwoSlopeNorm, I do not want to rescale the smaller negative side to use the entire colormap range, I just want the colorbar attached to the side of my graph to stop at -10.
This link also does not solve my issue, as it's the TwoSlopeNorm solution again.
After changing the ylim of the colorbar, the rectangle formed by the surrounding spines is too large. You can make this outline invisible. And then add a new rectangular border:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(0, 1, 61)
y = np.linspace(0, 1, 61)
C = np.linspace(-10, 50, 61)
M = np.abs(C).max() # used for vmin and vmax
fig, ax = plt.subplots(1, 1, figsize=(5, 3), dpi=150)
sc = ax.scatter(x, y, c=C, marker='o', edgecolor='k', vmin=-M, vmax=M, cmap=plt.cm.RdBu_r)
cbar = fig.colorbar(sc, ax=ax, label='$R - R_0$ (mm)')
cb_ymin = C.min()
cb_ymax = C.max()
cb_xmin, cb_xmax = cbar.ax.get_xlim()
cbar.ax.set_ylim(cb_ymin, cb_ymax)
cbar.outline.set_visible(False) # hide the surrounding spines, which are too large after set_ylim
cbar.ax.add_patch(plt.Rectangle((cb_xmin, cb_ymin), cb_xmax - cb_xmin, cb_ymax - cb_ymin,
fc='none', ec='black', clip_on=False))
plt.show()
Another approach until v3.5 is released is to make a custom colormap that does what you want (see also https://matplotlib.org/stable/tutorials/colors/colormap-manipulation.html#sphx-glr-tutorials-colors-colormap-manipulation-py)
import matplotlib.pyplot as plt
import numpy as np
import matplotlib.cm as cm
from matplotlib.colors import ListedColormap
fig, axs = plt.subplots(2, 1)
X = np.random.randn(32, 32) + 2
pc = axs[0].pcolormesh(X, vmin=-6, vmax=6, cmap='RdBu_r')
fig.colorbar(pc, ax=axs[0])
import matplotlib.pyplot as plt
import numpy as np
import matplotlib.cm as cm
from matplotlib.colors import ListedColormap
fig, axs = plt.subplots(2, 1)
X = np.random.randn(32, 32) + 2
pc = axs[0].pcolormesh(X, vmin=-6, vmax=6, cmap='RdBu_r')
fig.colorbar(pc, ax=axs[0])
def keep_center_colormap(vmin, vmax, center=0):
vmin = vmin - center
vmax = vmax - center
dv = max(-vmin, vmax) * 2
N = int(256 * dv / (vmax-vmin))
RdBu_r = cm.get_cmap('RdBu_r', N)
newcolors = RdBu_r(np.linspace(0, 1, N))
beg = int((dv / 2 + vmin)*N / dv)
end = N - int((dv / 2 - vmax)*N / dv)
newmap = ListedColormap(newcolors[beg:end])
return newmap
newmap = keep_center_colormap(-2, 6, center=0)
pc = axs[1].pcolormesh(X, vmin=-2, vmax=6, cmap=newmap)
fig.colorbar(pc, ax=axs[1])
plt.show()
How to set ticks to be on the opposite axes please? When I drop it the axes changes but not in this way. Thank you
import matplotlib as mpl
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
mpl.rcParams['legend.fontsize'] = 10
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)
ax.plot(x, y, z, label='parametric curve')
ax.legend()
plt.show()
How to set linewidth of axis in 3d plot in python? Is it somehow possible with mpl.rcParams?
Code:
import matplotlib as mpl
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
mpl.rcParams['legend.fontsize'] = 10
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)
ax.plot(x, y, z, label='parametric curve')
ax.legend()
plt.show()
Try this:
ax.plot_surface(X, Y, Z, linewidth=1)
I am still trying to understand how solve_ivp works against odeint, but just as I was getting the hang of it something happened.
I am trying to solve for the motion of a non linear pendulum. With odeint, everything works like a charm, on solve_ivp hoever something weird happens:
import numpy as np
from matplotlib import pyplot as plt
from scipy.integrate import solve_ivp, odeint
g = 9.81
l = 0.1
def f(t, r):
omega = r[0]
theta = r[1]
return np.array([-g / l * np.sin(theta), omega])
time = np.linspace(0, 10, 1000)
init_r = [0, np.radians(179)]
results = solve_ivp(f, (0, 10), init_r, method="RK45", t_eval=time) #??????
cenas = odeint(f, init_r, time, tfirst=True)
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.plot(results.t, results.y[1])
ax1.plot(time, cenas[:, 1])
plt.show()
What am I missing?
It is a numerical problem. The default relative and absolute tolerances of solve_ivp are 1e-3 and 1e-6, respectively. For many problems, these values are too big, and tighter error tolerances should be given. The default relative tolerance for odeint is 1.49e-8.
If you add the argument rtol=1e-8 to the solve_ivp call, the plots agree:
import numpy as np
from matplotlib import pyplot as plt
from scipy.integrate import solve_ivp, odeint
g = 9.81
l = 0.1
def f(t, r):
omega = r[0]
theta = r[1]
return np.array([-g / l * np.sin(theta), omega])
time = np.linspace(0, 10, 1000)
init_r = [0, np.radians(179)]
results = solve_ivp(f, (0, 10), init_r, method='RK45', t_eval=time, rtol=1e-8)
cenas = odeint(f, init_r, time, tfirst=True)
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.plot(results.t, results.y[1])
ax1.plot(time, cenas[:, 1])
plt.show()
Plot:
draw a graph of equation in the form of y=mx+b in python3.x
example y = 5x + 9
This is a very general question. Try to be more specific. It depends how you want to draw it.
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(0., 5., 0.2)
y = 5 * x + 9
plt.plot(x, y)
plt.show()
or
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(-1., 5., 0.2)
y = 5 * x + 9
fig, ax = plt.subplots()
ax.plot(x,y)
ax.grid(True, which='both')
ax.axhline(y=0, color='k')
ax.axvline(x=0, color='k')
These are very basic drawing. You can create more sophisticated graphs, but you will have to be more specific in your question.
You can define your y(x) function and then plot it as follows:
import matplotlib.pyplot as plt
def y(x):
return [5*i+9 for i in x]
x = range(0,10)
plt.plot(x,y(x))
plt.show()
This produces follwing graph:
With turtle
You can as well get a graph with turtle with following code for example:
from turtle import Turtle, Screen
def y(x):
return 5*x+9
def plotter(turtle, x_range):
turtle.penup()
for x in x_range:
turtle.goto(x, y(x))
turtle.pendown()
screen = Screen()
screen.setworldcoordinates(0, 0, 9, 60)
turtle = Turtle(visible=False)
x = range(0,10)
plotter(turtle, x)
screen.exitonclick()
Which produces: