How To Draw A Filled Arc In Matplotlib
Solution 1:
@jeanrjc's solution almost gets you there, but it adds a completely unnecessary white triangle, which will hide other objects as well (see figure below, version 1).
This is a simpler approach, which only adds a polygon of the arc:
Basically we create a series of points (points
) along the edge of the circle (from theta1
to theta2
). This is already enough, as we can set the close
flag in the Polygon
constructor which will add the line from the last to the first point (creating a closed arc).
import matplotlib.patches as mpatches
import matplotlib.pyplot as plt
import numpy as np
def arc_patch(center, radius, theta1, theta2, ax=None, resolution=50, **kwargs):
# make sure ax is not empty
if ax is None:
ax = plt.gca()
# generate the points
theta = np.linspace(np.radians(theta1), np.radians(theta2), resolution)
points = np.vstack((radius*np.cos(theta) + center[0],
radius*np.sin(theta) + center[1]))
# build the polygon and add it to the axes
poly = mpatches.Polygon(points.T, closed=True, **kwargs)
ax.add_patch(poly)
return poly
And then we apply it:
fig, ax = plt.subplots(1,2)
# @jeanrjc solution, which might hide other objects in your plot
ax[0].plot([-1,1],[1,-1], 'r', zorder = -10)
filled_arc((0.,0.3), 1, 90, 180, ax[0], 'blue')
ax[0].set_title('version 1')
# simpler approach, which really is just the arc
ax[1].plot([-1,1],[1,-1], 'r', zorder = -10)
arc_patch((0.,0.3), 1, 90, 180, ax=ax[1], fill=True, color='blue')
ax[1].set_title('version 2')
# axis settings
for a in ax:
a.set_aspect('equal')
a.set_xlim(-1.5, 1.5)
a.set_ylim(-1.5, 1.5)
plt.show()
Result (version 2):
Solution 2:
You can use fill_between
to achieve this
import matplotlib.patches as mpatches
import matplotlib.pyplot as plt
import numpy as np
fg, ax = plt.subplots(1, 1)
r=2.
yoff=-1
x=np.arange(-1.,1.05,0.05)
y=np.sqrt(r-x**2)+yoff
ax.fill_between(x,y,0)
ax.axis([-2, 2, -2, 2])
ax.set_aspect("equal")
fg.canvas.draw()
Play around with r and yoff to move the arc
EDIT:
OK, so you want to be able to plot arbitrary angles? You just need to find the equation of the chord, rather than using a flat line like above. Here's a function to do just that:
import matplotlib.patches as mpatches
import matplotlib.pyplot as plt
import numpy as np
fg, ax = plt.subplots(1, 1)
col='rgbkmcyk'
def filled_arc(center,r,theta1,theta2):
# Range of angles
phi=np.linspace(theta1,theta2,100)
# x values
x=center[0]+r*np.sin(np.radians(phi))
# y values. need to correct for negative values in range theta=90--270
yy = np.sqrt(r-x**2)
yy = [-yy[i] if phi[i] > 90 and phi[i] < 270 else yy[i] for i in range(len(yy))]
y = center[1] + np.array(yy)
# Equation of the chord
m=(y[-1]-y[0])/(x[-1]-x[0])
c=y[0]-m*x[0]
y2=m*x+c
# Plot the filled arc
ax.fill_between(x,y,y2,color=col[theta1/45])
# Lets plot a whole range of arcs
for i in [0,45,90,135,180,225,270,315]:
filled_arc([0,0],1,i,i+45)
ax.axis([-2, 2, -2, 2])
ax.set_aspect("equal")
fg.savefig('filled_arc.png')
And here's the output:
Solution 3:
Here's a simpler workaround. Use the hatch argument in your mpatches.Arc command. If you repeat symbols with the hatch argument it increases the density of the patterning. I find that if you use 6 dashes, '-', or 6 dots, '.' (others probably also work), then it solidly fills in the arc as desired. When I run this
import matplotlib.patches as mpatches
import matplotlib.pyplot as plt
plt.axes()
pac = mpatches.Arc([0, -2.5], 5, 5, 45, theta1=45, theta2=135, hatch = '......')
plt.gca().add_patch(pac)
pac.set_color('cyan')
plt.axis('equal')
plt.show()
I get this:
Arc filled with dense dot hatch and rotated 45 degrees just for show
Solution 4:
You can draw a wedge, and then hide part of it with a triangle:
import matplotlib.patches as mpatches
import matplotlib.pyplot as plt
import numpy as np
def filled_arc(center, radius, theta1, theta2, ax, color):
circ = mpatches.Wedge(center, radius, theta1, theta2, fill=True, color=color)
pt1 = (radius * (np.cos(theta1*np.pi/180.)) + center[0],
radius * (np.sin(theta1*np.pi/180.)) + center[1])
pt2 = (radius * (np.cos(theta2*np.pi/180.)) + center[0],
radius * (np.sin(theta2*np.pi/180.)) + center[1])
pt3 = center
pol = mpatches.Polygon([pt1, pt2, pt3], color=ax.get_axis_bgcolor(),
ec=ax.get_axis_bgcolor(), lw=2 )
ax.add_patch(circ)
ax.add_patch(pol)
and then you can call it:
fig, ax = plt.subplots(1,2)
filled_arc((0,0), 1, 45, 135, ax[0], "blue")
filled_arc((0,0), 1, 0, 40, ax[1], "blue")
and you get:
or:
fig, ax = plt.subplots(1, 1)
for i in range(0,360,45):
filled_arc((0,0), 1, i, i+45, ax, plt.cm.jet(i))
and you get:
HTH
Solution 5:
The command ax.get_axis_bgcolor()
needs to be replaced by ax.get_fc()
for newer matplotlib.
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