CODEWITHSUNDEEP
matplotlibgeometrymatplotlib-basemapcartopymetpy
Johndo
Johndo

Reputation: 1

How to plot a circle with a varying radius in each quadrant?

I'd like to draw a shape similar to the following example (from the Naval Research Laboratory TC page). The shape is defined by 4 radii, one for each quadrant.

sample

I have multiple track centers in latitude and longitude coordinates of which I plot using Basemap:

def m_plot_wind_speeds(x,y, mps):

    # There's a switch-like statement here to determine the color of the
    # line based on wind speed which I ignored. This is passed to the 
    # color kwarg in m.plot as cur_color. 

    m.plot(x,y, '.-', markersize=ms, linewidth=lw, color=cur_color, \
 mew=1.5, markerfacecolor='k')

m = Basemap(projection='cyl',area_thresh=1000, \
    llcrnrlat=southLat,urcrnrlat=northLat,llcrnrlon=westLong,urcrnrlon=eastLong,resolution='h')
parallels = np.arange(southLat,northLat+10,10.) # make latitude lines ever 5 degrees from 30N-50N
meridians = np.arange(westLong,eastLong+30,30.) # make longitude lines every 5 degrees from 95W to 70W

m.drawparallels(parallels,labels=[1,0,0,0],labelstyle="+/-", linewidth=0, fontsize=6)
m.drawmeridians(meridians,labels=[0,0,0,1],labelstyle="+/-", linewidth=0, fontsize=6)
m.drawcountries(linewidth=0.25)

m.bluemarble()

# data is a [10]x[~]x[10] list. There are 10 trajectories, each with
# varying lengths. Each trajectory has 10 attributes.

for traj in data:
    lat = []
    lon = []
    wind_speed=[]

    for i in traj:
        lat.append(float(i[1]))
        lon.append(float(i[0]))
        wind_speed.append(float(i[2]))

   for j,var in enumerate(traj):
        if j > 0:
            x,y = m([lon[j], lon[j-1]], [lat[j], lat[j-1]])
        else:
            x,y = m([lon[j], lon[j]],[lat[j], lat[j]])
        m_plot_wind_speeds(x,y,wind_speed[j])


        # TODO: Insert a function here that takes in a 4 radii and plots them 
        # in each quadrant.

Upvotes: 0

Views: 633

Answers (1)

JohanC
JohanC

Reputation: 80509

If you don't mind having lines towards the center, wedges can be used.

Alternatively, using arcs, hlines and vlines, the outline can be drawn. You'd still need either wedges or a specially crafted polygon to fill.

import numpy as np
from matplotlib import patches
import matplotlib.pyplot as plt

def draw_quadrants_arcs(xcenter, ycenter, radii, lw=2, ec='crimson', ax=None):
    ax = ax or plt.gca()
    for rad, theta in zip(radii, [0, 90, 180, 270]):
        arc = patches.Arc((xcenter, ycenter), 2*rad, 2*rad, theta1=theta, theta2=theta+90,
                              lw=lw, ec=ec, fc='none')
        ax.add_patch(arc)
    ax.hlines([ycenter, ycenter], [xcenter + radii[0], xcenter - radii[1]], [xcenter + radii[3], xcenter - radii[2]],
              lw=lw, colors=ec)
    ax.vlines([xcenter, xcenter], [ycenter + radii[0], ycenter - radii[2]], [ycenter + radii[1], ycenter - radii[3]],
              lw=lw, colors=ec)

def draw_quadrants_wedges(xcenter, ycenter, radii, lw=2, ec='crimson', fc='lime', alpha=1, ax=None):
    ax = ax or plt.gca()
    for rad, theta in zip(radii, [0, 90, 180, 270]):
        wedge = patches.Wedge((xcenter, ycenter), rad, theta, theta + 90,
                              lw=lw, ec=ec, fc=fc, alpha=alpha)
        ax.add_patch(wedge)

xcenter, ycenter = 6, 10
radii = [6, 2, 4, 7]
# only wedges
draw_quadrants_wedges(xcenter, ycenter, radii)
# only arcs
draw_quadrants_arcs(xcenter+12, ycenter, radii)
# wedges and arcs together
draw_quadrants_wedges(xcenter+24, ycenter, radii, ec='none', lw=0, fc='limegreen', alpha=0.3)
draw_quadrants_arcs(xcenter+24, ycenter, radii)

plt.xlim(0, 40)
plt.ylim(0, 20)
plt.gca().set_aspect('equal', 'box')
plt.show()

example plot

Upvotes: 1

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