我想包围某一点。圆的半径需要为5公里,但如何设置我的标记以使圆圈在地图上5公里?
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.basemap import Basemap
width, height = 400000, 320000
ax=plt.figure(figsize=(20,10))
lonA =[2.631547,2.861595,2.931014]
latA =[51.120983,51.209122,51.238868]
m= Basemap(width=width,height=height,projection='lcc',
resolution='h',lat_0=52.35,lon_0=4.5)
m.drawmapboundary(fill_color='turquoise')
m.fillcontinents(color='white',lake_color='aqua')
m.drawcountries(linestyle='--')
scatter2=m.scatter([], [], s=100, c='white', marker='o', label = 'Aurelia aurita', zorder=3, alpha=0.5, edgecolor='steelblue')
z,a = m(lonA[0:3], latA[0:3])
scatter2.set_offsets(np.c_[z,a])
plt.show()
答案 0 :(得分:0)
要以地图单位(米)绘制具有指定半径的圆,首先,我创建一个函数(genCircle2),它接受圆的输入参数并返回沿该圆的周长的点阵列。在下面的代码中,命令m(lon,lat)
用于计算(mx,my)以米为单位的地图投影坐标。
命令genCircle2(cx=mx, cy=my, rad=5000.)
计算圆绘图的点数。这是工作代码。
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.basemap import Basemap
def genCircle2(cx=0, cy=0, rad=1):
"""Generate points along perimeters of a circle"""
points = []
segs = 20
for ea in range(segs+1):
xi = cx + rad*np.cos(ea*2.*np.pi/segs)
yi = cy + rad*np.sin(ea*2.*np.pi/segs)
points.append([xi,yi])
return np.array(points)
width, height = 400000, 320000
ax = plt.figure(figsize=(12,10))
# long, lat
lonA = [2.631547, 2.861595, 2.931014]
latA = [51.120983, 51.209122, 51.238868]
# accompanying attributes, colors and ...
clrs = ['r', 'g', 'b']
m = Basemap(width=width, height=height, projection='lcc', \
resolution='i', lat_0=52.35, lon_0=4.5)
m.drawmapboundary(fill_color='turquoise')
m.fillcontinents(color='white', lake_color='aqua')
m.drawcountries(linestyle='--')
# plot circles at points defined by (lonA,latA)
for lon,lat,clr in zip(lonA, latA, clrs):
mx,my = m(lon,lat) # get map coordinates from (lon,lat)
cclpnts = genCircle2(cx=mx, cy=my, rad=5000.) # get points along circle's perimeter
m.plot(cclpnts[:,0], cclpnts[:,1], \
label='Aurelia aurita', color=clr, \
linewidth=0.75) # plot circle
plt.show()
结果图: