如何在basemap中绘制网格数据并找到特定方位角上的网格点

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提问于 2025-04-17 16:29

更新:我正在尝试绘制一些数据。我有一组从参考点测量得到的反方位角(baz),这些数据来自一个网格。我想找出在这个网格上,沿着反方位角所画的“大圆”会经过的所有点。为此,我会遍历网格上的每一个点,计算这个点和参考点之间的预期反方位角,然后与每个测量得到的反方位角进行比较。如果两者之间的差异很小(小于2度),我就会给这个点加权。最后,我把所有结果放在地图上。下面是我使用的代码,但结果看起来有点奇怪,有人知道我哪里出错了吗?或者有没有比我现在的方法更快的?

from matplotlib.colorbar import ColorbarBase
import matplotlib.cm as cm
import matplotlib.pyplot as plt
import numpy as np
from mpl_toolkits.basemap import Basemap
import mpl_toolkits.basemap.pyproj as pyproj


llcrnrlon = -30.0
llcrnrlat = 45.0
urcrnrlon = 0.0
urcrnrlat = 65.0
lon_0 = (urcrnrlon + llcrnrlon) / 2.
lat_0 = (urcrnrlat + llcrnrlat) / 2.

lat = 51.58661577  # reference point
lon = -9.18822525


# Generate random back-azimuths.
baz = zeros((20))
for i in xrange(len(baz)):
  baz[i] = random.randint(200,230)


####################################################################
## Set up the map background.
m = Basemap(llcrnrlon=llcrnrlon,llcrnrlat=llcrnrlat,urcrnrlon=urcrnrlon,urcrnrlat=urcrnrlat,
    resolution='i',projection='lcc',lon_0=lon_0,lat_0=lat_0)
m.drawcoastlines()
m.fillcontinents() 

# draw parallels
m.drawparallels(np.arange(10,70,10),labels=[1,0,0,0])
# draw meridians
m.drawmeridians(np.arange(-80, 25, 10),labels=[0,0,0,1])

# Plot station locations.
x, y = m(lon, lat)            # array ref points
m.plot(x,y,'ro', ms=5)


####################################################################
## Set up the grids etc.
glons = np.linspace(llcrnrlon, urcrnrlon, 100)
glats = np.linspace(llcrnrlat, urcrnrlat, 100)

# Convert to map coords.
xlons, ylats = m(glons, glats)

# create grid for pcolormesh.
grid_lon, grid_lat = np.meshgrid(xlons, ylats)

# create weights for pcolormesh.
weights = np.zeros(np.shape(grid_lon))

# create grid of lat-lon coords for baz calculation.
gln, glt = np.meshgrid(glons, glats)


####################################################################
## calculate baz from grid_lon, grid_lat to lon, lat. If less 
## than error weight grid point.

# method for BAZ calculation via pyproj.
def get_baz(lon1, lat1, lon2, lat2):
  g = pyproj.Geod(ellps='WGS84')
  az, baz, dist = g.inv(lon1, lat1, lon2, lat2)
  return baz

# BAZ calcultion for each point in grid.
ll=0
for mBAZ in baz:
  for i in xrange(len(gln)):
    for k in xrange(len(gln[i])):
      nbaz = get_baz(lon, lat, gln[i][k], glt[i][k])
      nbaz += 180
      if abs(nbaz - mBAZ) < 2:
    weights[i][k] = 1
  ll+=1


# plot grid.
m.pcolormesh(grid_lon, grid_lat, weights, cmap=plt.cm.YlOrBr)
plt.colorbar()
plt.show()

原始问题在下面,现在已经过时了。

我正在尝试绘制一些数据。我有一个数据集,给出了每个方向的一系列值(频率)。我想把它们绘制在一个网格上,这样网格上每个特定方位角的点就可以根据特定频率的功率进行加权。我用basemap创建了一个地图,并在上面绘制了一个网格,如下所示,

from mpl_toolkits.basemap import Basemap
import matplotlib.pyplot as plt
import numpy as np
from shoot import *

llcrnrlon = -20.0
llcrnrlat = 45.0
urcrnrlon = 10.0
urcrnrlat = 65.0
lon_0 = (urcrnrlon + llcrnrlon) / 2.
lat_0 = (urcrnrlat + llcrnrlat) / 2.

m = Basemap(llcrnrlon=llcrnrlon,llcrnrlat=llcrnrlat,urcrnrlon=urcrnrlon,urcrnrlat=urcrnrlat,
        resolution='i',projection='lcc',lon_0=lon_0,lat_0=lat_0)

## Set up the grid.
glons = np.linspace(-20,10,50)
glats = np.linspace(45, 65, 50)
xlons, ylats = m(glons, glats)
grid_lon, grid_lat = np.meshgrid(xlons, ylats) 
pwr = np.zeros((50,50))

m.drawcoastlines()
m.fillcontinents() 

# draw parallels
m.drawparallels(np.arange(10,70,10),labels=[1,0,0,0])
# draw meridians
m.drawmeridians(np.arange(-80, 25, 10),labels=[0,0,0,1])

lats = [54.8639587, 51.5641564]
lons = [-8.1778180, -9.2754284]

x, y = m(lons, lats)            # array ref points

# Plot station locations.
m.plot(x,y,'ro', ms=5)
m.pcolormesh(grid_lon, grid_lat, pwr)

然后我使用一些在这个不错的网站上找到的函数来绘制我想要的大圆。

glon1 = lons[0]
glat1 = lats[0]
azimuth = 280.
maxdist = 200.
great(m, glon1, glat1, azimuth, color='orange', lw=2.0)
plt.show()

不过,仅仅绘制这条线还不够,我还想找到大圆经过的网格点,以便给它们分配一个值。有人知道该怎么做吗?

1 个回答

0

你能具体说一下你指的是哪个交叉点吗?运行你的代码只返回了一行...

在这里输入图片描述

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