我想为地图上沿路径从一个位置移动到另一个位置的点设置动画。
例如，我使用大地变换绘制了一条从纽约到新德里的路径。例如。取自文档Adding data to the map
plt.plot([ny_lon, delhi_lon], [ny_lat, delhi_lat],
color='blue', linewidth=2, marker='o',
transform=ccrs.Geodetic(),
)
现在我想沿着这条路径移动一个点。
我的想法是沿着路径以某种方式获得一些（比如 50 个）点，并在每个帧的每个点上绘制一个标记。但我无法找到一种方法来获得路径上的点。
我在 classCRS 下找到了一个函数 transform_points，但我无法使用它，因为这给了我相同数量的点数，而不是两者之间的点数。
提前致谢！
【问题讨论】：
标签： cartopy
有几种方法可以解决这个问题。
如果您熟悉 matplotlib，我可能会从最基本的开始，但这种方法会间接使用 cartopy 的功能，因此更难配置/扩展。
Line2D 对象（从plt.plot 返回的东西）上有一个私有的_get_transformed_path 方法。生成的 TransformedPath 对象有一个 get_transformed_path_and_affine 方法，它基本上会给我们投影线（在正在绘制的 Axes 的坐标系中）。
In [1]: import cartopy.crs as ccrs
In [3]: import matplotlib.pyplot as plt
In [4]: ax = plt.axes(projection=ccrs.Robinson())
In [6]: ny_lon, ny_lat = -75, 43
In [7]: delhi_lon, delhi_lat = 77.23, 28.61
In [8]: [line] = plt.plot([ny_lon, delhi_lon], [ny_lat, delhi_lat],
...: color='blue', linewidth=2, marker='o',
...: transform=ccrs.Geodetic(),
...: )
In [9]: t_path = line._get_transformed_path()
In [10]: path_in_data_coords, _ = t_path.get_transformed_path_and_affine()
In [11]: path_in_data_coords.vertices
Out[11]:
array([[-6425061.82215208, 4594257.92617961],
[-5808923.84969279, 5250795.00604155],
[-5206753.88613758, 5777772.51828996],
[-4554622.94040482, 6244967.03723341],
[-3887558.58343227, 6627927.97123701],
[-3200922.19194864, 6932398.19937816],
[-2480001.76507805, 7165675.95095855],
[-1702269.5101901 , 7332885.72276795],
[ -859899.12295981, 7431215.78426759],
[ 23837.23431173, 7453455.61302756],
[ 889905.10635756, 7397128.77301289],
[ 1695586.66856764, 7268519.87627204],
[ 2434052.81300274, 7073912.54130764],
[ 3122221.22299409, 6812894.40443648],
[ 3782033.80448001, 6478364.28561403],
[ 4425266.18173684, 6062312.15662039],
[ 5049148.25986903, 5563097.6328901 ],
[ 5616318.74912886, 5008293.21452795],
[ 6213232.98764984, 4307186.23400115],
[ 6720608.93929235, 3584542.06839575],
[ 7034261.06659143, 3059873.62740856]])
我们可以将它与 matplotlib 的动画功能结合起来按照要求进行：
import cartopy.crs as ccrs
import matplotlib.animation as animation
import matplotlib.pyplot as plt
ax = plt.axes(projection=ccrs.Robinson())
ax.stock_img()
ny_lon, ny_lat = -75, 43
delhi_lon, delhi_lat = 77.23, 28.61
[line] = plt.plot([ny_lon, delhi_lon], [ny_lat, delhi_lat],
color='blue', linewidth=2, marker='o',
transform=ccrs.Geodetic(),
)
t_path = line._get_transformed_path()
path_in_data_coords, _ = t_path.get_transformed_path_and_affine()
# Draw the point that we want to animate.
[point] = plt.plot(ny_lon, ny_lat, marker='o', transform=ax.projection)
def animate_point(i):
verts = path_in_data_coords.vertices
i = i % verts.shape[0]
# Set the coordinates of the line to the coordinate of the path.
point.set_data(verts[i, 0], verts[i, 1])
ani = animation.FuncAnimation(
ax.figure, animate_point,
frames= path_in_data_coords.vertices.shape[0],
interval=125, repeat=True)
ani.save('point_ani.gif', writer='imagemagick')
plt.show()
在底层，cartopy 的 matplotlib 实现（如上所示）正在调用 project_geometry 方法。我们不妨直接使用它，因为使用 Shapely 几何图形通常比使用 matplotlib 路径更方便。
使用这种方法，我们只需定义一个形状匀称的几何图形，然后构建我们想要转换几何图形的源和目标坐标参考系：
target_cs.project_geometry(geometry, source_cs)
我们唯一需要注意的是结果可以是 MultiLineString（或更一般地说，任何多几何类型）。但是，在我们的简单案例中，我们不需要处理它（顺便提一下，第一个示例中返回的简单 Path 也是如此）。
产生与上面类似的情节的代码：
import cartopy.crs as ccrs
import matplotlib.animation as animation
import matplotlib.pyplot as plt
import numpy as np
import shapely.geometry as sgeom
ax = plt.axes(projection=ccrs.Robinson())
ax.stock_img()
ny_lon, ny_lat = -75, 43
delhi_lon, delhi_lat = 77.23, 28.61
line = sgeom.LineString([[ny_lon, ny_lat], [delhi_lon, delhi_lat]])
projected_line = ccrs.PlateCarree().project_geometry(line, ccrs.Geodetic())
# We only animate along one of the projected lines.
if isinstance(projected_line, sgeom.MultiLineString):
projected_line = projected_line.geoms[0]
ax.add_geometries(
[projected_line], ccrs.PlateCarree(),
edgecolor='blue', facecolor='none')
[point] = plt.plot(ny_lon, ny_lat, marker='o', transform=ccrs.PlateCarree())
def animate_point(i):
verts = np.array(projected_line.coords)
i = i % verts.shape[0]
# Set the coordinates of the line to the coordinate of the path.
point.set_data(verts[i, 0], verts[i, 1])
ani = animation.FuncAnimation(
ax.figure, animate_point,
frames=len(projected_line.coords),
interval=125, repeat=True)
ani.save('projected_line_ani.gif', writer='imagemagick')
plt.show()
该方法自然地推广到为任何类型的 matplotlib Arrrrtist 设置动画......在这种情况下，我对大圆分辨率进行了更多控制，并沿大圆制作了一个图像：
import cartopy.crs as ccrs
import matplotlib.animation as animation
import matplotlib.pyplot as plt
import numpy as np
import shapely.geometry as sgeom
ax = plt.axes(projection=ccrs.Mercator())
ax.stock_img()
line = sgeom.LineString([[-5.9845, 37.3891], [-82.3666, 23.1136]])
# Higher resolution version of Mercator. Same workaround as found in
# https://github.com/SciTools/cartopy/issues/8#issuecomment-326987465.
class HighRes(ax.projection.__class__):
@property
def threshold(self):
return super(HighRes, self).threshold / 100
projected_line = HighRes().project_geometry(line, ccrs.Geodetic())
# We only animate along one of the projected lines.
if isinstance(projected_line, sgeom.MultiLineString):
projected_line = projected_line.geoms[0]
# Add the projected line to the map.
ax.add_geometries(
[projected_line], ax.projection,
edgecolor='blue', facecolor='none')
def ll_to_extent(x, y, ax_size=(4000000, 4000000)):
"""
Return an image extent in centered on the given
point with the given width and height.
"""
return [x - ax_size[0] / 2, x + ax_size[0] / 2,
y - ax_size[1] / 2, y + ax_size[1] / 2]
# Image from https://pixabay.com/en/sailing-ship-boat-sail-pirate-28930/.
pirate = plt.imread('pirates.png')
img = ax.imshow(pirate, extent=ll_to_extent(0, 0), transform=ax.projection, origin='upper')
ax.set_global()
def animate_ship(i):
verts = np.array(projected_line.coords)
i = i % verts.shape[0]
# Set the extent of the image to the coordinate of the path.
img.set_extent(ll_to_extent(verts[i, 0], verts[i, 1]))
ani = animation.FuncAnimation(
ax.figure, animate_ship,
frames=len(projected_line.coords),
interval=125, repeat=False)
ani.save('arrrr.gif', writer='imagemagick')
plt.show()
这个答案的所有代码和图片都可以在https://gist.github.com/pelson/618a5f4ca003e56f06d43815b21848f6找到。
【讨论】：