Question

我正在寻找一种解决方法，在Lambert投影中将x和y轴刻度和标签添加到Cartopy地图。

我提出的解决方案只是一个近似值，对于较大的地图会产生更糟糕的结果：它涉及使用transform_points方法将所需的刻度位置转换为地图投影。为此，我使用y轴（或x轴）的中位数经度（或纬度）以及所需的纬度（或经度）刻度位置来计算地图投影坐标。请参阅下面的代码。

因此，我假设沿y轴的恒定经度（沿x轴的纬度），这是不正确的，因此导致偏差。（注意附加结果图中的差异：在set_extent中设置46°并产生刻度位置）。

有没有更准确的解决方案？有什么提示我怎么能解决这个问题呢？

感谢任何想法！

import matplotlib.pyplot as plt
import cartopy.crs as ccrs
import numpy as np

def main():
    #my desired Lambert projection:
    myproj = ccrs.LambertConformal(central_longitude=13.3333, central_latitude=47.5,
                                   false_easting=400000, false_northing=400000,
                                   secant_latitudes=(46, 49))

    arat = 1.1 #just some factor for the aspect ratio
    fig_len = 12
    fig_hig = fig_len/arat
    fig = plt.figure(figsize=(fig_len,fig_hig), frameon=True)
    ax = fig.add_axes([0.08,0.05,0.8,0.94], projection = myproj)

    ax.set_extent([10,16,46,49])
    #This is what is not (yet) working in Cartopy due to Lambert projection:
    #ax.gridlines(draw_labels=True) #TypeError: Cannot label gridlines on a LambertConformal plot.  Only PlateCarree and Mercator plots are currently supported.
    x_lons = [12,13,14] #want these longitudes as tick positions
    y_lats = [46, 47, 48, 49] #want these latitudes as tick positions
    tick_fs = 16
    #my workaround functions:
    cartopy_xlabel(ax,x_lons,myproj,tick_fs)
    cartopy_ylabel(ax,y_lats,myproj,tick_fs)

    plt.show()
    plt.close()

def cartopy_xlabel(ax,x_lons,myproj,tick_fs):    
    #transform the corner points of my map to lat/lon
    xy_bounds = ax.get_extent()
    ll_lonlat = ccrs.Geodetic().transform_point(xy_bounds[0],xy_bounds[2], myproj)
    lr_lonlat = ccrs.Geodetic().transform_point(xy_bounds[1],xy_bounds[2], myproj)
    #take the median value as my fixed latitude for the x-axis
    l_lat_median = np.median([ll_lonlat[1],lr_lonlat[1]]) #use this lat for transform on lower x-axis
    x_lats_helper = np.ones_like(x_lons)*l_lat_median

    x_lons = np.asarray(x_lons)
    x_lats_helper = np.asarray(x_lats_helper)
    x_lons_xy = myproj.transform_points(ccrs.Geodetic(), x_lons,x_lats_helper)
    x_lons_xy = list(x_lons_xy[:,0]) #only lon pos in xy are of interest     
    x_lons = list(x_lons)

    x_lons_labels =[]
    for j in xrange(len(x_lons)):
        if x_lons[j]>0:
            ew=r'$^\circ$E'
        else:
            ew=r'$^\circ$W'
        x_lons_labels.append(str(x_lons[j])+ew)
    ax.set_xticks(x_lons_xy)
    ax.set_xticklabels(x_lons_labels,fontsize=tick_fs)

def cartopy_ylabel(ax,y_lats,myproj,tick_fs):        
    xy_bounds = ax.get_extent()
    ll_lonlat = ccrs.Geodetic().transform_point(xy_bounds[0],xy_bounds[2], myproj)
    ul_lonlat = ccrs.Geodetic().transform_point(xy_bounds[0],xy_bounds[3], myproj)
    l_lon_median = np.median([ll_lonlat[0],ul_lonlat[0]]) #use this lon for transform on left y-axis
    y_lons_helper = np.ones_like(y_lats)*l_lon_median

    y_lats = np.asarray(y_lats)    
    y_lats_xy = myproj.transform_points(ccrs.Geodetic(), y_lons_helper, y_lats)
    y_lats_xy = list(y_lats_xy[:,1]) #only lat pos in xy are of interest 

    y_lats = list(y_lats)

    y_lats_labels =[]
    for j in xrange(len(y_lats)):
        if y_lats[j]>0:
            ew=r'$^\circ$N'
        else:
            ew=r'$^\circ$S'
        y_lats_labels.append(str(y_lats[j])+ew)
    ax.set_yticks(y_lats_xy)
    ax.set_yticklabels(y_lats_labels,fontsize=tick_fs)

if __name__ == '__main__': main()

enter image description here

Answer 1

本笔记本详细介绍了我的（非常粗略）解决方法：http://nbviewer.ipython.org/gist/ajdawson/dd536f786741e987ae4e

笔记本需要使用cartopy＆gt; = 0.12。

我所做的就是找到相应网格线与地图边界的交点。我假设地图边界始终是矩形的，我只能标记底部和左侧。希望这可以作为建立的东西。

Answer 2

我自己没有尝试过，但我注意到salem package docs有能力处理其他投影的网格线与他们自己的绘图工具，这不会改变matplotlib的轴的投影。

Answer 3

自cartopy v0.18.0开始，任何cartopy投影现在都支持标签网格线。 https://twitter.com/QuLogic/status/1257148289838911488

Answer 4

不幸的是，仍然使用 0.18 版本，我无法在所有投影中标记轴网格。我不得不修改 this github repo 上提出的解决方案才能为我工作。这是我的工作解决方法：

def gridlines_with_labels(ax, top=True, bottom=True, left=True,
                      right=True, **kwargs):
"""
Like :meth:`cartopy.mpl.geoaxes.GeoAxes.gridlines`, but will draw
gridline labels for arbitrary projections.
Parameters
----------
ax : :class:`cartopy.mpl.geoaxes.GeoAxes`
    The :class:`GeoAxes` object to which to add the gridlines.
top, bottom, left, right : bool, optional
    Whether or not to add gridline labels at the corresponding side
    of the plot (default: all True).
kwargs : dict, optional
    Extra keyword arguments to be passed to :meth:`ax.gridlines`.
Returns
-------
:class:`cartopy.mpl.gridliner.Gridliner`
    The :class:`Gridliner` object resulting from ``ax.gridlines()``.
Example
-------
>>> import matplotlib.pyplot as plt
>>> import cartopy.crs as ccrs
>>> plt.figure(figsize=(10, 10))
>>> ax = plt.axes(projection=ccrs.Orthographic(-5, 53))
>>> ax.set_extent([-10.0, 0.0, 50.0, 56.0], crs=ccrs.PlateCarree())
>>> ax.coastlines('10m')
>>> gridlines_with_labels(ax)
>>> plt.show()
"""

# Add gridlines
gridliner = ax.gridlines(**kwargs)

ax.tick_params(length=0)

# Get projected extent
xmin, xmax, ymin, ymax = ax.get_extent()

# Determine tick positions
sides = {}
N = 500
if bottom:
    sides['bottom'] = np.stack([np.linspace(xmin, xmax, N),
                                np.ones(N) * ymin])
if top:
    sides['top'] = np.stack([np.linspace(xmin, xmax, N),
                            np.ones(N) * ymax])
if left:
    sides['left'] = np.stack([np.ones(N) * xmin,
                              np.linspace(ymin, ymax, N)])
if right:
    sides['right'] = np.stack([np.ones(N) * xmax,
                               np.linspace(ymin, ymax, N)])

# Get latitude and longitude coordinates of axes boundary at each side
# in discrete steps
gridline_coords = {}
for side, values in sides.items():
    gridline_coords[side] = ccrs.PlateCarree().transform_points(
        ax.projection, values[0], values[1])

lon_lim, lat_lim = gridliner._axes_domain()
ticklocs = {
    'x': gridliner.xlocator.tick_values(lon_lim[0], lon_lim[1]),
    'y': gridliner.ylocator.tick_values(lat_lim[0], lat_lim[1])
}

# Compute the positions on the outer boundary where
coords = {}
for name, g in gridline_coords.items():
    if name in ('bottom', 'top'):
        compare, axis = 'x', 0
    else:
        compare, axis = 'y', 1
    coords[name] = np.array([
        sides[name][:, np.argmin(np.abs(
            gridline_coords[name][:, axis] - c))]
        for c in ticklocs[compare]
    ])

# Create overlay axes for top and right tick labels
ax_topright = ax.figure.add_axes(ax.get_position(), frameon=False)
ax_topright.tick_params(
    left=False, labelleft=False,
    right=right, labelright=right,
    bottom=False, labelbottom=False,
    top=top, labeltop=top,
    length=0
)
ax_topright.set_xlim(ax.get_xlim())
ax_topright.set_ylim(ax.get_ylim())

for side, tick_coords in coords.items():
    if side in ('bottom', 'top'):
        axis, idx = 'x', 0
    else:
        axis, idx = 'y', 1

    _ax = ax if side in ('bottom', 'left') else ax_topright

    ticks = tick_coords[:, idx]

    valid = np.logical_and(
        ticklocs[axis] >= gridline_coords[side][0, idx],
        ticklocs[axis] <= gridline_coords[side][-1, idx])

    if side in ('bottom', 'top'):
        _ax.set_xticks(ticks[valid])
        _ax.set_xticklabels([LONGITUDE_FORMATTER.format_data(t)
                             for t in ticklocs[axis][valid]])
    else:
        _ax.set_yticks(ticks[valid])
        _ax.set_yticklabels([LATITUDE_FORMATTER.format_data(t)
                             for t in np.asarray(ticklocs[axis])[valid]])

return gridliner

Cartopy：轴标签 - 解决方法

4 个答案: