Cartopy:轴标签 - 解决方法 [英] Cartopy: axis label - workaround
问题描述
我正在寻找一种解决方法，以将x和y轴刻度和标签添加到Lambert投影中的Cartopy地图中.
I am looking for a workaround to add x and y axis ticks and labels to a Cartopy map in Lambert projection.
我提出的解决方案只是一个近似值，它会为较大的地图产生更糟糕的结果: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.
The solution I have come up with is just an approximation which will yield worse results for larger maps: It involves transforming desired tick locations to map projection using the transform_points method. For this I use the median longitude (or latitude) of my y axis (or x axis) together with the desired latitudes (or longitudes) tick positions to compute map projection coordinates. See code below.
因此，我假设沿着y轴的经度是恒定的(沿着x轴的纬度)是不正确的，因此会导致偏差.(请注意所附结果图中的差异:set_extent中的46°设置和结果刻度位置).
Thus, I am assuming constant longitudes along the y-axis (latitudes along the x-axis), which is not correct and hence leads to deviations. (Note the difference in the attached resulting figure: 46° set in set_extent and resulting tick position).
有没有更准确的解决方案?有什么提示我可以如何解决这个问题吗?
Are there any solutions out there which are more accurate? Any hints how I could approach this problem otherwise?
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()
推荐答案
我的(相当粗糙的)解决方法在这个笔记本中有详细说明:http://nbviewer.ipython.org/gist/ajdawson/dd536f786741e987ae4e
My (quite crude) work-around to this is detailed in this notebook: http://nbviewer.ipython.org/gist/ajdawson/dd536f786741e987ae4e
笔记本电脑需要的Cartopy> = 0.12.
The notebook requires cartopy >= 0.12.
我所要做的就是找到适当的网格线与地图边界的交点.我假设地图边界总是矩形的，我只能标记底部和左侧.希望这可能会有所帮助.
All I've done is find the intersection of the appropriate gridline with the map boundary. I've assumed the map boundary will always be rectangular, and I can only label the bottom and left sides. Hopefully this might be useful as something to build on.
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