from mpl_toolkits.basemap import Basemap
import matplotlib.pyplot as plt
import numpy as np
m = Basemap(projection='cyl',resolution='c',area_thresh=10,llcrnrlon=-180,urcrnrlon=180,\
llcrnrlat=-90,urcrnrlat=90)
m.etopo()
实际上,我不知道如何提供显示比例尺所需的lat,lon,lat0和lon0参数。如何提供?
map.drawmapscale(????,barstyle='simple',units='km',fontsize=9,labelstyle='simple',fontcolor='k')
中的教程
http://matplotlib.org/basemap/api/basemap_api.html将其描述如下:
drawmapscale(lon, lat, lon0, lat0, length, barstyle='simple', units='km', fontsize=9, yoffset=None, labelstyle='simple', fontcolor='k', fillcolor1='w', fillcolor2='k', ax=None, format='%d', zorder=None)
如果有人可以帮助我,我将不胜感激。
答案 0 :(得分:5)
drawmapscale
似乎不支持Basemap
个projection='cyl'
个实例(可能还有其他人;我只检查了projection='cyl'
和projection='moll'
):< / p>
In [7]: m = Basemap(projection='cyl',resolution='c',area_thresh=10,llcrnrlon=-180,\
urcrnrlon=180, llcrnrlat=-90,urcrnrlat=90)
In [8]: m.etopo()
Out[8]: <matplotlib.image.AxesImage at 0x10a899e90>
In [10]: m.drawmapscale(50, -75, 0, 0, 400)
这会导致以下错误:
ValueError: cannot draw map scale for projection='cyl'
但drawmapscale
似乎确实适用于其他预测。以Mollweide为例:
In [11]: m = Basemap(projection='moll', lon_0=0)
In [12]: m.etopo()
Out[12]: <matplotlib.image.AxesImage at 0x10c299450>
In [13]: m.drawmapscale(50, -75, 0, 0, 400)
Out[13]:
[<matplotlib.lines.Line2D at 0x11d2e41d0>,
<matplotlib.lines.Line2D at 0x109cd4d90>,
<matplotlib.lines.Line2D at 0x11d2e4750>,
<matplotlib.text.Text at 0x11d2e4d90>,
<matplotlib.text.Text at 0x11d2e5610>]
不幸的是,Basemap API似乎没有提到任何关于它不适用于所有预测的内容。但here似乎是一种潜在的解决方法。