我有一些示例代码来制作极坐标轮廓图:
import numpy as np
import matplotlib.pyplot as plt
azimuths = np.radians(np.linspace(0, 180, 90))
zeniths = np.arange(50, 5050, 50)
r, theta = np.meshgrid(zeniths, azimuths)
values = 90.0+5.0*np.random.random((len(azimuths), len(zeniths)))
fig, ax = plt.subplots(subplot_kw=dict(projection='polar'))
ax.set_theta_zero_location("W")
pp = plt.contourf(theta, r, values, label='tmp')
cbar = plt.colorbar(pp, orientation='vertical')
cbar.ax.set_ylabel('scale label')
plt.show()
这给了我类似的东西:
...但我想要更像这样的东西:
......中间有空间,只显示0到180度。有没有人知道这样做的便捷方法?
答案 0 :(得分:1)
我不确定方便这是怎么回事,但这是一个可以解决的问题(取自here):
import numpy as np
import mpl_toolkits.axisartist.floating_axes as floating_axes
from matplotlib.projections import PolarAxes
from mpl_toolkits.axisartist.grid_finder import FixedLocator, MaxNLocator, \
DictFormatter
import matplotlib.pyplot as plt
tr = PolarAxes.PolarTransform()
degree_ticks = lambda d: (d*np.pi/180, "%d$^\\circ$"%(360-d))
angle_ticks = map(degree_ticks, np.linspace(180, 360, 5))
grid_locator1 = FixedLocator([v for v, s in angle_ticks])
tick_formatter1 = DictFormatter(dict(angle_ticks))
tick_formatter2 = DictFormatter(dict(zip(np.linspace(1000, 6000, 6),
map(str, np.linspace(0, 5000, 6)))))
grid_locator2 = MaxNLocator(5)
gh = floating_axes.GridHelperCurveLinear(tr,
extremes=(2*np.pi, np.pi, 1000, 6000),
grid_locator1=grid_locator1,
grid_locator2=grid_locator2,
tick_formatter1=tick_formatter1,
tick_formatter2=tick_formatter2)
fig = plt.figure()
ax = floating_axes.FloatingSubplot(fig, 111, grid_helper=gh)
fig.add_subplot(ax)
azimuths = np.radians(np.linspace(180, 360, 90)) # added 180 degrees
zeniths = np.arange(1050, 6050, 50) # added 1000
r, theta = np.meshgrid(zeniths, azimuths)
values = 90.0+5.0*np.random.random((len(azimuths), len(zeniths)))
aux_ax = ax.get_aux_axes(tr)
aux_ax.patch = ax.patch
ax.patch.zorder = 0.9
aux_ax.contourf(theta, r, values) # use aux_ax instead of ax
plt.show()
请注意(为了获得原点附近的空间),您需要在半径方向上将所有数据点移动1000,并在θ方向上移动pi(以获得下半球)。 这会产生: