如何在Matplotlib中的极轴上使用对数刻度

Question

我正在尝试创建一个在径向轴上具有对数刻度的极坐标图，但是我一直遇到错误。下面是一些示例代码和错误。在笛卡尔坐标系中似乎工作正常，有人知道发生了什么吗？

import matplotlib.pyplot as plt
import numpy as np
import matplotlib.cm as cm

bazbins = np.linspace(0, 2*np.pi, 360)
fbins = np.logspace(np.log10(0.05), np.log10(0.5), 101)
theta, r = np.meshgrid(bazbins, fbins) 

# Set up plot window
fig, ax = plt.subplots(figsize=(12,9))#, subplot_kw=dict(projection='polar'))

# polar
ax.set_theta_zero_location('N')
ax.set_theta_direction(-1)
ax.set_rscale('log')

# carthesia
#ax.set_yscale('log')

# Plot data
#ax.pcolormesh(theta, r, r)
plt.gca().invert_yaxis()
ax.contourf(theta, r, r)
ax.set_ylim((0.0, 0.5))
plt.show()

Tkinter回调中的异常：

Traceback (most recent call last):
  File "/usr/lib/python2.7/lib-tk/Tkinter.py", line 1413, in __call__
    return self.func(*args)
  File "/usr/lib/pymodules/python2.7/matplotlib/backends/backend_tkagg.py", line 236, in resize
    self.show()
  File "/usr/lib/pymodules/python2.7/matplotlib/backends/backend_tkagg.py", line 239, in draw
    FigureCanvasAgg.draw(self)
  File "/usr/lib/pymodules/python2.7/matplotlib/backends/backend_agg.py", line 421, in draw
    self.figure.draw(self.renderer)
  File "/usr/lib/pymodules/python2.7/matplotlib/artist.py", line 55, in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "/usr/lib/pymodules/python2.7/matplotlib/figure.py", line 898, in draw
    func(*args)
  File "/usr/lib/pymodules/python2.7/matplotlib/artist.py", line 55, in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "/usr/lib/pymodules/python2.7/matplotlib/axes.py", line 1997, in draw
    a.draw(renderer)
  File "/usr/lib/pymodules/python2.7/matplotlib/artist.py", line 55, in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "/usr/lib/pymodules/python2.7/matplotlib/axis.py", line 1045, in draw
    tick.draw(renderer)
  File "/usr/lib/pymodules/python2.7/matplotlib/artist.py", line 55, in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "/usr/lib/pymodules/python2.7/matplotlib/axis.py", line 239, in draw
    self.label1.draw(renderer)
  File "/usr/lib/pymodules/python2.7/matplotlib/artist.py", line 55, in draw_wrapper
    draw(artist, renderer, *args, **kwargs)
  File "/usr/lib/pymodules/python2.7/matplotlib/text.py", line 591, in draw
    ismath=ismath)
  File "/usr/lib/pymodules/python2.7/matplotlib/backends/backend_agg.py", line 156, in draw_text
    return self.draw_mathtext(gc, x, y, s, prop, angle)
  File "/usr/lib/pymodules/python2.7/matplotlib/backends/backend_agg.py", line 145, in draw_mathtext
    x = int(x) + ox
ValueError: cannot convert float NaN to integer

Answer 1

当前的matplotlib和对数极坐标图存在更多问题。

例如，尝试在matplotlib示例中为极坐标图的半径添加一个较小的值，然后使用set_rlim(0)和对其set_rscale('log')进行绘制（如此处的注释中所建议）。所有低于0.1的值都会得到一些特殊处理。这会影响r轴上的刻度（注意完全错位的10e-2和10e-3）以及绘制的数据：

该行为似乎没有记录。我最终手动完成了对数转换（上述系列中的第三幅图）。对于遇到此线程的其他人，这是我的代码：

import numpy as np
import matplotlib.pyplot as plt

def scatter_polar_mpl(ax, theta, r):
    ax.scatter(theta, r)
    ax.set_rlim(0)
    ax.set_title('polar matplotlib')

def scatter_logpolar_mpl(ax, theta, r):
    ax.scatter(theta, r)
    ax.set_rlim(0)
    ax.set_rscale('log')
    ax.set_title('log-polar matplotlib')

def scatter_logpolar(ax, theta, r_, bullseye=0.3, **kwargs):
    min10 = np.log10(np.min(r_))
    max10 = np.log10(np.max(r_))
    r = np.log10(r_) - min10 + bullseye
    ax.scatter(theta, r, **kwargs)
    l = np.arange(np.floor(min10), max10)
    ax.set_rticks(l - min10 + bullseye) 
    ax.set_yticklabels(["1e%d" % x for x in l])
    ax.set_rlim(0, max10 - min10 + bullseye)
    ax.set_title('log-polar manual')
    return ax

r = np.arange(0, 3.0, 0.01) + 0.001

theta = 2 * np.pi * r

ax = plt.subplots(1, 3, subplot_kw=dict(polar=True))[1].flatten()
scatter_polar_mpl(ax[0], theta, r)
scatter_logpolar_mpl(ax[1], theta, r)
scatter_logpolar(ax[2], theta, r)

plt.show()

Answer 2

这似乎是 matplotlib 中的一个错误。set_rlim您不必在(or )set_ylim之前调用set_rscale(或) 方法set_yscale。另外，必须以 0 或 0.0 作为下限来调用set_rlimor 。set_ylim下限的其他值也会导致崩溃。其他后端也会出现此问题（我已经确认了 gtkagg 和 pdf 后端的问题）。

我针对此问题提交了一份错误报告，可以在此处找到。如果这个问题影响了您，请转到错误报告页面并发表评论，让 matplotlib 开发人员知道这个问题对用户很重要。