如何使用Matplotlib处理渐近线/不连续性

Question

当绘制具有不连续性/渐近线/奇点/等的图形时，是否有任何自动方法可以防止Matplotlib在“中断”中“加入点”？ （请参阅下面的代码/图片） 我读到Sage有一个看起来很好的[detect_poles]工具，但我真的希望它能与Matplotlib一起工作。

import matplotlib.pyplot as plt 
import numpy as np
from sympy import sympify, lambdify
from sympy.abc import x

fig = plt.figure(1) 
ax = fig.add_subplot(111) 

# set up axis 
ax.spines['left'].set_position('zero') 
ax.spines['right'].set_color('none') 
ax.spines['bottom'].set_position('zero') 
ax.spines['top'].set_color('none') 
ax.xaxis.set_ticks_position('bottom') 
ax.yaxis.set_ticks_position('left') 

# setup x and y ranges and precision
xx = np.arange(-0.5,5.5,0.01) 

# draw my curve 
myfunction=sympify(1/(x-2))
mylambdifiedfunction=lambdify(x,myfunction,'numpy')
ax.plot(xx, mylambdifiedfunction(xx),zorder=100,linewidth=3,color='red') 

#set bounds 
ax.set_xbound(-1,6)
ax.set_ybound(-4,4) 

plt.show()

Answer 1

通过使用masked arrays，您可以避免绘制曲线的选定区域。

要删除x = 2处的奇点：

import matplotlib.numerix.ma as M    # for older versions, prior to .98
#import numpy.ma as M                # for newer versions of matplotlib
from pylab import *

figure()

xx = np.arange(-0.5,5.5,0.01) 
vals = 1/(xx-2)        
vals = M.array(vals)
mvals = M.masked_where(xx==2, vals)

subplot(121)
plot(xx, mvals, linewidth=3, color='red') 
xlim(-1,6)
ylim(-5,5) 

这条简单的曲线可能会更清楚地排除哪些点：

xx = np.arange(0,6,.2) 
vals = M.array(xx)
mvals = M.masked_where(vals%2==0, vals)
subplot(122)
plot(xx, mvals, color='b', linewidth=3)
plot(xx, vals, 'rx')
show()

Answer 2

这可能不是您正在寻找的优雅解决方案，但如果只想在大多数情况下获得结果，您可以将绘制数据的大小值分别“剪切”到+∞和-∞ 。 Matplotlib没有绘制这些。当然，你必须小心，不要让你的分辨率太低或剪裁阈值太高。

utol = 100.
ltol = -100.
yy = 1/(xx-2)
yy[yy>utol] = np.inf
yy[yy<ltol] = -np.inf

ax.plot(xx, yy, zorder=100, linewidth=3, color='red') 

Answer 3

不，我认为没有内置的方法告诉matplotlib忽略这些 点。毕竟，它只是连接点而对功能一无所知 或者点之间会发生什么。

但是，您可以使用sympy找到极点，然后将功能的连续部分拼接在一起。在这里，一些公认的丑陋代码确实如此：

from pylab import *
from sympy import solve
from sympy.abc import x
from sympy.functions.elementary.complexes import im

xmin = -0.5
xmax = 5.5
xstep = 0.01

# solve for 1/f(x)=0 -- we will have poles there
discontinuities = sort(solve(1/(1/(x-2)),x))

# pieces from xmin to last discontinuity
last_b = xmin
for b in discontinuities:
    # check that this discontinuity is inside our range, also make sure it's real
    if b<last_b or b>xmax or im(b):
      continue
    xi = np.arange(last_b, b, xstep)
    plot(xi, 1./(xi-2),'r-')
    last_b = b

# from last discontinuity to xmax
xi = np.arange(last_b, xmax, xstep)
plot(xi, 1./(xi-2),'r-')

xlim(xmin, xmax)
ylim(-4,4)
show()

example http://i43.tinypic.com/30mvbzb.jpg