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/任何图形的图形时,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有一个看上去很好的[detect_poles]工具,但我真的希望它与Matplotlib一起使用.

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

通过使用蒙板阵列,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

这可能不是你要找的优雅的解决方案,但如果只是想在大多数情况下的结果,你可以"夹"您绘制的数据的大小值+?和-?分别.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matplotlib可以忽略这些观点.毕竟,它只是连接点,对功能或点之间发生的事情一无所知.

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.这里有一些令人难以置信的丑陋代码,它确实如此:

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()

例如http://i43.tinypic.com/30mvbzb.jpg