Question

我有点问题。我有一个程序可以为潜在的东西绘制波函数，当我在绘图中使用选项（使用pylab）时，它看起来很好' - '例如：http://img41.imageshack.us/img41/8798/59138635.png

如果我使用'o'，我会得到： http://img16.imageshack.us/img16/3741/22378006.png

你看它看起来很难看：\

是否有一种简单的方法可以使圆圈更加间隔，或者这取决于代码的细节？

代码是：

from math import *
from scipy.special import *
from pylab import *
from scipy.linalg import *

firebrick=(178./255.,34./255.,34./255.)
indianred=(176./255.,23./255.,31./255.)
steelblue=(70./255.,130./255.,180./255.)
slategray1=(198./255.,226./255.,255./255.)
slategray4=(108./255.,123./255.,139./255.)
lavender=(230./255.,230./255.,230./255.)
cobalt=(61./255.,89./255.,171./255.)
midnightblue=(25./255.,25./255.,112./255.)
forestgreen=(34./255.,139./255.,34./255.)

#grid
Nmesh=512
L=4.0
dx=L/Nmesh
Xmax=L
x=arange(-L,L+0.0001,dx)
Npts=len(x)
numwav=2   #number of wave function that is being drawn

V=zeros([Npts],float)
for i in range(Npts):
    V[i]=x[i]**4

a=zeros([2,Npts-2],float)
wave=zeros([Npts],float)

wave1=zeros([Npts],float)
encor=3.0/4*(3.0/4)**(1.0/3)

#numerical solution
for i in range(1,Npts-1,1):
    a[0,i-1]= 1.0/dx**2+V[i]     #diagonal elements
    a[1,i-1]=-1.0/dx**2/2        #the elements below the diagonal
a[1,Npts-3]=-99.0                #element is not used
eig,vec=eig_banded(a,lower=1)    #routine that diagonalizes the tridiagonal matrix

for i in range(1,Npts-1,1):
    wave[i]=vec[i-1,numwav]
wave[0]=0.0             #wave function has the value zero on the first point on the grid
wave[Npts-1]=0.0        #wave function has the value zero on the last point on the grid

wave=150*wave+eig[numwav]

#potential graph
line=plt.plot(x,V)
plt.setp(line,color='firebrick',linewidth=2)

#plot of the selected level and wave function
plt.axhline(y=eig[numwav],linewidth=2,color='steelblue')

#plot of the points of the wave function
plt.plot(x,wave,"b-",linewidth=2,color='forestgreen')

plt.xlabel('x',size=16)
plt.ylabel('V(x)',size=16)
plt.axis([-4.0,4.0,-5.0,16.0]) #x and y axes range
plt.grid(True)
plt.show()

Answer 1

在进一步调查之后，我有一个更好（但不同）的答案。 Matplotlib提供markevery关键字以允许放置标记的步幅。所以我建议，如果你想要20个左右的可见度点，在绿线之上：

stride = max( int(len(x) / 20), 1)
plt.plot(x,wave,"-o",color='forestgreen', markevery=stride)

如果您只想要标记，我之前的回答可以正常工作，但如果您想要两条线和标记，这样做会更好。

Answer 2

它看起来很丑的原因是你的网格间距太小，无法用标记绘图。如何在行中添加标记只是将它们添加到每10个（或其他）点：

plt.plot(x[::10],wave[::10],"o",color='forestgreen')

matplotlib：使用标记时的丑陋情节

2 个答案: