什麼是sympy 1.0的matplotlib的正確版本？

Question

我試圖在Pycharm中使用Sympy（1.0）的繪圖模塊，但遇到如下所示的錯誤。我想這是由matplotlib（2.0.2）和Sympy（1.0）之間的版本不兼容造成的。有人有線索嗎？在此先感謝〜

Traceback (most recent call last): 
    File "/home/leizh/PycharmProjects/Learn_python/Smoothness_Bilinear_Quadrilateral_Elmt.py", line 49, in <module> 
    plot_parametric(cos(u),sin(u),(u,-5,5)) 
    File "/home/leizh/.local/lib/python3.5/site-packages/sympy/plotting/plot.py", line 1415, in plot_parametric 
    plots.show() 
    File "/home/leizh/.local/lib/python3.5/site-packages/sympy/plotting/plot.py", line 184, in show 
    self._backend = self.backend(self) 
    File "/home/leizh/.local/lib/python3.5/site-packages/sympy/plotting/plot.py", line 1056, in __new__ 
    return MatplotlibBackend(parent) 
    File "/home/leizh/.local/lib/python3.5/site-packages/sympy/plotting/plot.py", line 868, in __init__ 
    self.cm = self.matplotlib.cm 
AttributeError: 'NoneType' object has no attribute 'cm' 

該代碼旨在計算雙線性四邊形元素的映射。

from sympy import * 
from sympy.plotting import * 

xi = Symbol("xi") 
eta = Symbol("eta") 

#Shape functions in reference element 
def Ni(xi,eta,i): 
    references_vertices = {1:[-1,-1],2:[1,-1],3:[1,1],4:[-1,1]} 
    xiv = references_vertices[i][0] 
    etav = references_vertices[i][1] 
    return Rational(1,4)*(1+xiv*xi)*(1+etav*eta) 

#Give a specific element in physical space with an angle >= 180 degree 
physical_vertices = {1:[-1,-1],2:[1,-1],3:[1,1],4:[0,0]} 

#Interpolation for (x,y) in terms of (xi,eta) 
def mapping(xi,eta,vertices): 
    x = 0 
    y = 0 
    for i in vertices: 
     xv = vertices[i][0] 
     yv = vertices[i][1] 
     x += Ni(xi,eta,i)*xv 
     y += Ni(xi,eta,i)*yv 
    return [x,y] 

#mapping (xi, eta) -> (x, y) 
xy = mapping(xi,eta,physical_vertices) 
print("x and y") 
print(factor(xy[0])) 
print(factor(xy[1])) 

#Jacobian 
jac = [] 
jac.append([xy[0].diff(xi),xy[0].diff(eta)]) 
jac.append([xy[1].diff(xi),xy[1].diff(eta)]) 

print("Jacobian Matrix") 
print(factor(jac)) 

#The determinant of Jacobian 
det_jac = jac[0][0]*jac[1][1]-jac[0][1]*jac[1][0] 
print(factor(det_jac)) 

#Plot 
plot3d_parametric_surface(xy[0], xy[1], det_jac,(xi,-1,1),(eta,-1,1)) 
det_jac.subs([(xi,1),(eta,-1)]) 

#test 
u = symbols('u') 
plot(u**2,(u,-1,1)) 
plot_parametric(cos(u),sin(u),(u,-5,5)) 

Answer 1

我已經能夠用matplotlib 2.0.2，sympy 1.0和python 3.4.6重現你的問題。然而，使用matplotlib 2.0.2，sympy 1.0和python 3.5.3工作得很好。請注意，我正在使用不同的計算機，但每次都使用全新的虛擬環境。所以這裏不應該有其他問題。我建議升級到python 3.5.x.

今後請提供「最小」的工作例子重現你的錯誤，例如：

import sympy as sym 

u = sym.symbols('u') 

sym.plotting.plot(sym.sin(u), (u,-5,5)) 

編輯：有2臺計算機之間的差別：一個使用qt4agg後端（DID不工作），其他使用tkagg（不工作）。所以在sympy和matplotlib中使用哪個後端似乎存在問題。