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我試圖評估一個函數(另一個的二階導數),但Sympy似乎有困難做到這一點......?Sympy無法區分與變量
from sympy import *
from sympy import Symbol
# Symbols
theta = Symbol('theta')
phi = Symbol('phi')
phi0 = Symbol('phi0')
H0 = Symbol('H0')
# Constants
a = 0.05
t = 100*1e-9
b = 0.05**2/(8*pi*1e-7)
c = 0.001/(4*pi*1e-7)
phi0 = 60*pi/180
H0 = -0.03/(4*pi*1e-7)
def m(theta,phi):
return Matrix([[sin(theta)*cos(phi), sin(theta)*cos(phi), cos(phi)]])
def h(phi0):
return Matrix([[cos(phi0), sin(phi0), 0]])
def k(theta,phi,phi0):
return m(theta,phi).dot(h(phi0))
def F(theta,phi,phi0,H0):
return -(t*a*H0)*k(theta,phi,phi0)+b*t*(cos(theta)**2)+c*t*(sin(2*theta)**2)+t*sin(theta)**4*sin(2*phi)**2
def F_theta(theta,phi,phi0,H0):
return simplify(diff(F(theta,phi,phi0,H0),theta))
def F_thetatheta(theta,phi,phi0,H0):
return simplify(diff(F_theta(theta,phi,phi0,H0),theta))
print F_thetatheta(theta,phi,phi0,H0), F_thetatheta(pi/2,phi,phi0,H0)
如下面看到的那樣,一般功能進行評價,但是當我嘗試更換由PI/2或另一個值THETA,這是行不通的。
(4.0e-7*pi*sin(theta)**4*cos(2*phi)**2 - 4.0e-7*pi*sin(theta)**4 + 0.00125*sin(theta)**2 - 0.0001875*sqrt(3)*sin(theta)*cos(phi) - 0.0001875*sin(theta)*cos(phi) + 1.2e-6*pi*cos(2*phi)**2*cos(theta)**4 - 1.2e-6*pi*cos(2*phi)**2*cos(theta)**2 - 1.2e-6*pi*cos(theta)**4 + 1.2e-6*pi*cos(theta)**2 + 0.004*cos(2*theta)**2 - 0.002625)/pi
Traceback (most recent call last):
File "Test.py", line 46, in <module>
print F_thetatheta(theta,phi,phi0,H0), F_thetatheta(pi/2,phi,phi0,H0)
File "Test.py", line 29, in F_thetatheta
return simplify(diff(F_theta(theta,phi,phi0,H0),theta))
File "Test.py", line 27, in F_theta
return simplify(diff(F(theta,phi,phi0,H0),theta))
File "/usr/lib64/python2.7/site-packages/sympy/core/function.py", line 1418, in diff
return Derivative(f, *symbols, **kwargs)
File "/usr/lib64/python2.7/site-packages/sympy/core/function.py", line 852, in __new__
Can\'t differentiate wrt the variable: %s, %s''' % (v, count)))
ValueError:
Can't differentiate wrt the variable: pi/2, 1
是的,抱歉,我現在明白了錯誤。我應該閱讀關於sympy的教程,我一定會需要它!但是你的建議實際上並不奏效。這相反:'F_thetatheta(theta,phi,phi0,H0).subs(theta,pi/2)' – aymenbh
@ user1816760 Ops,你說的沒錯,我沒有意識到phi還在最後的表達,在這種情況下,你需要像你一樣替換theta。如果你同時評估theta和phi,這將起作用:'F_thetatheta(theta,phi,phi0,H0).evalf(subs = {theta:pi/2,phi:0})'。無論如何,你最終需要'evalf'來獲得一個數值而不是確切的數值(我會修復我的答案)。 – jorgeca