我正在尝试用numpy做傅里叶系列。我正在尝试编写函数,因为它们已经定义为here
但我在定义a0时遇到了麻烦。
# "M1(t)" function definition.
def M1(t, *args):
tau, M0 = args
omega = 2 * np.pi / tau
return (2 * M0 + M0 * np.sin(omega * t - 2 / 3 * np.pi) +
M0 * np.sin(omega * t - 4/ 3 * np.pi))
# "M2(t)" function definition.
def M2(t, *args):
tau, M0 = args
omega = 2 * np.pi / tau
return (3 * M0 + M0 * np.sin(omega * t) + M0 * np.sin(omega * t - 2 / 3 * np.pi) +
M0 * np.sin(omega * t - 4/ 3 * np.pi))
def a0(tau, *args):
# limits of integrals; a = lower of 1st integral;
# b = higher of 1st and lower od 2nd integral;
# c = higher of 2nd integral
a, b, c = 0, tau / 2, tau
i1, err1 = quad(M1, a, b, *args)
i2, err2 = quad(M2, b, c, *args)
return 2 / tau * (i1 + i2)
当我运行此代码时,我收到以下错误:
TypeError: integer argument expected, got float
根据要求,追溯错误:
Traceback (most recent call last):
File "C:/Users/Alex/Documents/Faks/Magisterij/1. letnik/VD/2. seminar/periodicno_vzbujanje.py", line 86, in <module>
a0 = a0(parameters[0], *parameters)
File "C:/Users/Alex/Documents/Faks/Magisterij/1. letnik/VD/2. seminar/periodicno_vzbujanje.py", line 41, in a0
i1, err1 = quad(M1, a, b, *args)
File "C:\Users\Alex\Anaconda3\lib\site-packages\scipy\integrate\quadpack.py", line 315, in quad
points)
File "C:\Users\Alex\Anaconda3\lib\site-packages\scipy\integrate\quadpack.py", line 380, in _quad
return _quadpack._qagse(func,a,b,args,full_output,epsabs,epsrel,limit)
TypeError: integer argument expected, got float
额外的问题:如何将组合功能传递给四元组?例如:M1(t, *args) * np.cos(omega * t)
?我是否必须将其定义为新功能然后将其传递或更快?因为我认为输入4个额外功能是多余的。
更新
我意识到我一直都错误地传递了错误的论据。
我将i2, err2 = quad(M2, b, c, *args)
更改为i2, err2 = quad(M2, b, c, args)
。但是现在我收到以下错误:
ValueError: not enough values to unpack (expected 2, got 1)
。
答案 0 :(得分:1)
使用Map
M1
In [202]: M1(0,1,1)
Out[202]: 1.9999999999999996
In [203]: integrate.quad(M1,0,1,(1,1))
Out[203]: (2.0, 2.220446049250313e-14)
In [204]: M1(0,*(1,1))
Out[204]: 1.9999999999999996
元组看起来应该与args
语法传递给M1
。