scipy curve_fit for multiple independent variables

[scidam]

I've seen on this optimization problem several times, but couldn't understand what you really want.
You have F = func1, and you try to fit it as follows:

F(xdata, parameters) = ydata

(xdata, ydata) are known; parameters should be determined (e.g. in least squares sense).

If the problem could be formulated as I just did, you could try least_squares utility function
to solve it.

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xdata = [214.737191559, -5.64912101538e-36, 36.1372453686, 189.459700978, 233.562136902, 201.230228832, -5.59364882619e-36, -36.3232002416, -188.192199081, -212.837139143, -232.342545403, -200.699429716]
ydata = [-5.88273617837e-37, -211.536123799, -186.67108047, -35.9497006815, 200.282998159, 232.085860035, 213.44274878, 187.945919272, 35.7227474297, -6.00785257974e-37, -199.746844708, -230.856058666]
 
xdata = np.array(xdata)
ydata = np.array(ydata)
X = (xdata, ydata)
def func1(Z):
    x,y = X
    a, b, c = Z  
#     x = np.array(X[0])
#     y = np.array(X[1])
     
    n  = 8
#     % A  = ydata
#     % B  = -xdata
#     % C  = xdata. - ydata
#     % H  = zdata
     
    g = np.subtract(x,y)
    I_0 = np.subtract(x,y)   # x-y = C
    I_1 = np.multiply(I_0,c) # c(x-y) = cC
    I_2 = np.multiply(b,-x)   #b(-x) = bB
    I_3 = np.multiply(a,y)  # aA
     
     
    I3_0 = np.subtract(I_1,I_2) # cC-bB
    I3_1 = np.subtract(I_3,I_1) # aA-cC
    I3_2 = np.subtract(I_2,I_3) # bB-aA
     
    I3_00 = np.multiply(I3_0,I3_1) # (cC-bB)(aA-cC)
    I3_01 = np.multiply(I3_00,I3_2) # (cC-bB)(aA-cC)(bB-aA)
     
    I3 = np.divide(I3_01,54) # (cC-bB)(aA-cC)(bB-aA)/54
     
    I2_0 = np.power((I3_1),2)  # (aA-cC)^2
    I2_1 = np.power((I3_0),2)  # (cC-bB)^2
    I2_2 = np.power((I3_2),2)  # (bB-aA)^2
     
    I2_00 = np.add(I2_0,I2_1)  # (aA-cC)^2 + (cC-bB)^2
    I2_01 = np.add(I2_00,I2_2) # (aA-cC)^2 + (cC-bB)^2 + (bB-aA)^2
     
    I2 = np.divide(I2_01,54)  # ((aA-cC)^2 + (cC-bB)^2 + (bB-aA)^2)/54
     
    th_0 = np.divide(I3,(np.power(I2,(3/2))))  # I3/(I2^(3/2))
     
#     print(th_0)
     
    th = np.arccos(np.clip((th_0),-1,1))  # arccos(I3/(I2^(3/2)))
     
#     print(th)
     
    ans_0 = np.divide(np.add((2*th),(np.pi)),6)   # (2*th + pi)/6
    ans_1 = np.divide(np.add((2*th),(3*np.pi)),6) # (2*th + 3*pi)/6
    ans_2 = np.divide(np.add((2*th),(5*np.pi)),6) # (2*th + 5*pi)/6
     
    ans_00 = np.multiply(np.cos(ans_0),2)  # 2*cos((2*th + pi)/6)
    ans_11 = np.multiply(np.cos(ans_1),2)  # 2*cos((2*th + 3*pi)/6)
    ans_22 = np.multiply(np.cos(ans_2),2)  # 2*cos((2*th + 5*pi)/6)
     
    ans_000 = np.power(np.absolute(ans_00),n)  # (abs(2*cos((2*th + pi)/6)))^n
    ans_111 = np.power(np.absolute(ans_11),n)  # (abs(2*cos((2*th + 3*pi)/6)))^n
    ans_222 = np.power(np.absolute(ans_22),n)  # (abs(2*cos((2*th + 5*pi)/6)))^n
     
    ans_0000 = np.add((np.power(np.absolute(ans_00),n)),(np.power(np.absolute(ans_11),n))) # (abs(2*cos((2*th + pi)/6)))^n + (abs(2*cos((2*th + 3*pi)/6)))^n 
    ans_1111 = np.add((ans_0000),(np.power(np.absolute(ans_22),n)))  # (abs(2*cos((2*th + pi)/6)))^n + (abs(2*cos((2*th + 3*pi)/6)))^n + (abs(2*cos((2*th + 5*pi)/6)))^n
     
    sna_0 = np.power(np.multiply(3,I2),(n/2))  # (3*I2)^(n/2) !!
    sna_1 = 2*(np.power(190,n)) # 2*(sigma^n) !!
     
    sna_00 = np.multiply(sna_0,ans_1111)
    sna_11 = np.subtract(sna_00,sna_1)
     
    return sna_11 - y
 
# initial guesses for a,b,c:
a, b, c = 1, 2, 3
p0      = np.array([a, b, c])
popt = least_squares(func1, p0)
print(popt)

The only thing I couldn't understand is why are you using np.substract, np.multiply, etc. instead of just -, or * symbols?

In any case, the objective function could be rewritten in more friendly and understandable form.