I want to minimize a function in order to obtain some parameters' value of : a,e,I,Omega,om,tp.
I use this "module" : docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.minimize.html.
My function has 13 parameters:
I imported : from scipy.optimize import minimize. Then I try to minimize it.And the error occurs:
Would someone help to solve this problem?
PS: I started python one week ago that may explain this syntax of the program, however I'm willing to improve myself.
from numpy import *
import numpy as np
import scipy as sp
from scipy.optimize import minimize
import matplotlib.pyplot as plt
from pylab import *
from os import chdir
chdir("/Users/benjaminjaillant/Desktop")
def Chi_VLT(a,e,I,tp,Omega,om,Mbh,R0,Vr_bh,alpha_bh,V_alp_bh,delta_bh,V_del_bh):
return sum(((Vr_etoile(t_vr_VLT*365*24*3600,a,e,I,tp,om,Mbh,Vr_bh)/1000)-vr_VLT)**2/vr_error_VLT**2) + sum(((alpha_etoile_IR(t_orbit_VLT*365*24*3600,a,e,I,tp,Omega,om,Mbh,alpha_bh,V_alp_bh,R0)*206264806.246)-Ra_VLT)**2/Ra_error_VLT**2) + sum(((delta_etoile_IR(t_orbit_VLT*365*24*3600,a,e,I,tp,Omega,om,Mbh,delta_bh,V_del_bh,R0)*206264806.246)-Dec_VLT)**2/Dec_error_VLT**2)
x0 = [1.5e14,0.8,2.5,63.10e9,4,1,8.5e36,2.5e20,2000,1.3e-8,-10e-18,2e-9,1.5e-17]
res = minimize(Chi_VLT, x0 , method='nelder-mead',options={'xtol': 1e-4,'maxiter':50 ,'disp': True})
print res.message
print res.x
I guess you are messing with whole thing here.
your function scipy.optimize.minimize takes two required positional argument,
fun and x0.
you need ndarray as x0
In your case your fun Chi_VLT requires 13 arguments, you need to pass that using args=(tuple, containing, 13, items)
Then only you will be able to minimize your fun.
The minimize routine expects an ndarray, say guess, as initial function arguments and accepts an additional tuple of arguments x0 which might constitute the coefficients in your cost function. If you rewrite CHI_VLT to take as first arg an ndarray and then the remaining arguments accordingly it should work.
res = minimize(CHI_VLT, guess, args=x0,...)
I need help making a program that calculates the Gaussian function f(x)=1/(sqrt(2*pi)s)*exp[-.5*((x-m)/s)**2] when m=0, s=2, and x=1.
Would it be just:
def Gaussian(m,s,x):
return 1/(sqrt(2*pi)s)*exp[-.5*((x-m)/s)**2]
print Gaussian(0,2,1)
I think you are missing from math import sqrt, from math import exp and from math import pi unless you didn't show it in your code.
I'm using python's numdifftools library to perform derivatives. However, a few tests prove the library is highly inaccurate:
import numpy as np
from numdifftools import Derivative
# Result should be 1/2 or 0.5
Derivative(np.log, 1)(2.0)
>>> array(0.5493061443340549)
Is there a way to fix this inaccuracy?
Using numdifftools 0.9.16 and numpy 1.9.3 the following code gives an exact result:
import numpy as np
from numdifftools import Derivative
# Result should be 1/2 or 0.5
Derivative(np.log)(2.0)
Output:
array(0.5000000000000238)
Issue spotted.
Derivative(np.log, 1)(2.0)
gives the wrong answer. The n should be explicitly stated:
Derivative(np.log, n=1)(2.0)
>>> array(0.5000000000000234)
I am new to Python, and I keep getting the following error
..., line 27, in <module>
eq=(p**2+2)/p/sqrt(p**2+4)
AttributeError: sqrt
I tried to add math.sqrt or numpy.sqrt but neither of these work. Does anyone know where I'm going wrong?
My code is:
from numpy import *
from matplotlib import *
from pylab import *
from sympy import Symbol
from sympy.solvers import solve
p=Symbol('p')
eq=(p**2+2)/p/sqrt(p**2+4)
solve(eq,1.34,set=True)
sqrt is defined in the math module, import it this way. This should remove the errors!
from math import sqrt
You are using a sympy symbol: either you wanted to do numerical sqrt (in which case use numpy.sqrt on an actual number) or you wanted symbolic sqrt (in which case use sympy.sqrt). Each of the imports replaces the definition of sqrt in the current namespace, from math, sympy or numpy. It's best to be explicit and not use "import *".
I suspect from the line which follows, you want sympy.sqrt here.
In a project using SciPy and NumPy, should I use scipy.pi, numpy.pi, or math.pi?
>>> import math
>>> import numpy as np
>>> import scipy
>>> math.pi == np.pi == scipy.pi
True
So it doesn't matter, they are all the same value.
The only reason all three modules provide a pi value is so if you are using just one of the three modules, you can conveniently have access to pi without having to import another module. They're not providing different values for pi.
One thing to note is that not all libraries will use the same meaning for pi, of course, so it never hurts to know what you're using. For example, the symbolic math library Sympy's representation of pi is not the same as math and numpy:
import math
import numpy
import scipy
import sympy
print(math.pi == numpy.pi)
> True
print(math.pi == scipy.pi)
> True
print(math.pi == sympy.pi)
> False
If we look its source code, scipy.pi is precisely math.pi; in fact, it's defined as
import math as _math
pi = _math.pi
In their source codes, math.pi is defined to be equal to 3.14159265358979323846 and numpy.pi is defined to be equal to 3.141592653589793238462643383279502884; both are well above the 15 digit accuracy of a float in Python, so it doesn't matter which one you use.
That said, if you're not already using numpy or scipy, importing them just for np.pi or scipy.pi would add unnecessary dependency while math is a Python standard library, so there's not dependency issues. For example, for pi in tensorflow code in python, one could use tf.constant(math.pi).