缺少 1 个必需的位置 arguments：

Question

I am facing some issues with my python code for a few days now. I am trying to optimize two unknown parameters by solving a set of simultaneous nonlinear coupled ODEs. I have coupled the solver_ivp with the minimization routine. I would be extremely grateful if someone could help me with this. The code is as follows. Thank you.

import numpy as np
import scipy as sp
from scipy.integrate import solve_ivp
import scipy.optimize as opt
from scipy.optimize import curve_fit
from scipy.optimize import least_squares 
from scipy.optimize import minimize
import matplotlib.pylab as plt
import pandas as pd

df = pd.read_csv('dep_data_CMB3.csv')
#print(df.columns)

xdat1 = np.asarray(df['t3_1'], dtype = 'float')
ydat1 = np.asarray(df['CMB3_1'], dtype = 'float')
#ydat2 = np.asarray(df['y2'], dtype = 'float')
#ydat3 = np.asarray(df['y3'], dtype = 'float')

xdat1 = np.asarray(df['t3_1'], dtype = 'float')
ydat1 = np.asarray(df['CMB3_1'], dtype = 'float')
#ydat2 = np.asarray(df['y2'], dtype = 'float')
#ydat3 = np.asarray(df['y3'], dtype = 'float')

plt.plot(xdat1, ydat1, 'ob', markersize = 2)
#plt.plot(xdat2, ydat2, 'og')
#plt.plot(xdat3, ydat3, 'or')
plt.xscale('log')
plt.show()

def func(tspan, ko, ki, phi, H):
dphidt = ko * (1 - phi) * H
dHdt = -ki * phi * H
    return [dphidt, dHdt]

H0 = [0.04818]
#H5 = [0.08083]
#H7 = [0.11242]
phi0 = 0.0
#phi5 = [0.0]
#phi7 = [0.0]
k0 = 0.0125
k1 = 0.00051
k = [k0, k1]
tspan = [0, max(xdat1)]

def newfunc(ko, ki, x, y, H):
tspan = [0, max(xdat1)]
y0 = [0, 0.04818]
result = solve_ivp(func, tspan, y0, method = 'RK45', args = (k0, k1, H0), dense_output = True)
resultnew = result.sol(tspan).T
ephi = y - resultnew[:,0]
sumephi = np.sum((ephi**2))
#eH = H0 - resultnew[:,1]
#sumeH = np.sum((eH**2))
    return sumephi

wguess = (k0, k1)
arguments = (xdat1, ydat1, H0)
res = minimize(newfunc, wguess, args = arguments, method = 'bfgs')

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call 
last)
<ipython-input-11-95e9b490310a> in <module>
      1 wguess = (k0, k1)
      2 arguments = (xdat1, ydat1, H0)
----> 3 res = minimize(newfunc, wguess, args = arguments, method = 
'bfgs')

~\Anaconda3\lib\site-packages\scipy\optimize\_minimize.py in 
minimize(fun, x0, args, method, jac, hess, hessp, bounds, 
constraints, tol, callback, options)
612         return _minimize_cg(fun, x0, args, jac, callback, 
**options)
613     elif meth == 'bfgs':
--> 614         return _minimize_bfgs(fun, x0, args, jac, callback, 
**options)
615     elif meth == 'newton-cg':
616         return _minimize_newtoncg(fun, x0, args, jac, hess, 
hessp, callback,

~\Anaconda3\lib\site-packages\scipy\optimize\optimize.py in 
_minimize_bfgs(fun, x0, args, jac, callback, gtol, norm, eps, 
maxiter, disp, return_all, finite_diff_rel_step, **unknown_options)
1133         maxiter = len(x0) * 200
1134 
-> 1135     sf = _prepare_scalar_function(fun, x0, jac, args=args, 
epsilon=eps,
1136                                   
finite_diff_rel_step=finite_diff_rel_step)
1137 

~\Anaconda3\lib\site-packages\scipy\optimize\optimize.py in 
_prepare_scalar_function(fun, x0, jac, args, bounds, epsilon, 
finite_diff_rel_step, hess)
259     # ScalarFunction caches. Reuse of fun(x) during grad
260     # calculation reduces overall function evaluations.
--> 261     sf = ScalarFunction(fun, x0, args, grad, hess,
262                         finite_diff_rel_step, bounds, 
epsilon=epsilon)
263 

~\Anaconda3\lib\site- 
packages\scipy\optimize\_differentiable_functions.py in 
__init__(self, fun, x0, args, grad, hess, finite_diff_rel_step, 
finite_diff_bounds, epsilon)
134 
135         self._update_fun_impl = update_fun
--> 136         self._update_fun()
137 
138         # Gradient evaluation

~\Anaconda3\lib\site- 
packages\scipy\optimize\_differentiable_functions.py in 
_update_fun(self)
224     def _update_fun(self):
225         if not self.f_updated:
--> 226             self._update_fun_impl()
227             self.f_updated = True
228 

~\Anaconda3\lib\site- 
packages\scipy\optimize\_differentiable_functions.py in update_fun()
131 
132         def update_fun():
--> 133             self.f = fun_wrapped(self.x)
134 
135         self._update_fun_impl = update_fun

~\Anaconda3\lib\site- 
packages\scipy\optimize\_differentiable_functions.py in 
fun_wrapped(x)
128         def fun_wrapped(x):
129             self.nfev += 1
--> 130             return fun(x, *args)
131 
132         def update_fun():

TypeError: newfunc() missing 1 required positional argument: 'H'

Answer 1

There is a number of questions with your code (and how you posted it):

• there are several unused imports;
• there is a bunch of commented out code that apparently has no bearing on the problem, but there's also some code that doesn't do anything (like the assignments to xdat1 and ydat1 , which are repeated a few lines down;
• func has a parameter that's not used, tspan ;
• newfunc has parameter ki , k0 and H , but doesn't use them, instead it uses the global k0 , k1 and H0 ; it also has a completely unused x ;

All of this indicates a very careless approach to coding which, even if it results in code that runs, rarely results in code that runs correctly. Try taking one step at a time. Only add code you know will work, or if you just try something and you think it works, try to find out why it works, so you don't inadvertently break it later.

Also, with regard to the question:

• indentation is completely gone, so that has to be guessed at;
• your code reads a.csv, but you didn't provide a sample of the data (like the header and a few lines of data, so people can try your code)

This is your code with the superfluous lines removed and some arbitrary test data, which reproduces the error (some of the errors left in, since it's unclear what the impact is without further analysis):

import numpy as np
from scipy.integrate import solve_ivp
from scipy.optimize import minimize
import matplotlib.pylab as plt
import pandas as pd

df = pd.DataFrame([{'t3_1': 1., 'CMB3_1': 2.}, {'t3_1': 3., 'CMB3_1': 4.}])

xdat1 = np.asarray(df['t3_1'], dtype='float')
ydat1 = np.asarray(df['CMB3_1'], dtype='float')

plt.plot(xdat1, ydat1, 'ob', markersize=2)
plt.xscale('log')
plt.show()


def func(tspan, ko, ki, phi, H):
    dphidt = ko * (1 - phi) * H
    dHdt = -ki * phi * H
    return [dphidt, dHdt]


H0 = [0.04818]
k0 = 0.0125
k1 = 0.00051
k = [k0, k1]
tspan = [0, max(xdat1)]


def newfunc(ko, ki, x, y, H):
    tspan = [0, max(xdat1)]
    y0 = [0, 0.04818]
    result = solve_ivp(func, tspan, y0, method='RK45', args=(k0, k1, H0), dense_output=True)
    resultnew = result.sol(tspan).T
    ephi = y - resultnew[:, 0]
    sumephi = np.sum((ephi ** 2))
    return sumephi


wguess = (k0, k1)
arguments = (xdat1, ydat1, H0)
res = minimize(newfunc, np.array(wguess), args=arguments, method='bfgs')

One problem is clear: newfunc is supposed to be minimized, but you provide the tuple (k0, k1) as a guess, while minimize expects an ndarray - a cast fixes that.

But then the actual error you see: TypeError: newfunc() missing 1 required positional argument: 'H' . This also makes sense, minimize will call newfunc with an array and arguments ( fun(x, *args) -> float ), as per the documentation of scipy.optimize.minimize . However, 1 array + 3 arguments makes 4, while newfunc expects 5 arguments. Perhaps you expected the elements of the tuple to be passed separately (it seems that way, from the naming), but changing newfunc to accommodate that causes new problems; here's the latter part of your code with the superfluous variables removed, and the passed variables used:

def newfunc(a, y, H):
    ko, ki = a
    tspan = [0, max(xdat1)]
    y0 = [0, 0.04818]
    result = solve_ivp(func, tspan, y0, method='RK45', args=(ko, ki, H), dense_output=True)
    resultnew = result.sol(tspan).T
    ephi = y - resultnew[:, 0]
    sumephi = np.sum((ephi ** 2))
    return sumephi


wguess = (k0, k1)
arguments = (ydat1, H0)
res = minimize(newfunc, np.array(wguess), args=arguments, method='bfgs')

However, this results in a new error: TypeError: can't multiply sequence by non-int of type 'numpy.float64' . Again, not a surprise, since your line: dHdt = -ki * phi * H takes two floats and tries to multiply them with a list ( H ).

Replacing H with a single value causes further errors. Since it's not clear what your code is even supposed to do without further analysis and a deeper understanding of what your code is even for, I think you should take the approach of rewriting it - do it a small part at a time, ensuring what you write works before adding more and ensuring you know what what you write means.

You'll note that several people voted down your question - this likely due to them grabbing a copy of your code and trying to run it like myself (or just reading it here), and concluding that you have overlooked some very basic problems in code that has many such problems.

I think my question answers your question why you're getting that error message, but fixing your program will require some work on your part.