In curvefit: TypeError: only size-1 arrays can be converted to Python scalars

Question

This error is recieved in jupyter, which is basicly python for as far I'm aware. I think I'm aware what this error means, however I don't see the problem with the code. Thanks in advance! 

import matplotlib.pyplot as plt

import numpy as np

import scipy.optimize 

from math import *

import matplotlib.patches as mpatches

X = [0,-0.0000959936,-0.000194899,-0.000273446,-0.000430558,-0.000366546,-0.000573153,-0.000657563,-0.000768193,0.000066904,0.000148355,0.000232717,

0.000285083,0.000427648,0.000401461,0.000526588,0.0000465422,-0.0000250891,0.0000436333,-0.0000261799,-0.0000610866,0.0000930847,-0.000101811]

Y = [0.96,0.47,0.11,0.25,0.15,0.12,0.13,0.11,0.11,0.78,0.16,0.19,0.13,0.14,0.16,0.13,0.6,2.02,1.91,1.4,0.94,0.68,0.97]

onzekerheidY = [0.0001,0.000001,-0.000101,-0.000182,-0.000344,-0.000278,-0.000491,-0.000578,-0.000692,0.000169,0.000253,0.00034,

0.000394,0.000541,0.000514,0.000643,0.000148,0.000074125,0.000145,0.000073,0.000037,0.000196,-0.000005]

def f(x, I0, theta, a, d):

    k = 9666438.934

    N = 1

    return I0 * (  ( (np.sin(k*(d/2)*np.sin(x)))/(k*(d/2)*np.sin(x)) * (np.sin(k*a*np.sin(x) *(N/2)))/(N*np.sin((k/2)*a*sin(x))) )   )^2

waarde,matrix = scipy.optimize.curve_fit(f,X,Y)

plt.plot(X,Y,'p',label='data')

a = np.linspace(-0.0300,0.0300,1000)

b = b = f(a, waarde[0], waarde[1],waarde[2],waarde[3])

plt.plot(a,b,'r',label= 'fit')

plt.errorbar(X,Y,onzekerheidY,linestyle='None',label = 'onzekerheden')

plt.grid()

plt.legend(loc = 'lower left')

plt.title('Fit van $R_{TM}$ in functie van $\Theta$ $_i$')

plt.xlabel('$\Theta$ $_i$')

plt.ylabel('$R_{TM}$')

And this is the error code:

TypeError                                 Traceback (most recent call last)
<ipython-input-9-b80cfce3b4ff> in <module>()
      4     return I0*( (np.sin(k*(d/2)*np.sin(x)))/(k*(d/2)*np.sin(x)) * (np.sin(k*a*np.sin(x) *(N/2)))/(N*np.sin((k/2)*a*sin(x)))    )^2
      5 
----> 6 waarde,matrix = scipy.optimize.curve_fit(f,X,Y)
      7 
      8 plt.plot(X,Y,'p',label='data')

C:\Users\boent\Anaconda2\lib\site-packages\scipy\optimize\minpack.pyc in curve_fit(f, xdata, ydata, p0, sigma, absolute_sigma, check_finite, bounds, method, jac, **kwargs)
    749         # Remove full_output from kwargs, otherwise we're passing it in twice.
    750         return_full = kwargs.pop('full_output', False)
--> 751         res = leastsq(func, p0, Dfun=jac, full_output=1, **kwargs)
    752         popt, pcov, infodict, errmsg, ier = res
    753         cost = np.sum(infodict['fvec'] ** 2)

C:\Users\boent\Anaconda2\lib\site-packages\scipy\optimize\minpack.pyc in leastsq(func, x0, args, Dfun, full_output, col_deriv, ftol, xtol, gtol, maxfev, epsfcn, factor, diag)
    381     if not isinstance(args, tuple):
    382         args = (args,)
--> 383     shape, dtype = _check_func('leastsq', 'func', func, x0, args, n)
    384     m = shape[0]
    385     if n > m:

C:\Users\boent\Anaconda2\lib\site-packages\scipy\optimize\minpack.pyc in _check_func(checker, argname, thefunc, x0, args, numinputs, output_shape)
     25 def _check_func(checker, argname, thefunc, x0, args, numinputs,
     26                 output_shape=None):
---> 27     res = atleast_1d(thefunc(*((x0[:numinputs],) + args)))
     28     if (output_shape is not None) and (shape(res) != output_shape):
     29         if (output_shape[0] != 1):

C:\Users\boent\Anaconda2\lib\site-packages\scipy\optimize\minpack.pyc in func_wrapped(params)
    461     if transform is None:
    462         def func_wrapped(params):
--> 463             return func(xdata, *params) - ydata
    464     elif transform.ndim == 1:
    465         def func_wrapped(params):

<ipython-input-9-b80cfce3b4ff> in f(x, I0, theta, a, d)
      2     k = 9666438.934
      3     N = 1
----> 4     return I0*( (np.sin(k*(d/2)*np.sin(x)))/(k*(d/2)*np.sin(x)) * (np.sin(k*a*np.sin(x) *(N/2)))/(N*np.sin((k/2)*a*sin(x)))    )^2
      5 
      6 waarde,matrix = scipy.optimize.curve_fit(f,X,Y)

TypeError: only size-1 arrays can be converted to Python scalars

Answer 1

The error message you encountered, "TypeError: only size-1 arrays can be converted to Python scalars," is occurring because of the "^" operator used in your code. In Python, the caret "^" is the bitwise XOR operator, not the exponentiation operator. Therefore, when you use "^2" in your return statement, it tries to perform the bitwise XOR operation between the expression and the integer 2, which causes the error.

To fix this issue, you should use the correct exponentiation operator, which is "**" in Python. Replace "^2" with "**2" in your return statement:

return I0 * ((np.sin(k*(d/2)*np.sin(x))) / (k*(d/2)*np.sin(x)) * (np.sin(k*a*np.sin(x)*(N/2))) / (N*np.sin((k/2)*a*np.sin(x))))**2

By making this change, the code should run without the TypeError.

To fix this issue, you should use the correct exponentiation operator, which is "**" in Python. Replace "^2" with "**2" in your return statement:

 

Answer 2

The encountered error, "TypeError: only size-1 arrays can be converted to Python scalars," stems from the utilization of the caret operator "^" in your code. It is important to note that in Python, the caret symbol "^" functions as the bitwise XOR operator, rather than the exponentiation operator. Consequently, when attempting to employ "^2" in your return statement, it interprets it as a bitwise XOR operation between the expression and the integer 2, resulting in the raised error.

To rectify this issue, it is imperative to replace the caret operator "^" with the appropriate exponentiation operator, which is denoted by "**" in Python:

return I0 * ((np.sin(k*(d/2)*np.sin(x))) / (k*(d/2)*np.sin(x)) * (np.sin(k*a*np.sin(x)*(N/2))) / (N*np.sin((k/2)*a*np.sin(x))))**2

By implementing this adjustment, the code will execute seamlessly without invoking the TypeError.

Answer 3

The caret operator "" is used in your code, which is the cause of the encountered error, "TypeError: only size-1 arrays can be converted to Python scalars." It's crucial to keep in mind that the caret symbol in Python represents the bitwise XOR operator rather than the exponentiation operator. As a result, it raises an error when you try to use "2" in your return statement because it sees it as a bitwise XOR operation between the expression and the integer 2.

The appropriate exponentiation operator, represented in Python by the symbol "**," must be used in place of the caret operator "" in order to resolve this problem:

return I0 * ((np.sin(k*(d/2)*np.sin(x))) / (k*(d/2)*np.sin(x)) * (np.sin(k*a*np.sin(x)*(N/2))) / (N*np.sin((k/2)*a*np.sin(x))))**2

By making this modification, the code will run without encountering the TypeError.