function() 需要 4 个位置参数，但 Python 3.5 中的 5 个是 given_def 函数

Question

我正在尝试从我的数据中拟合指数函数。我对将数学函数拟合到我的数据还不是很有经验。下面是我现在的代码。

import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit

my_x = (4,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40)
my_y = (0.022172333,0.020881,0.017729,0.021641333,0.02479,0.030755667,0.037235,0.048389,0.068451,0.06898974,0.161409,0.242802333,0.316012667,0.440762333,0.569118333,0.7016839,0.832527333)

def myfunc(x,a,b,c):
    return a*np.exp(b*x)+c


p=[my_x,0.0045,0.1262,0] #pre-determined a=0.0045, b=0.1262, c=0 according to excel
popt, pcov = curve_fit(myfunc,my_x,my_y, p0=p)


plt.plot (my_x,myfunc(my_x, *popt))

我得到的错误消息如下。

return function(xdata, *params) - ydata

TypeError: myfunc() takes 4 positional arguments but 5 were given

我认为我不太明白这个错误信息在说什么。谁能帮助我了解导致此错误的原因以及如何改进我的代码？

Answer 1

该myfunc函数接收四个参数x，a、b和c。

错误消息说您提供了 5 个参数。这是因为您解压了poptwho 可能包含四个元素（因此解压时包含四个参数）。

在这一行：

plt.plot (my_x,myfunc(my_x, *popt))

该myfunc功能接收my_x+四个参数。它有五个参数，但函数只需要四个。

更多关于拆箱的信息在这里。