Python / curve_fit：无法通过带有初始猜测的数组 [英] Python/curve_fit: cannot pass array with init guess

问题描述

我具有此函数来计算某种多项式：

I have this function to compute some sort of polynomial:

def pipoly(df,pj):
    n=np.size(pj)
    p=pj[0]
    for j in range(1,n):
        p+=pj[j]*df**j
    return p

pj 应该是一个数组包含多项式系数的初始猜测；因此，多项式的阶数由第一行中的函数本身确定。 df 是标量变量。此函数传递给scipy.optimize的 curve_fit 为

pj is supposed to be an array that contains the initial guesses of the coefficients of the polynomial; the degree of the polynomial is hence determined by the function itself in the first line. df is a scalar variable. This function is passed to scipy.optimize's curve_fit as

parfit,covfig=curve_fit(pipoly,[f-f0[j] for f in f_df[if0[j]:if0[i]]],
                            pmode_xp[ph][if0[j]:if0[i]],
                            p0=([pmode0[ph][-1],(pmode_xp[ph][if0[i]]-pmode_xp[ph][if0[j]])/df]))

函数名称后的前两个参数是数组（二维数组的一维切片），我已经确认它们具有相同的长度。在 pipoly 之后的第三个参数应该是一个元组，它带有 pj 的初始猜测，我之前已经打印了出来： [0.4586590267346888，0.7419930843896957] 。那么，为什么Python会抱怨 TypeError：pipoly（）接受2个位置参数，但是却给了3个？而且，如果我删除了 p0 参数，我会被告知 pj 被视为标量，因此不能索引。我该如何清楚地暗示 pj 是一个数组？

The first two arguments after the name of the function are arrays (1D slices of 2D arrays), and I have confirmed that they have the same length. The third argument after pipoly is supposed to be a tuple with the initial guesses for pj, which I printed out before: [0.4586590267346888, 0.7419930843896957]. So why is Python complaining that TypeError: pipoly() takes 2 positional arguments but 3 were given? And if I remove the p0 argument, I'm told that the pj is considered a scalar and can therefore not have an index. How do I make it clear to pipoly that pj is to be an array?

推荐答案

您的语句：

pj 应该是一个数组包含
多项式的系数；

pj is supposed to be an array that contains the coefficients of the polynomial;

是错误的。根据 curve_fit（）文档：

scipy.optimize.curve_fit（f，xdata，ydata，p0 = None，sigma = None，absolute_sigma = False，check_finite = True，bounds =（-inf，inf），方法=无，jac =无，** kwargs）[源代码]使用非线性最小二乘法将函数f拟合到数据。

scipy.optimize.curve_fit(f, xdata, ydata, p0=None, sigma=None, absolute_sigma=False, check_finite=True, bounds=(-inf, inf), method=None, jac=None, **kwargs)[source] Use non-linear least squares to fit a function, f, to data.

假设 ydata = f（xdata，* params）+ eps

这意味着要由 curve_fit（）使用的 pipoly（）函数取数量等于您多项式的参数数量加一个（变量，这是第一个参数）的参数。

错误：

That means that your pipoly() function, to be used by curve_fit() must take a number of arguments equal to the number of parameters of you polynomial plus one (the variable, which is the first argument).
The error:

TypeError：pipoly（）接受2个位置参数，但给了3个位置参数？

TypeError: pipoly() takes 2 positional arguments but 3 were given?

告诉您 pipoly 会收到3个参数，因为您可能正在测试线性政治学，因此这3个参数分别是自变量和2个参数（ [f-f0 [ j]表示f_df [if0 [j0：if0 [i]]] 中的f是2个长度的列表）。

编写时，它需要2

is telling you that pipoly receives 3 arguments because you probably were testing a linear polinomyal, so the three arguments were the independent variable and two parameter (the [f-f0[j] for f in f_df[if0[j]:if0[i]]] stuff is a 2-length list).
As you write it, instead it takes 2 argument only.

您可以通过在 pj 之前添加星号来轻松解决问题：

You can easily solve your problem by adding an asterisk before pj:

def pipoly(df,*pj):
    n=len(pj) #len() is sufficient here, but np.size() works too.
    p=pj[0]
    for j in range(1,n):
        p+=pj[j]*df**j
    return p

这样，您的函数将接受可变数量的参数。 此处详细说明其含义和在Python函数参数中使用星号。

This way your function accepts a variable number of arguments. Here more on the meaning and the use of the asterisk in python function parameters.

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