如何在Python中进行指数和对数曲线拟合?我发现只有多项式拟合

发布于 09-13 12:17 字数 147 浏览 13 评论 0原文

我有一组数据,我想比较哪一行最能描述它(不同阶的多项式,指数或对数)。

我使用 Python 和 Numpy,对于多项式拟合,有一个函数 polyfit()。但我没有发现这样的指数和对数拟合函数。

有吗?或者另外如何解决?

I have a set of data and I want to compare which line describes it best (polynomials of different orders, exponential or logarithmic).

I use Python and Numpy and for polynomial fitting there is a function polyfit(). But I found no such functions for exponential and logarithmic fitting.

Are there any? Or how to solve it otherwise?

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评论(7

狼亦尘2024-09-20 12:17:35

要拟合 y = A + B log x,只需将 y 拟合到 (记录x)。

>>> x = numpy.array([1, 7, 20, 50, 79])
>>> y = numpy.array([10, 19, 30, 35, 51])
>>> numpy.polyfit(numpy.log(x), y, 1)
array([ 8.46295607,  6.61867463])
# y ≈ 8.46 log(x) + 6.62

为了拟合y = AeBx,两边取对数得到logy = log A + Bx。因此适合(log y)与x

请注意,将 (log y) 拟合为线性将强调 y 的小值,从而导致大 y 出现较大偏差。这是因为 polyfit(线性回归)的工作原理是最小化 ΣiY)2< /sup> = Σi (YiŶ< em>i)2。当Yi = log yi时,残基 ΔYi = Δ(log yi< /sub>) ≈ Δyi / |yi|。因此,即使 polyfit 对于较大的 y 做出了非常糟糕的决定,“除以-|y|”因子将对其进行补偿,导致 polyfit 倾向于较小的值。

这可以通过为每个条目赋予与y成比例的“权重”来缓解。 polyfit 通过 w 关键字参数支持加权最小二乘。

>>> x = numpy.array([10, 19, 30, 35, 51])
>>> y = numpy.array([1, 7, 20, 50, 79])
>>> numpy.polyfit(x, numpy.log(y), 1)
array([ 0.10502711, -0.40116352])
#    y ≈ exp(-0.401) * exp(0.105 * x) = 0.670 * exp(0.105 * x)
# (^ biased towards small values)
>>> numpy.polyfit(x, numpy.log(y), 1, w=numpy.sqrt(y))
array([ 0.06009446,  1.41648096])
#    y ≈ exp(1.42) * exp(0.0601 * x) = 4.12 * exp(0.0601 * x)
# (^ not so biased)

请注意,Excel、LibreOffice 和大多数科学计算器通常使用未加权(有偏差)公式来计算指数回归/趋势线。如果您希望结果与这些平台兼容,甚至不要包含权重如果它能提供更好的结果。


现在,如果您可以使用 scipy,则可以使用 scipy.optimize.curve_fit 无需转换即可拟合任何模型。

对于 y = A + B log x 结果与转换方法相同:

>>> x = numpy.array([1, 7, 20, 50, 79])
>>> y = numpy.array([10, 19, 30, 35, 51])
>>> scipy.optimize.curve_fit(lambda t,a,b: a+b*numpy.log(t),  x,  y)
(array([ 6.61867467,  8.46295606]), 
 array([[ 28.15948002,  -7.89609542],
        [ -7.89609542,   2.9857172 ]]))
# y ≈ 6.62 + 8.46 log(x)

对于 y = AeBx,但是,我们可以获得更好的拟合,因为它计算 Δ(log y) 直接。但我们需要提供一个初始化猜测,以便 curve_fit 能够达到所需的局部最小值。

>>> x = numpy.array([10, 19, 30, 35, 51])
>>> y = numpy.array([1, 7, 20, 50, 79])
>>> scipy.optimize.curve_fit(lambda t,a,b: a*numpy.exp(b*t),  x,  y)
(array([  5.60728326e-21,   9.99993501e-01]),
 array([[  4.14809412e-27,  -1.45078961e-08],
        [ -1.45078961e-08,   5.07411462e+10]]))
# oops, definitely wrong.
>>> scipy.optimize.curve_fit(lambda t,a,b: a*numpy.exp(b*t),  x,  y,  p0=(4, 0.1))
(array([ 4.88003249,  0.05531256]),
 array([[  1.01261314e+01,  -4.31940132e-02],
        [ -4.31940132e-02,   1.91188656e-04]]))
# y ≈ 4.88 exp(0.0553 x). much better.

指数比较回归

For fitting y = A + B log x, just fit y against (log x).

>>> x = numpy.array([1, 7, 20, 50, 79])
>>> y = numpy.array([10, 19, 30, 35, 51])
>>> numpy.polyfit(numpy.log(x), y, 1)
array([ 8.46295607,  6.61867463])
# y ≈ 8.46 log(x) + 6.62

For fitting y = AeBx, take the logarithm of both side gives log y = log A + Bx. So fit (log y) against x.

Note that fitting (log y) as if it is linear will emphasize small values of y, causing large deviation for large y. This is because polyfit (linear regression) works by minimizing ∑iY)2 = ∑i (YiŶi)2. When Yi = log yi, the residues ΔYi = Δ(log yi) ≈ Δyi / |yi|. So even if polyfit makes a very bad decision for large y, the "divide-by-|y|" factor will compensate for it, causing polyfit favors small values.

This could be alleviated by giving each entry a "weight" proportional to y. polyfit supports weighted-least-squares via the w keyword argument.

>>> x = numpy.array([10, 19, 30, 35, 51])
>>> y = numpy.array([1, 7, 20, 50, 79])
>>> numpy.polyfit(x, numpy.log(y), 1)
array([ 0.10502711, -0.40116352])
#    y ≈ exp(-0.401) * exp(0.105 * x) = 0.670 * exp(0.105 * x)
# (^ biased towards small values)
>>> numpy.polyfit(x, numpy.log(y), 1, w=numpy.sqrt(y))
array([ 0.06009446,  1.41648096])
#    y ≈ exp(1.42) * exp(0.0601 * x) = 4.12 * exp(0.0601 * x)
# (^ not so biased)

Note that Excel, LibreOffice and most scientific calculators typically use the unweighted (biased) formula for the exponential regression / trend lines. If you want your results to be compatible with these platforms, do not include the weights even if it provides better results.


Now, if you can use scipy, you could use scipy.optimize.curve_fit to fit any model without transformations.

For y = A + B log x the result is the same as the transformation method:

>>> x = numpy.array([1, 7, 20, 50, 79])
>>> y = numpy.array([10, 19, 30, 35, 51])
>>> scipy.optimize.curve_fit(lambda t,a,b: a+b*numpy.log(t),  x,  y)
(array([ 6.61867467,  8.46295606]), 
 array([[ 28.15948002,  -7.89609542],
        [ -7.89609542,   2.9857172 ]]))
# y ≈ 6.62 + 8.46 log(x)

For y = AeBx, however, we can get a better fit since it computes Δ(log y) directly. But we need to provide an initialize guess so curve_fit can reach the desired local minimum.

>>> x = numpy.array([10, 19, 30, 35, 51])
>>> y = numpy.array([1, 7, 20, 50, 79])
>>> scipy.optimize.curve_fit(lambda t,a,b: a*numpy.exp(b*t),  x,  y)
(array([  5.60728326e-21,   9.99993501e-01]),
 array([[  4.14809412e-27,  -1.45078961e-08],
        [ -1.45078961e-08,   5.07411462e+10]]))
# oops, definitely wrong.
>>> scipy.optimize.curve_fit(lambda t,a,b: a*numpy.exp(b*t),  x,  y,  p0=(4, 0.1))
(array([ 4.88003249,  0.05531256]),
 array([[  1.01261314e+01,  -4.31940132e-02],
        [ -4.31940132e-02,   1.91188656e-04]]))
# y ≈ 4.88 exp(0.0553 x). much better.

comparison of exponential regression

音盲2024-09-20 12:17:35

您还可以使用 scipy.optimize 中的 curve_fit 将一组数据拟合到您喜欢的任何函数。例如,如果您想拟合指数函数(来自文档 ):

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

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

x = np.linspace(0,4,50)
y = func(x, 2.5, 1.3, 0.5)
yn = y + 0.2*np.random.normal(size=len(x))

popt, pcov = curve_fit(func, x, yn)

然后如果你想绘图,你可以这样做:

plt.figure()
plt.plot(x, yn, 'ko', label="Original Noised Data")
plt.plot(x, func(x, *popt), 'r-', label="Fitted Curve")
plt.legend()
plt.show()

(注意:绘图时 popt 前面的 * 会将术语展开到 func 所期望的 abc。)

You can also fit a set of a data to whatever function you like using curve_fit from scipy.optimize. For example if you want to fit an exponential function (from the documentation):

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

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

x = np.linspace(0,4,50)
y = func(x, 2.5, 1.3, 0.5)
yn = y + 0.2*np.random.normal(size=len(x))

popt, pcov = curve_fit(func, x, yn)

And then if you want to plot, you could do:

plt.figure()
plt.plot(x, yn, 'ko', label="Original Noised Data")
plt.plot(x, func(x, *popt), 'r-', label="Fitted Curve")
plt.legend()
plt.show()

(Note: the * in front of popt when you plot will expand out the terms into the a, b, and c that func is expecting.)

苏大泽ㄣ2024-09-20 12:17:35

我在这方面遇到了一些麻烦,所以让我非常明确,以便像我这样的菜鸟能够理解。

假设我们有一个数据文件或类似的文件,

# -*- coding: utf-8 -*-

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

"""
Generate some data, let's imagine that you already have this. 
"""
x = np.linspace(0, 3, 50)
y = np.exp(x)

"""
Plot your data
"""
plt.plot(x, y, 'ro',label="Original Data")

"""
brutal force to avoid errors
"""    
x = np.array(x, dtype=float) #transform your data in a numpy array of floats 
y = np.array(y, dtype=float) #so the curve_fit can work

"""
create a function to fit with your data. a, b, c and d are the coefficients
that curve_fit will calculate for you. 
In this part you need to guess and/or use mathematical knowledge to find
a function that resembles your data
"""
def func(x, a, b, c, d):
    return a*x**3 + b*x**2 +c*x + d

"""
make the curve_fit
"""
popt, pcov = curve_fit(func, x, y)

"""
The result is:
popt[0] = a , popt[1] = b, popt[2] = c and popt[3] = d of the function,
so f(x) = popt[0]*x**3 + popt[1]*x**2 + popt[2]*x + popt[3].
"""
print "a = %s , b = %s, c = %s, d = %s" % (popt[0], popt[1], popt[2], popt[3])

"""
Use sympy to generate the latex sintax of the function
"""
xs = sym.Symbol('\lambda')    
tex = sym.latex(func(xs,*popt)).replace('

结果是:
a = 0.849195983017 , b = -1.18101681765, c = 2.24061176543, d = 0.816643894816

原始数据和拟合函数

, '') plt.title(r'$f(\lambda)= %s

结果是:
a = 0.849195983017 , b = -1.18101681765, c = 2.24061176543, d = 0.816643894816

原始数据和拟合函数

%(tex),fontsize=16) """ Print the coefficients and plot the funcion. """ plt.plot(x, func(x, *popt), label="Fitted Curve") #same as line above \/ #plt.plot(x, popt[0]*x**3 + popt[1]*x**2 + popt[2]*x + popt[3], label="Fitted Curve") plt.legend(loc='upper left') plt.show()

结果是:
a = 0.849195983017 , b = -1.18101681765, c = 2.24061176543, d = 0.816643894816

原始数据和拟合函数

I was having some trouble with this so let me be very explicit so noobs like me can understand.

Lets say that we have a data file or something like that

# -*- coding: utf-8 -*-

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

"""
Generate some data, let's imagine that you already have this. 
"""
x = np.linspace(0, 3, 50)
y = np.exp(x)

"""
Plot your data
"""
plt.plot(x, y, 'ro',label="Original Data")

"""
brutal force to avoid errors
"""    
x = np.array(x, dtype=float) #transform your data in a numpy array of floats 
y = np.array(y, dtype=float) #so the curve_fit can work

"""
create a function to fit with your data. a, b, c and d are the coefficients
that curve_fit will calculate for you. 
In this part you need to guess and/or use mathematical knowledge to find
a function that resembles your data
"""
def func(x, a, b, c, d):
    return a*x**3 + b*x**2 +c*x + d

"""
make the curve_fit
"""
popt, pcov = curve_fit(func, x, y)

"""
The result is:
popt[0] = a , popt[1] = b, popt[2] = c and popt[3] = d of the function,
so f(x) = popt[0]*x**3 + popt[1]*x**2 + popt[2]*x + popt[3].
"""
print "a = %s , b = %s, c = %s, d = %s" % (popt[0], popt[1], popt[2], popt[3])

"""
Use sympy to generate the latex sintax of the function
"""
xs = sym.Symbol('\lambda')    
tex = sym.latex(func(xs,*popt)).replace('

the result is:
a = 0.849195983017 , b = -1.18101681765, c = 2.24061176543, d = 0.816643894816

Raw data and fitted function

, '') plt.title(r'$f(\lambda)= %s

the result is:
a = 0.849195983017 , b = -1.18101681765, c = 2.24061176543, d = 0.816643894816

Raw data and fitted function

%(tex),fontsize=16) """ Print the coefficients and plot the funcion. """ plt.plot(x, func(x, *popt), label="Fitted Curve") #same as line above \/ #plt.plot(x, popt[0]*x**3 + popt[1]*x**2 + popt[2]*x + popt[3], label="Fitted Curve") plt.legend(loc='upper left') plt.show()

the result is:
a = 0.849195983017 , b = -1.18101681765, c = 2.24061176543, d = 0.816643894816

Raw data and fitted function

梦年海沫深2024-09-20 12:17:35

这是使用来自 scikit 学习

给定

import numpy as np

import matplotlib.pyplot as plt

from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import FunctionTransformer


np.random.seed(123)

# General Functions
def func_exp(x, a, b, c):
    """Return values from a general exponential function."""
    return a * np.exp(b * x) + c


def func_log(x, a, b, c):
    """Return values from a general log function."""
    return a * np.log(b * x) + c


# Helper
def generate_data(func, *args, jitter=0):
    """Return a tuple of arrays with random data along a general function."""
    xs = np.linspace(1, 5, 50)
    ys = func(xs, *args)
    noise = jitter * np.random.normal(size=len(xs)) + jitter
    xs = xs.reshape(-1, 1)                                  # xs[:, np.newaxis]
    ys = (ys + noise).reshape(-1, 1)
    return xs, ys
transformer = FunctionTransformer(np.log, validate=True)

代码

拟合指数数据

# Data
x_samp, y_samp = generate_data(func_exp, 2.5, 1.2, 0.7, jitter=3)
y_trans = transformer.fit_transform(y_samp)             # 1

# Regression
regressor = LinearRegression()
results = regressor.fit(x_samp, y_trans)                # 2
model = results.predict
y_fit = model(x_samp)

# Visualization
plt.scatter(x_samp, y_samp)
plt.plot(x_samp, np.exp(y_fit), "k--", label="Fit")     # 3
plt.title("Exponential Fit")

在此处输入图像描述

拟合日志数据

# Data
x_samp, y_samp = generate_data(func_log, 2.5, 1.2, 0.7, jitter=0.15)
x_trans = transformer.fit_transform(x_samp)             # 1

# Regression
regressor = LinearRegression()
results = regressor.fit(x_trans, y_samp)                # 2
model = results.predict
y_fit = model(x_trans)

# Visualization
plt.scatter(x_samp, y_samp)
plt.plot(x_samp, y_fit, "k--", label="Fit")             # 3
plt.title("Logarithmic Fit")

在此处输入图像描述


详细信息

常规步骤

  1. 将日志操作应用于数据值(xy 或两者)
  2. 将数据回归到线性化模型
  3. 通过“反转”任何日志操作(使用 np.exp())并拟合原始数据

假设我们的数据遵循指数趋势,一般方程+可能是:

在此处输入图像描述

我们可以线性化后一个方程(例如 y = 截距 + 斜率* x) 通过获取日志

在此处输入图像描述

给定线性方程++和回归参数,我们可以

  • 通过截距计算:A (ln(A))
  • B 通过斜率 (B)

线性化技术总结

Relationship |  Example   |     General Eqn.     |  Altered Var.  |        Linearized Eqn.  
-------------|------------|----------------------|----------------|------------------------------------------
Linear       | x          | y =     B * x    + C | -              |        y =   C    + B * x
Logarithmic  | log(x)     | y = A * log(B*x) + C | log(x)         |        y =   C    + A * (log(B) + log(x))
Exponential  | 2**x, e**x | y = A * exp(B*x) + C | log(y)         | log(y-C) = log(A) + B * x
Power        | x**2       | y =     B * x**N + C | log(x), log(y) | log(y-C) = log(B) + N * log(x)

+注意:当噪声为小且C=0。谨慎使用。

++注意:虽然更改 x 数据有助于线性化指数数据,但更改 y 数据有助于线性化对数数据> 数据。

Here's a linearization option on simple data that uses tools from scikit learn.

Given

import numpy as np

import matplotlib.pyplot as plt

from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import FunctionTransformer


np.random.seed(123)

# General Functions
def func_exp(x, a, b, c):
    """Return values from a general exponential function."""
    return a * np.exp(b * x) + c


def func_log(x, a, b, c):
    """Return values from a general log function."""
    return a * np.log(b * x) + c


# Helper
def generate_data(func, *args, jitter=0):
    """Return a tuple of arrays with random data along a general function."""
    xs = np.linspace(1, 5, 50)
    ys = func(xs, *args)
    noise = jitter * np.random.normal(size=len(xs)) + jitter
    xs = xs.reshape(-1, 1)                                  # xs[:, np.newaxis]
    ys = (ys + noise).reshape(-1, 1)
    return xs, ys
transformer = FunctionTransformer(np.log, validate=True)

Code

Fit exponential data

# Data
x_samp, y_samp = generate_data(func_exp, 2.5, 1.2, 0.7, jitter=3)
y_trans = transformer.fit_transform(y_samp)             # 1

# Regression
regressor = LinearRegression()
results = regressor.fit(x_samp, y_trans)                # 2
model = results.predict
y_fit = model(x_samp)

# Visualization
plt.scatter(x_samp, y_samp)
plt.plot(x_samp, np.exp(y_fit), "k--", label="Fit")     # 3
plt.title("Exponential Fit")

enter image description here

Fit log data

# Data
x_samp, y_samp = generate_data(func_log, 2.5, 1.2, 0.7, jitter=0.15)
x_trans = transformer.fit_transform(x_samp)             # 1

# Regression
regressor = LinearRegression()
results = regressor.fit(x_trans, y_samp)                # 2
model = results.predict
y_fit = model(x_trans)

# Visualization
plt.scatter(x_samp, y_samp)
plt.plot(x_samp, y_fit, "k--", label="Fit")             # 3
plt.title("Logarithmic Fit")

enter image description here


Details

General Steps

  1. Apply a log operation to data values (x, y or both)
  2. Regress the data to a linearized model
  3. Plot by "reversing" any log operations (with np.exp()) and fit to original data

Assuming our data follows an exponential trend, a general equation+ may be:

enter image description here

We can linearize the latter equation (e.g. y = intercept + slope * x) by taking the log:

enter image description here

Given a linearized equation++ and the regression parameters, we could calculate:

  • A via intercept (ln(A))
  • B via slope (B)

Summary of Linearization Techniques

Relationship |  Example   |     General Eqn.     |  Altered Var.  |        Linearized Eqn.  
-------------|------------|----------------------|----------------|------------------------------------------
Linear       | x          | y =     B * x    + C | -              |        y =   C    + B * x
Logarithmic  | log(x)     | y = A * log(B*x) + C | log(x)         |        y =   C    + A * (log(B) + log(x))
Exponential  | 2**x, e**x | y = A * exp(B*x) + C | log(y)         | log(y-C) = log(A) + B * x
Power        | x**2       | y =     B * x**N + C | log(x), log(y) | log(y-C) = log(B) + N * log(x)

+Note: linearizing exponential functions works best when the noise is small and C=0. Use with caution.

++Note: while altering x data helps linearize exponential data, altering y data helps linearize log data.

ˇ宁静的妩媚2024-09-20 12:17:35

好吧,我想你总是可以使用:

np.log   -->  natural log
np.log10 -->  base 10
np.log2  -->  base 2

稍微修改 IanVS 的答案

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

def func(x, a, b, c):
  #return a * np.exp(-b * x) + c
  return a * np.log(b * x) + c

x = np.linspace(1,5,50)   # changed boundary conditions to avoid division by 0
y = func(x, 2.5, 1.3, 0.5)
yn = y + 0.2*np.random.normal(size=len(x))

popt, pcov = curve_fit(func, x, yn)

plt.figure()
plt.plot(x, yn, 'ko', label="Original Noised Data")
plt.plot(x, func(x, *popt), 'r-', label="Fitted Curve")
plt.legend()
plt.show()

这会产生下图:

在此处输入图像描述

Well I guess you can always use:

np.log   -->  natural log
np.log10 -->  base 10
np.log2  -->  base 2

Slightly modifying IanVS's answer:

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

def func(x, a, b, c):
  #return a * np.exp(-b * x) + c
  return a * np.log(b * x) + c

x = np.linspace(1,5,50)   # changed boundary conditions to avoid division by 0
y = func(x, 2.5, 1.3, 0.5)
yn = y + 0.2*np.random.normal(size=len(x))

popt, pcov = curve_fit(func, x, yn)

plt.figure()
plt.plot(x, yn, 'ko', label="Original Noised Data")
plt.plot(x, func(x, *popt), 'r-', label="Fitted Curve")
plt.legend()
plt.show()

This results in the following graph:

enter image description here

轻许诺言2024-09-20 12:17:35

我们在解决这两个问题的同时演示了 lmfit 的功能。

给定

import lmfit

import numpy as np

import matplotlib.pyplot as plt


%matplotlib inline
np.random.seed(123)
# General Functions
def func_log(x, a, b, c):
    """Return values from a general log function."""
    return a * np.log(b * x) + c


# Data
x_samp = np.linspace(1, 5, 50)
_noise = np.random.normal(size=len(x_samp), scale=0.06)
y_samp = 2.5 * np.exp(1.2 * x_samp) + 0.7 + _noise
y_samp2 = 2.5 * np.log(1.2 * x_samp) + 0.7 + _noise

代码

方法 1 - lmfit 模型

拟合指数数据

regressor = lmfit.models.ExponentialModel()                # 1    
initial_guess = dict(amplitude=1, decay=-1)                # 2
results = regressor.fit(y_samp, x=x_samp, **initial_guess)
y_fit = results.best_fit    

plt.plot(x_samp, y_samp, "o", label="Data")
plt.plot(x_samp, y_fit, "k--", label="Fit")
plt.legend()

在此处输入图像描述

方法2 - 自定义模型

拟合日志数据

regressor = lmfit.Model(func_log)                          # 1
initial_guess = dict(a=1, b=.1, c=.1)                      # 2
results = regressor.fit(y_samp2, x=x_samp, **initial_guess)
y_fit = results.best_fit

plt.plot(x_samp, y_samp2, "o", label="Data")
plt.plot(x_samp, y_fit, "k--", label="Fit")
plt.legend()

在此处输入图像描述


详细信息

  1. 选择回归类
  2. 提供尊重函数域的命名初始猜测

您可以从回归器确定推断参数目的。示例:

regressor.param_names
# ['decay', 'amplitude']

进行预测,请使用 ModelResult.eval()方法。

model = results.eval
y_pred = model(x=np.array([1.5]))

注意:ExponentialModel()遵循衰减函数,它接受两个参数,其中一个为负数。

输入图片此处描述

另请参阅 ExponentialGaussianModel(),它接受 更多参数

通过>安装库pip 安装 lmfit

We demonstrate features of lmfit while solving both problems.

Given

import lmfit

import numpy as np

import matplotlib.pyplot as plt


%matplotlib inline
np.random.seed(123)
# General Functions
def func_log(x, a, b, c):
    """Return values from a general log function."""
    return a * np.log(b * x) + c


# Data
x_samp = np.linspace(1, 5, 50)
_noise = np.random.normal(size=len(x_samp), scale=0.06)
y_samp = 2.5 * np.exp(1.2 * x_samp) + 0.7 + _noise
y_samp2 = 2.5 * np.log(1.2 * x_samp) + 0.7 + _noise

Code

Approach 1 - lmfit Model

Fit exponential data

regressor = lmfit.models.ExponentialModel()                # 1    
initial_guess = dict(amplitude=1, decay=-1)                # 2
results = regressor.fit(y_samp, x=x_samp, **initial_guess)
y_fit = results.best_fit    

plt.plot(x_samp, y_samp, "o", label="Data")
plt.plot(x_samp, y_fit, "k--", label="Fit")
plt.legend()

enter image description here

Approach 2 - Custom Model

Fit log data

regressor = lmfit.Model(func_log)                          # 1
initial_guess = dict(a=1, b=.1, c=.1)                      # 2
results = regressor.fit(y_samp2, x=x_samp, **initial_guess)
y_fit = results.best_fit

plt.plot(x_samp, y_samp2, "o", label="Data")
plt.plot(x_samp, y_fit, "k--", label="Fit")
plt.legend()

enter image description here


Details

  1. Choose a regression class
  2. Supply named, initial guesses that respect the function's domain

You can determine the inferred parameters from the regressor object. Example:

regressor.param_names
# ['decay', 'amplitude']

To make predictions, use the ModelResult.eval() method.

model = results.eval
y_pred = model(x=np.array([1.5]))

Note: the ExponentialModel() follows a decay function, which accepts two parameters, one of which is negative.

enter image description here

See also ExponentialGaussianModel(), which accepts more parameters.

Install the library via > pip install lmfit.

渡你暖光2024-09-20 12:17:35

Wolfram 有一个用于拟合指数的封闭式解决方案。他们也有类似的解决方案来拟合 对数幂律

我发现这比 scipy 的 curve_fit 效果更好。特别是当您没有“接近零”的数据时。这是一个示例:

import numpy as np
import matplotlib.pyplot as plt

# Fit the function y = A * exp(B * x) to the data
# returns (A, B)
# From: https://mathworld.wolfram.com/LeastSquaresFittingExponential.html
def fit_exp(xs, ys):
    S_x2_y = 0.0
    S_y_lny = 0.0
    S_x_y = 0.0
    S_x_y_lny = 0.0
    S_y = 0.0
    for (x,y) in zip(xs, ys):
        S_x2_y += x * x * y
        S_y_lny += y * np.log(y)
        S_x_y += x * y
        S_x_y_lny += x * y * np.log(y)
        S_y += y
    #end
    a = (S_x2_y * S_y_lny - S_x_y * S_x_y_lny) / (S_y * S_x2_y - S_x_y * S_x_y)
    b = (S_y * S_x_y_lny - S_x_y * S_y_lny) / (S_y * S_x2_y - S_x_y * S_x_y)
    return (np.exp(a), b)


xs = [33, 34, 35, 36, 37, 38, 39, 40, 41, 42]
ys = [3187, 3545, 4045, 4447, 4872, 5660, 5983, 6254, 6681, 7206]

(A, B) = fit_exp(xs, ys)

plt.figure()
plt.plot(xs, ys, 'o-', label='Raw Data')
plt.plot(xs, [A * np.exp(B *x) for x in xs], 'o-', label='Fit')

plt.title('Exponential Fit Test')
plt.xlabel('X')
plt.ylabel('Y')
plt.legend(loc='best')
plt.tight_layout()
plt.show()

在此处输入图像描述

Wolfram has a closed form solution for fitting an exponential. They also have similar solutions for fitting a logarithmic and power law.

I found this to work better than scipy's curve_fit. Especially when you don't have data "near zero". Here is an example:

import numpy as np
import matplotlib.pyplot as plt

# Fit the function y = A * exp(B * x) to the data
# returns (A, B)
# From: https://mathworld.wolfram.com/LeastSquaresFittingExponential.html
def fit_exp(xs, ys):
    S_x2_y = 0.0
    S_y_lny = 0.0
    S_x_y = 0.0
    S_x_y_lny = 0.0
    S_y = 0.0
    for (x,y) in zip(xs, ys):
        S_x2_y += x * x * y
        S_y_lny += y * np.log(y)
        S_x_y += x * y
        S_x_y_lny += x * y * np.log(y)
        S_y += y
    #end
    a = (S_x2_y * S_y_lny - S_x_y * S_x_y_lny) / (S_y * S_x2_y - S_x_y * S_x_y)
    b = (S_y * S_x_y_lny - S_x_y * S_y_lny) / (S_y * S_x2_y - S_x_y * S_x_y)
    return (np.exp(a), b)


xs = [33, 34, 35, 36, 37, 38, 39, 40, 41, 42]
ys = [3187, 3545, 4045, 4447, 4872, 5660, 5983, 6254, 6681, 7206]

(A, B) = fit_exp(xs, ys)

plt.figure()
plt.plot(xs, ys, 'o-', label='Raw Data')
plt.plot(xs, [A * np.exp(B *x) for x in xs], 'o-', label='Fit')

plt.title('Exponential Fit Test')
plt.xlabel('X')
plt.ylabel('Y')
plt.legend(loc='best')
plt.tight_layout()
plt.show()

enter image description here

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