我有一组数据,我想比较哪行最能描述它(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()
。但是我发现没有这样的函数用于指数和对数拟合。
有吗?否则如何解决?
对于拟合y = A + B日志x,只需将y对(log x)进行拟合。
>>> 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
对于拟合y = Ae Bx,取双方的对数给出对数y = log A + Bx] >。因此,对x适合(log y)。
请注意,拟合(log y
)就像是线性的一样,将强调y的较小值,从而导致较大的[[y产生较大的偏差。这是因为polyfit
(线性回归)通过最小化∑ i]((ΔY)2 = ∑ i( Y i-Ŷ i)2。当Y i = log y i时,残基ΔY i = Δ(log y i)≈Δy i / | y i |。因此,即使polyfit
对大y做出了非常糟糕的决定,“ |除以|| y |”因数将对其进行补偿,导致polyfit
偏爱较小的值。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对指数回归/趋势线使用未加权(有偏)公式。如果您希望结果与这些平台兼容,即使它提供了更好的结果。
现在,如果可以使用scipy,则可以使用scipy.optimize.curve_fit
来拟合任何模型而无需进行转换。
y
=A
+ B log x,结果与转换方法相同:scipy.optimize.curve_fit
对于y =Bx,由于它可以直接计算Δ(log y),因此我们可以得到更好的拟合度。但是我们需要提供一个初始猜测,以便Ae
>>> 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)
可以达到所需的局部最小值。curve_fit
>>> 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.
然后,如果要绘制,则可以执行:
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)
((注:在绘制时,在
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()
前面的*
会将术语扩展为popt
期望的a
,b
和c
。)
我对此有一些麻烦,所以让我非常明确,这样像我这样的菜鸟就可以理解。让我们说我们有一个数据文件或类似的东西
func
结果是:a = 0.849195983017,b = -1.18101681765,c = 2.24061176543,d = 0.816643894816
“>好吧,我想您可以随时使用:# -*- 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('$', '') plt.title(r'$f(\lambda)= %s$' %(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()
略微修改np.log --> natural log np.log10 --> base 10 np.log2 --> base 2
:
IanVS's answer这将导致下图:
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()
给出
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 # Data 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
代码适合指数数据
transformer = FunctionTransformer(np.log, validate=True)
x_samp, y_samp = generate_data(func_exp, 2.5, 1.2, 0.7, jitter=3)
y_trans = transformer.fit_transform(y_samp) # 1
model = LinearRegression().fit(x_samp, y_trans) # 2
y_fit = model.predict(x_samp)
plt.scatter(x_samp, y_samp)
plt.plot(x_samp, np.exp(y_fit), "k--", label="Fit") # 3
plt.title("Exponential Fit")
适合日志数据
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
model = LinearRegression().fit(x_trans, y_samp) # 2
y_fit = model.predict(x_trans)
plt.scatter(x_samp, y_samp)
plt.plot(x_samp, y_fit, "k--", label="Fit") # 3
plt.title("Logarithmic Fit")
详细信息
一般步骤
y
并适合原始数据进行绘图]np.exp()
我们可以通过取来线性化后一个方程式(例如y =截距+斜率* x):
给出线性化方程
++
和回归参数,我们可以计算:ln(A)
通过斜率(B
)] >>+B
++注:更改x数据有助于线性化指数
数据,而更改y数据有助于线性化log数据。# -*- 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('$', '')
plt.title(r'$f(\lambda)= %s$' %(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()
给出