Python 使用scipy曲线拟合指数曲线(拟合曲线与实际曲线不匹配)
我尝试使用Python 使用scipy曲线拟合指数曲线(拟合曲线与实际曲线不匹配),python,scipy,curve-fitting,exponential,data-fitting,Python,Scipy,Curve Fitting,Exponential,Data Fitting,我尝试使用curve\u fit(scipy.optimize)拟合指数曲线,但拟合曲线与实际曲线完全不同。 现在我正在使用以下代码: X=[0.0, 9.0, 18.0, 27.0, 36.0, 45.0, 54.0] Y=[0.090316199, -0.078157925, -0.350137315, -0.695193468, -1.106773689, -1.60467115, -2.196169408] #plot Y against X fig = plt.figure(num=
curve\u fit(scipy.optimize)
拟合指数曲线,但拟合曲线与实际曲线完全不同。
现在我正在使用以下代码:
X=[0.0, 9.0, 18.0, 27.0, 36.0, 45.0, 54.0]
Y=[0.090316199, -0.078157925, -0.350137315, -0.695193468, -1.106773689, -1.60467115, -2.196169408]
#plot Y against X
fig = plt.figure(num=None, figsize=(9, 7),facecolor='w', edgecolor='k')
ax=fig.add_subplot(111)
ax.scatter(X,Y)
#fit using curve_fit
popt, pcov = curve_fit(func, X, Y,maxfev=10000)
#compute Y_estiamted using fitted parameters
Y_estimated=[popt[0]*np.exp(i+popt[1])+popt[2] for i in X]
#plot Y_estiamted against X
ax.scatter(X,Y_estimated, c='r')
def func(x,a,b,c):
return a*(np.exp(x+b))+c
蓝色曲线是真实曲线,红色曲线是拟合曲线
如您所见,拟合的红色曲线与真实的蓝色曲线根本不匹配。任何帮助都将不胜感激 我认为问题在于模型函数。如果将其更改为以下函数:
def func(x, a, b, c, d):
return a * (np.exp(d*(x + b))) + c
然后它会找到一个很好的匹配:
我更改了代码中的一些内容:
def func(x, a, b, c, d):
return a * (np.exp(d*(x + b))) + c
X = [0.0, 9.0, 18.0, 27.0, 36.0, 45.0, 54.0]
Y = [0.090316199, -0.078157925, -0.350137315, -0.695193468, -1.106773689, -1.60467115, -2.196169408]
# plot Y against X
fig = plt.figure(num=None, figsize=(9, 7), facecolor='w', edgecolor='k')
ax = fig.add_subplot(111)
ax.scatter(X, Y)
# fit using curve_fit
popt, pcov = curve_fit(func, X, Y, maxfev=10000)
# compute Y_estiamted using fitted parameters
x = np.linspace(min(X), max(X), 100)
Y_estimated = func(x, *popt)
# plot Y_estiamted against X
ax.plot(x, Y_estimated, c='r')
我很好地拟合了一个渐近指数型方程,它只有一个形状参数和一个小的偏移量,“1.0-pow(a,x)+b”。下面是一个图形化的Python装配工,它将这个等式与您的数据一起使用
我认为这可能会有帮助。@PéterLeéh代码中使用的方程无法匹配数据的形状,因此初始参数估计是不相关的。您之前链接到的答案在这种情况下没有帮助。请参阅我对这个问题的答案,它有一个只有两个参数的方程。@JamesPhillips的答案也是正确的(有更好的代码示例)。问题是他为什么选择指数方法作为模型,特别是对于这个数据集。谢谢!但是你怎么知道使用np.power而不是np.exp呢?我通过开源Python曲线拟合网站zunzun.com上的“function finder”运行了发布的数据,寻找具有两个或更少参数的方程。这个方程式是最好的结果之一。
import numpy, scipy, matplotlib
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
# ignore warnings within curve_fit() routine
import warnings
warnings.filterwarnings("ignore")
X=[0.0, 9.0, 18.0, 27.0, 36.0, 45.0, 54.0]
Y=[0.090316199, -0.078157925, -0.350137315, -0.695193468, -1.106773689, -1.60467115, -2.196169408]
# alias data to match previous example
xData = numpy.array(X, dtype=float)
yData = numpy.array(Y, dtype=float)
def func(x, a, b): # Asymptotic Exponential A equation with offset from zunzun.com
return 1.0 - numpy.power(a, x) + b
# these are the same as the scipy defaults
initialParameters = numpy.array([1.0, 1.0])
# curve fit the test data
fittedParameters, pcov = curve_fit(func, xData, yData, initialParameters)
modelPredictions = func(xData, *fittedParameters)
absError = modelPredictions - yData
SE = numpy.square(absError) # squared errors
MSE = numpy.mean(SE) # mean squared errors
RMSE = numpy.sqrt(MSE) # Root Mean Squared Error, RMSE
Rsquared = 1.0 - (numpy.var(absError) / numpy.var(yData))
print('Parameters:', fittedParameters)
print('RMSE:', RMSE)
print('R-squared:', Rsquared)
print()
##########################################################
# graphics output section
def ModelAndScatterPlot(graphWidth, graphHeight):
f = plt.figure(figsize=(graphWidth/100.0, graphHeight/100.0), dpi=100)
axes = f.add_subplot(111)
# first the raw data as a scatter plot
axes.plot(xData, yData, 'D')
# create data for the fitted equation plot
xModel = numpy.linspace(min(xData), max(xData))
yModel = func(xModel, *fittedParameters)
# now the model as a line plot
axes.plot(xModel, yModel)
axes.set_xlabel('X Data') # X axis data label
axes.set_ylabel('Y Data') # Y axis data label
plt.show()
plt.close('all') # clean up after using pyplot
graphWidth = 800
graphHeight = 600
ModelAndScatterPlot(graphWidth, graphHeight)