为什么 scipy.optimize.curce 拟合函数不能正确拟合数据点,为什么要给出较大的 pfit 值?
why scipy.optimize.curce fit function is not fitting the data points correctly and why giving large values of pfit?
在下面的代码中,为什么拟合函数会给出很大的 pfit 值,以及为什么它不能正确拟合数据点。我的拟合函数有什么问题吗?
L = np.array([12,24,36,48])
Ec_L =np.array([-2.21173697, -2.01880398, -1.96508108, -2.0691906 ])
def ff(L,a,v,Ec):
return (a*L**(-1.0/v))+Ec
x_data = 1.0/L
y_data = Ec_L
plt.scatter(x_data, y_data, marker='.', color='orange')
pfit,pcov = optimize.curve_fit(ff,x_data,y_data)
print("pfit: ",pfit) #pfit: [ 563.99154975 4377.13071157 -566.48046716]
print(pcov)
plt.plot(x_data, ff(L,*pfit), marker='.', color='red')
您在测试中使用 L
,但在试穿中使用 1/L
;我不知道你的意图,但如果你改为使用
plt.plot(x_data, ff(1/L,*pfit), marker='.', color='red')
合身度看起来没那么差:
您的数据似乎非常适合二次方程,并且似乎位于抛物线上。这是一个使用您的数据和二阶多项式方程的图形多项式拟合器,可以在代码顶部更改多项式阶数。
import numpy, matplotlib
import matplotlib.pyplot as plt
L = [12,24,36,48]
Ec_L = [-2.21173697, -2.01880398, -1.96508108, -2.0691906 ]
# rename to match previous example code
xData = numpy.array(L, dtype=float)
yData = numpy.array(Ec_L, dtype=float)
polynomialOrder = 2 # example quadratic equation
# curve fit the test data
fittedParameters = numpy.polyfit(xData, yData, polynomialOrder)
print('Fitted Parameters:', fittedParameters)
# predict a single value
print('Single value prediction:', numpy.polyval(fittedParameters, 3.0))
# Use polyval to find model predictions
modelPredictions = numpy.polyval(fittedParameters, xData)
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('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 = numpy.polyval(fittedParameters, xModel)
# now the model as a line plot
axes.plot(xModel, yModel)
axes.set_title('numpy polyfit() quadratic example') # add a title
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)
在下面的代码中,为什么拟合函数会给出很大的 pfit 值,以及为什么它不能正确拟合数据点。我的拟合函数有什么问题吗?
L = np.array([12,24,36,48])
Ec_L =np.array([-2.21173697, -2.01880398, -1.96508108, -2.0691906 ])
def ff(L,a,v,Ec):
return (a*L**(-1.0/v))+Ec
x_data = 1.0/L
y_data = Ec_L
plt.scatter(x_data, y_data, marker='.', color='orange')
pfit,pcov = optimize.curve_fit(ff,x_data,y_data)
print("pfit: ",pfit) #pfit: [ 563.99154975 4377.13071157 -566.48046716]
print(pcov)
plt.plot(x_data, ff(L,*pfit), marker='.', color='red')
您在测试中使用 L
,但在试穿中使用 1/L
;我不知道你的意图,但如果你改为使用
plt.plot(x_data, ff(1/L,*pfit), marker='.', color='red')
合身度看起来没那么差:
您的数据似乎非常适合二次方程,并且似乎位于抛物线上。这是一个使用您的数据和二阶多项式方程的图形多项式拟合器,可以在代码顶部更改多项式阶数。
import numpy, matplotlib
import matplotlib.pyplot as plt
L = [12,24,36,48]
Ec_L = [-2.21173697, -2.01880398, -1.96508108, -2.0691906 ]
# rename to match previous example code
xData = numpy.array(L, dtype=float)
yData = numpy.array(Ec_L, dtype=float)
polynomialOrder = 2 # example quadratic equation
# curve fit the test data
fittedParameters = numpy.polyfit(xData, yData, polynomialOrder)
print('Fitted Parameters:', fittedParameters)
# predict a single value
print('Single value prediction:', numpy.polyval(fittedParameters, 3.0))
# Use polyval to find model predictions
modelPredictions = numpy.polyval(fittedParameters, xData)
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('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 = numpy.polyval(fittedParameters, xModel)
# now the model as a line plot
axes.plot(xModel, yModel)
axes.set_title('numpy polyfit() quadratic example') # add a title
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)