带约束的简单线性回归

Question

我开发了一种算法来循环遍历 15 个变量并为每个变量生成一个简单的 OLS。然后算法再循环 11 次以产生相同的 15 个 OLS 回归，但 X 变量的滞后每次增加 1。我选择具有最高 r^2 的自变量，并使用 3,4 或 5 个变量的最佳滞后

即

Y_t+1 - Y_t = B ( X_t+k - X_t) + e

我的数据集如下所示：

Regression = pd.DataFrame(np.random.randint(low=0, high=10, size=(100, 6)), 
                columns=['Y', 'X1', 'X2', 'X3', 'X4','X5'])

到目前为止我拟合的 OLS 回归使用以下代码：

Y = Regression['Y']
X = Regression[['X1','X2','X3']]

Model = sm.OLS(Y,X).fit()
predictions = Model.predict(X)

Model.summary()

问题在于，使用 OLS，您可以获得负系数（我就是这样做的）。我很感激通过以下方式限制此模型的帮助：

sum(B_i) = 1

B_i >= 0

Answer 1

这很好用，

from scipy.optimize import minimize

# Define the Model
model = lambda b, X: b[0] * X[:,0] + b[1] * X[:,1] + b[2] * X[:,2]

# The objective Function to minimize (least-squares regression)
obj = lambda b, Y, X: np.sum(np.abs(Y-model(b, X))**2)

# Bounds: b[0], b[1], b[2] >= 0
bnds = [(0, None), (0, None), (0, None)]

# Constraint: b[0] + b[1] + b[2] - 1 = 0
cons = [{"type": "eq", "fun": lambda b: b[0]+b[1]+b[2] - 1}]

# Initial guess for b[1], b[2], b[3]:
xinit = np.array([0, 0, 1])

res = minimize(obj, args=(Y, X), x0=xinit, bounds=bnds, constraints=cons)

print(f"b1={res.x[0]}, b2={res.x[1]}, b3={res.x[2]}")

#Save the coefficients for further analysis on goodness of fit

beta1 = res.x[0]

beta2 = res.x[1]

beta3 = res.x[2]

Answer 2

根据 cmets，这是一个使用 scipy 的差分进化模块来确定有界参数估计的示例。该模块在内部使用拉丁超立方体算法来确保对参数空间的彻底搜索，并需要搜索范围，尽管这些范围可能很大。默认情况下，differential_evolution 模块将在内部以使用边界对 curve_fit() 的调用结束 - 这可以禁用 - 并确保最终拟合的参数不受限制，此示例稍后调用 curve_fit 而不通过边界。您可以从打印的结果中看到，对differential_evolution 的调用显示第一个参数以-0.185 为界，而后来调用curve_fit() 的结果并非如此。在您的情况下，您可以将下限设为零，以便参数不是负数，但如果代码导致参数处于或非常接近界限，则这不是最佳的，如本例所示。

import numpy, scipy, matplotlib
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy.optimize import differential_evolution
import warnings

xData = numpy.array([19.1647, 18.0189, 16.9550, 15.7683, 14.7044, 13.6269, 12.6040, 11.4309, 10.2987, 9.23465, 8.18440, 7.89789, 7.62498, 7.36571, 7.01106, 6.71094, 6.46548, 6.27436, 6.16543, 6.05569, 5.91904, 5.78247, 5.53661, 4.85425, 4.29468, 3.74888, 3.16206, 2.58882, 1.93371, 1.52426, 1.14211, 0.719035, 0.377708, 0.0226971, -0.223181, -0.537231, -0.878491, -1.27484, -1.45266, -1.57583, -1.61717])
yData = numpy.array([0.644557, 0.641059, 0.637555, 0.634059, 0.634135, 0.631825, 0.631899, 0.627209, 0.622516, 0.617818, 0.616103, 0.613736, 0.610175, 0.606613, 0.605445, 0.603676, 0.604887, 0.600127, 0.604909, 0.588207, 0.581056, 0.576292, 0.566761, 0.555472, 0.545367, 0.538842, 0.529336, 0.518635, 0.506747, 0.499018, 0.491885, 0.484754, 0.475230, 0.464514, 0.454387, 0.444861, 0.437128, 0.415076, 0.401363, 0.390034, 0.378698])


def func(t, n_0, L, offset): #exponential curve fitting function
    return n_0*numpy.exp(-L*t) + offset


# function for genetic algorithm to minimize (sum of squared error)
def sumOfSquaredError(parameterTuple):
    warnings.filterwarnings("ignore") # do not print warnings by genetic algorithm
    val = func(xData, *parameterTuple)
    return numpy.sum((yData - val) ** 2.0)


def generate_Initial_Parameters():
    # min and max used for bounds
    maxX = max(xData)
    minX = min(xData)
    maxY = max(yData)
    minY = min(yData)

    parameterBounds = []
    parameterBounds.append([-0.185, maxX]) # seach bounds for n_0
    parameterBounds.append([minX, maxX]) # seach bounds for L
    parameterBounds.append([0.0, maxY]) # seach bounds for Offset

    # "seed" the numpy random number generator for repeatable results
    result = differential_evolution(sumOfSquaredError, parameterBounds, seed=3)
    return result.x

# by default, differential_evolution completes by calling
# curve_fit() using parameter bounds
geneticParameters = generate_Initial_Parameters()
print('fit with parameter bounds (note the -0.185)')
print(geneticParameters)
print()

# second call to curve_fit made with no bounds for comparison
fittedParameters, pcov = curve_fit(func, xData, yData, geneticParameters)

print('re-fit with no parameter bounds')
print(fittedParameters)
print()

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()
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)

Answer 3

我不知道您可以轻松地限制系数，我有两种解决方案，

1- 使用导致负系数的时间序列的倒数 (1/x)。这将要求您首先进行正态回归，然后反转具有负面关系的回归。获取权重并执行 wi/sum(wi)。

2-您似乎正在处理时间序列，使用对数差异(np.log(ts).diff().dropna()) 作为输入并获取权重。如有必要，将其除以权重之和，然后通过 np.exp(predicted_ts.cumsum()) 恢复您的估计。