如何进行约束线性回归 - scikit 学习？

Question

我正在尝试使用一些约束来执行线性回归主题以获得一定的预测。 我想让模型预测一半的线性预测，而后半部分的线性预测接近前半部分的最后一个值，使用非常窄的范围（使用约束），类似于图中的绿线。

完整代码：

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
from sklearn.model_selection import train_test_split
pd.options.mode.chained_assignment = None  # default='warn'
data = [5.269, 5.346, 5.375, 5.482, 5.519, 5.57, 5.593999999999999, 5.627000000000001, 5.724, 5.818, 5.792999999999999, 5.817, 5.8389999999999995, 5.882000000000001, 5.92, 6.025, 6.064, 6.111000000000001, 6.1160000000000005, 6.138, 6.247000000000001, 6.279, 6.332000000000001, 6.3389999999999995, 6.3420000000000005, 6.412999999999999, 6.442, 6.519, 6.596, 6.603, 6.627999999999999, 6.76, 6.837000000000001, 6.781000000000001, 6.8260000000000005, 6.849, 6.875, 6.982, 7.018, 7.042000000000001, 7.068, 7.091, 7.204, 7.228, 7.261, 7.3420000000000005, 7.414, 7.44, 7.516, 7.542000000000001, 7.627000000000001, 7.667000000000001, 7.821000000000001, 7.792999999999999, 7.756, 7.871, 8.006, 8.078, 7.916, 7.974, 8.074, 8.119, 8.228, 7.976, 8.045, 8.312999999999999, 8.335, 8.388, 8.437999999999999, 8.456, 8.227, 8.266, 8.277999999999999, 8.289, 8.299, 8.318, 8.332, 8.34, 8.349, 8.36, 8.363999999999999, 8.368, 8.282, 8.283999999999999]
time = range(1,85,1)   
x=int(0.7*len(data))
df = pd.DataFrame(list(zip(*[time, data])))
df.columns = ['time', 'data']
# print df
x=int(0.7*len(df))
train = df[:x]
valid = df[x:]
models = []
names = []
tr_x_ax = []
va_x_ax = []
pr_x_ax = []
tr_y_ax = []
va_y_ax = []
pr_y_ax = []
time_model = []
models.append(('LR', LinearRegression()))

for name, model in models:
    x_train=df.iloc[:, 0][:x].values
    y_train=df.iloc[:, 1][:x].values
    x_valid=df.iloc[:, 0][x:].values
    y_valid=df.iloc[:, 1][x:].values

    model = LinearRegression()
    # poly = PolynomialFeatures(5)
    x_train= x_train.reshape(-1, 1)
    y_train= y_train.reshape(-1, 1)
    x_valid = x_valid.reshape(-1, 1)
    y_valid = y_valid.reshape(-1, 1)
    # model.fit(x_train,y_train)
    model.fit(x_train,y_train.ravel())
    # score = model.score(x_train,y_train.ravel())
    # print 'score', score
    preds = model.predict(x_valid)
    tr_x_ax.extend(train['data'])
    va_x_ax.extend(valid['data'])
    pr_x_ax.extend(preds)

    valid['Predictions'] = preds
    valid.index = df[x:].index
    train.index = df[:x].index
    plt.figure(figsize=(5,5))
    # plt.plot(train['data'],label='data')
    # plt.plot(valid[['Close', 'Predictions']])
    x = valid['data']
    # print x
    # plt.plot(valid['data'],label='validation')
    plt.plot(valid['Predictions'],label='Predictions before',color='orange')



y =range(0,58)
y1 =range(58,84)
for index, item in enumerate(pr_x_ax):
    if index >13:
        pr_x_ax[index] = pr_x_ax[13]
pr_x_ax = list([float(i) for i in pr_x_ax])
va_x_ax = list([float(i) for i in va_x_ax])
tr_x_ax = list([float(i) for i in tr_x_ax])
plt.plot(y,tr_x_ax,  label='train' , color='red',  linewidth=2)
plt.plot(y1,va_x_ax,  label='validation1' , color='blue',  linewidth=2)
plt.plot(y1,pr_x_ax,  label='Predictions after' , color='green',  linewidth=2)
plt.xlabel("time")
plt.ylabel("data")
plt.xticks(rotation=45)
plt.legend()
plt.show()

如果你看到这个图：

label：Predictions before，模型预测它没有任何约束（我不需要这个结果）。

标签：Predictions after，模型在约束范围内预测它，但这是在模型预测之后并且所有值都等于 index = 71 , item 8.56 的最后一个值。

我在第 64 行中使用了循环 for index, item in enumerate(pr_x_ax):，曲线是从 71 秒到 85 秒的直线，如您所见，以便向您展示我需要模型如何工作。

我可以建立模型给出相同的结果而不是 for 循环吗？？？

请多多指教

Answer 1

我希望通过绘制绿线在您的问题中，您确实希望经过训练的模型能够预测向右的线性水平转向。但是当前训练的模型只画了一条橙色的直线。

对于任何算法和类型的任何训练模型来说，为了学习行为模型中的一些异常变化，至少需要具有该异常变化的一些样本。或者至少观察数据中的一些隐藏含义应该指向有这种不寻常的变化。

换句话说，为了让您的模型学习右转绿线，模型应该在训练数据集中具有右转点。但是您将train = df[:int(0.7 * len(df))] 的第一个（最左边）70% 的数据作为训练数据，并且该训练数据没有这样的右转，并且该训练数据看起来接近一条直线。

因此，您需要以不同的方式将数据重新采样到训练和验证中 - 从X 的整个范围内随机抽取 70% 的样本，其余的用于验证。这样在您的训练数据中也包括右转的样本。

第二件事是LinearRegression 模型总是只用一条直线对预测进行建模，这条直线不能右转。为了右转，您需要一些更复杂的模型。

模型右转的一种方法是分段线性，即有几条连接的直线。我在sklearn里面没有找到现成的分段线性模型，只用了其他的pip模型。所以我决定实现我自己的简单类PieceWiseLinearRegression，它使用np.piecewise() 和scipy.optimize.curve_fit() 来对分段线性函数进行建模。

下一张图显示了应用上面提到的两个东西的结果，代码在后面，以不同的方式重新采样数据集并建模分段线性函数。您当前的线性模型LR 仍然只使用一条直线进行预测，而我的分段线性PWLR2 橙色线由两段组成并正确预测右转：

为了清楚地看到一张PWLR2 图表，我也做了下一张图片：

我的类PieceWiseLinearRegression 在创建对象时只接受一个参数n - 用于预测的线性段数。以上图片使用n = 2。

import sys, numpy as np, pandas as pd
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
np.random.seed(0)

class PieceWiseLinearRegression:
    @classmethod
    def nargs_func(cls, f, n):
        return eval('lambda ' + ', '.join([f'a{i}'for i in range(n)]) + ': f(' + ', '.join([f'a{i}'for i in range(n)]) + ')', locals())
        
    @classmethod
    def piecewise_linear(cls, n):
        condlist = lambda xs, xa: [(lambda x: (
            (xs[i] <= x if i > 0 else np.full_like(x, True, dtype = np.bool_)) &
            (x < xs[i + 1] if i < n - 1 else np.full_like(x, True, dtype = np.bool_))
        ))(xa) for i in range(n)]
        funclist = lambda xs, ys: [(lambda i: (
            lambda x: (
                (x - xs[i]) * (ys[i + 1] - ys[i]) / (
                    (xs[i + 1] - xs[i]) if abs(xs[i + 1] - xs[i]) > 10 ** -7 else 10 ** -7 * (-1, 1)[xs[i + 1] - xs[i] >= 0]
                ) + ys[i]
            )
        ))(j) for j in range(n)]
        def f(x, *pargs):
            assert len(pargs) == (n + 1) * 2, (n, pargs)
            xs, ys = pargs[0::2], pargs[1::2]
            xa = x.ravel().astype(np.float64)
            ya = np.piecewise(x = xa, condlist = condlist(xs, xa), funclist = funclist(xs, ys)).ravel()
            #print('xs', xs, 'ys', ys, 'xa', xa, 'ya', ya)
            return ya
        return cls.nargs_func(f, 1 + (n + 1) * 2)
        
    def __init__(self, n):
        self.n = n
        self.f = self.piecewise_linear(self.n)

    def fit(self, x, y):
        from scipy import optimize
        self.p, self.e = optimize.curve_fit(self.f, x, y, p0 = [j for i in range(self.n + 1) for j in (np.amin(x) + i * (np.amax(x) - np.amin(x)) / self.n, 1)])
        #print('p', self.p)
        
    def predict(self, x):
        return self.f(x, *self.p)

data = [5.269, 5.346, 5.375, 5.482, 5.519, 5.57, 5.593999999999999, 5.627000000000001, 5.724, 5.818, 5.792999999999999, 5.817, 5.8389999999999995, 5.882000000000001, 5.92, 6.025, 6.064, 6.111000000000001, 6.1160000000000005, 6.138, 6.247000000000001, 6.279, 6.332000000000001, 6.3389999999999995, 6.3420000000000005, 6.412999999999999, 6.442, 6.519, 6.596, 6.603, 6.627999999999999, 6.76, 6.837000000000001, 6.781000000000001, 6.8260000000000005, 6.849, 6.875, 6.982, 7.018, 7.042000000000001, 7.068, 7.091, 7.204, 7.228, 7.261, 7.3420000000000005, 7.414, 7.44, 7.516, 7.542000000000001, 7.627000000000001, 7.667000000000001, 7.821000000000001, 7.792999999999999, 7.756, 7.871, 8.006, 8.078, 7.916, 7.974, 8.074, 8.119, 8.228, 7.976, 8.045, 8.312999999999999, 8.335, 8.388, 8.437999999999999, 8.456, 8.227, 8.266, 8.277999999999999, 8.289, 8.299, 8.318, 8.332, 8.34, 8.349, 8.36, 8.363999999999999, 8.368, 8.282, 8.283999999999999]
time = list(range(1, 85))
df = pd.DataFrame(list(zip(time, data)), columns = ['time', 'data'])

choose_train = np.random.uniform(size = (len(df),)) < 0.8
choose_valid = ~choose_train

x_all = df.iloc[:, 0].values
y_all = df.iloc[:, 1].values
x_train = df.iloc[:, 0][choose_train].values
y_train = df.iloc[:, 1][choose_train].values
x_valid = df.iloc[:, 0][choose_valid].values
y_valid = df.iloc[:, 1][choose_valid].values
x_all_lin = np.linspace(np.amin(x_all), np.amax(x_all), 500)

models = []
models.append(('LR', LinearRegression()))
models.append(('PWLR2', PieceWiseLinearRegression(2)))
        
for imodel, (name, model) in enumerate(models):
    model.fit(x_train[:, None], y_train)
    x_all_lin_pred = model.predict(x_all_lin[:, None])
    plt.plot(x_all_lin, x_all_lin_pred, label = f'pred {name}')

plt.plot(x_train, y_train, label='train')
plt.plot(x_valid, y_valid, label='valid')
plt.xlabel('time')
plt.ylabel('data')
plt.legend()
plt.show()