Python - 滚动窗口 OLS 回归估计 [英] Python - Rolling window OLS Regression estimation
问题描述
为了进行评估，我在
我想在t+1处做Y的后向预测(即根据前一个值预测Y的下一个值即 p(Y)t+1 通过包括均方误差 (MSE) - 例如，如果我们查看第 5 行，X<的值/code> 为 2，Y 的值为 10.假设预测值 (p(Y)t+1) 为 6，因此 mse 将是 (10-6)^2.我们如何使用 statsmodels 或 scikit-learn 来实现 >pd.stats.ols.MovingOLS 在 Pandas 版本 0.20.0 中被删除，因为我找不到任何参考?
这里是使用 statsmodels 进行滚动 OLS 的大纲，应该适用于您的数据.只需使用 df=pd.read_csv('estimated_pred.csv') 而不是我随机生成的 df:
将pandas导入为pd将 numpy 导入为 np将 statsmodels.api 导入为 sm#随机数据#df=pd.DataFrame(np.random.normal(size=(500,3)),columns=['time','X','Y'])df=pd.read_csv('estimated_pred.csv')df=df.dropna() #取消注释这一行以删除nans窗口 = 5df['a']=None #constantdf['b1']=无#beta1df['b2']=无#beta2对于我在范围内(窗口，len(df)):temp=df.iloc[i-window:i,:]RollOLS=sm.OLS(temp.loc[:,'Y'],sm.add_constant(temp.loc[:,['time','X']])).fit()df.iloc[i,df.columns.get_loc('a')]=RollOLS.params[0]df.iloc[i,df.columns.get_loc('b1')]=RollOLS.params[1]df.iloc[i,df.columns.get_loc('b2')]=RollOLS.params[2]#鉴于PRIOR行的估计参数，以下行为您提供一行中的预测值df['预测']=df['a'].shift(1)+df['b1'].shift(1)*df['time']+df['b2'].shift(1)*df['X']我存储常数和 beta，但有多种方法可以进行预测...您可以使用拟合模型对象，我的是 RollOLS 和 .predict() 方法，或者自己乘以我在最后一行中所做的(在这种情况下更容易这样做，因为变量的数量是固定和已知的，您可以一次性完成简单的列数学运算).
使用 sm 进行预测时，它看起来像这样:
predict_x=np.random.normal(size=(20,2))RollOLS.predict(sm.add_constant(predict_x))但请记住，如果您按顺序运行上述代码，则预测值将仅使用最后一个窗口的模型.如果您想使用不同的模型，那么您可以随时保存这些模型，或者在 for 循环中预测值.请注意，您还可以使用 RollOLS.fittedvalues 获得拟合值，因此，如果您正在平滑数据拉取并为循环中的每次迭代保存 RollOLS.fittedvalues[-1].
为了帮助了解如何使用您自己的数据，这里是滚动回归循环运行后我的 df 的尾部:
时间 X Y a b1 b2495 0.662463 0.771971 0.643008 -0.0235751 0.037875 0.0907694496 -0.127879 1.293141 0.404959 0.00314073 0.0441054 0.113387497 -0.006581 -0.824247 0.226653 0.0105847 0.0439867 0.118228498 1.870858 0.920964 0.571535 0.0123463 0.0428359 0.11598499 0.724296 0.537296 -0.411965 0.00104044 0.055003 0.118953For my evaluation, I have a dataset found in this link (https://drive.google.com/drive/folders/0B2Iv8dfU4fTUMVFyYTEtWXlzYkk) as in the following format. The third column (Y) in my dataset is my true value - that's what I wanted to predict (estimate).
time X Y
0.000543 0 10
0.000575 0 10
0.041324 1 10
0.041331 2 10
0.041336 3 10
0.04134 4 10
...
9.987735 55 239
9.987739 56 239
9.987744 57 239
9.987749 58 239
9.987938 59 239
I want to run a rolling of for example 5 window OLS regression estimation, and I have tried it with the following script.
# /usr/bin/python -tt
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
df = pd.read_csv('estimated_pred.csv')
model = pd.stats.ols.MovingOLS(y=df.Y, x=df[['X']],
window_type='rolling', window=5, intercept=True)
df['Y_hat'] = model.y_predict
print(df['Y_hat'])
print (model.summary)
df.plot.scatter(x='X', y='Y', s=0.1)
The summary of the regression analysis is shown below.
-------------------------Summary of Regression Analysis-------------------------
Formula: Y ~ <X> + <intercept>
Number of Observations: 5
Number of Degrees of Freedom: 2
R-squared: -inf
Adj R-squared: -inf
Rmse: 0.0000
F-stat (1, 3): nan, p-value: nan
Degrees of Freedom: model 1, resid 3
-----------------------Summary of Estimated Coefficients------------------------
Variable Coef Std Err t-stat p-value CI 2.5% CI 97.5%
--------------------------------------------------------------------------------
X 0.0000 0.0000 1.97 0.1429 0.0000 0.0000
intercept 239.0000 0.0000 14567091934632472.00 0.0000 239.0000 239.0000
---------------------------------End of Summary---------------------------------
I want to do a backward prediction of Y at t+1 (i.e. predict the next value of Y according to the previous value i.e. p(Y)t+1 by including the mean squared error (MSE) - for example, if we look at row 5, the value of X is 2 and the value of Y is 10. Let's say the prediction value (p(Y)t+1) is 6 and therefore the mse will be (10-6)^2. How can we do this using either statsmodels or scikit-learn for pd.stats.ols.MovingOLS was removed in Pandas version 0.20.0 and since I can't find any reference?
Here is an outline of doing rolling OLS with statsmodels and should work for your data. simply use df=pd.read_csv('estimated_pred.csv') instead of my randomly generated df:
import pandas as pd
import numpy as np
import statsmodels.api as sm
#random data
#df=pd.DataFrame(np.random.normal(size=(500,3)),columns=['time','X','Y'])
df=pd.read_csv('estimated_pred.csv')
df=df.dropna() #uncomment this line to drop nans
window = 5
df['a']=None #constant
df['b1']=None #beta1
df['b2']=None #beta2
for i in range(window,len(df)):
temp=df.iloc[i-window:i,:]
RollOLS=sm.OLS(temp.loc[:,'Y'],sm.add_constant(temp.loc[:,['time','X']])).fit()
df.iloc[i,df.columns.get_loc('a')]=RollOLS.params[0]
df.iloc[i,df.columns.get_loc('b1')]=RollOLS.params[1]
df.iloc[i,df.columns.get_loc('b2')]=RollOLS.params[2]
#The following line gives you predicted values in a row, given the PRIOR row's estimated parameters
df['predicted']=df['a'].shift(1)+df['b1'].shift(1)*df['time']+df['b2'].shift(1)*df['X']
I store the constant and betas, but there are a number of ways to approach predicting... you can use your fitted model object mine is RollOLS and the .predict() method, or multiply it yourself which I did in the final line (easier to do this way in this case because number of variables is fixed and known and you can do simple column math all in one go).
to do predictions with sm though as you go it would look like this:
predict_x=np.random.normal(size=(20,2))
RollOLS.predict(sm.add_constant(predict_x))
but keep in mind, if you ran the above code in sequence the predicted values would be using the model of the last window only. if you want to use a different model then you can save those as you go, or predict values within the for loop. Note you can also get fitted values with RollOLS.fittedvalues, and so if you are smoothing data pull and save RollOLS.fittedvalues[-1] for each iteration in the loop.
To help see how to use for your own data here is the tail of my df after the rolling regression loop is run:
time X Y a b1 b2
495 0.662463 0.771971 0.643008 -0.0235751 0.037875 0.0907694
496 -0.127879 1.293141 0.404959 0.00314073 0.0441054 0.113387
497 -0.006581 -0.824247 0.226653 0.0105847 0.0439867 0.118228
498 1.870858 0.920964 0.571535 0.0123463 0.0428359 0.11598
499 0.724296 0.537296 -0.411965 0.00104044 0.055003 0.118953
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