我认为问题在于参数window_type='rolling' and window=30干脆什么都不做。首先，我将向您展示原因，最后我将提供一个用于滚动窗口上的线性回归的设置。
1.你的函数的问题：
由于您尚未提供一些示例数据，因此这里有一个函数，它返回具有一些随机数的所需大小的数据帧：
# Function to build synthetic data
import numpy as np
import pandas as pd
import statsmodels.api as sm
from collections import OrderedDict
def sample(rSeed, periodLength, colNames):
np.random.seed(rSeed)
date = pd.to_datetime("1st of Dec, 1999")
cols = OrderedDict()
for col in colNames:
cols[col] = np.random.normal(loc=0.0, scale=1.0, size=periodLength)
dates = date+pd.to_timedelta(np.arange(periodLength), 'D')
df = pd.DataFrame(cols, index = dates)
return(df)
Output:
X1 X2
2018-12-01 -1.085631 -1.294085
2018-12-02 0.997345 -1.038788
2018-12-03 0.282978 1.743712
2018-12-04 -1.506295 -0.798063
2018-12-05 -0.578600 0.029683
.
.
.
2019-01-17 0.412912 -1.363472
2019-01-18 0.978736 0.379401
2019-01-19 2.238143 -0.379176
现在，尝试：
rolling_beta = sm.OLS(df['X2'], df['X1'], window_type='rolling', window=30).fit()
rolling_beta.params
Output:
X1 -0.075784
dtype: float64
这至少也代表了输出的结构，这意味着您期望对每个样本窗口进行估计，但您得到的是单个估计。因此，我在网上和 statsmodels 文档中查找了一些使用相同函数的其他示例，但我无法找到实际有效的具体示例。我确实发现了一些关于不久前如何弃用此功能的讨论。然后我使用一些虚假的参数输入测试了相同的函数：
rolling_beta = sm.OLS(df['X2'], df['X1'], window_type='amazing', window=3000000).fit()
rolling_beta.params
Output:
X1 -0.075784
dtype: float64
正如您所看到的，估计值是相同的，并且不会为虚假输入返回任何错误消息。所以我建议你看一下下面的函数。这是我为了执行滚动回归估计而整理的东西。
2.pandas 数据框滚动窗口的回归函数
df = sample(rSeed = 123, colNames = ['X1', 'X2', 'X3'], periodLength = 50)
def RegressionRoll(df, subset, dependent, independent, const, win, parameters):
"""
RegressionRoll takes a dataframe, makes a subset of the data if you like,
and runs a series of regressions with a specified window length, and
returns a dataframe with BETA or R^2 for each window split of the data.
Parameters:
===========
df: pandas dataframe
subset: integer - has to be smaller than the size of the df
dependent: string that specifies name of denpendent variable
inependent: LIST of strings that specifies name of indenpendent variables
const: boolean - whether or not to include a constant term
win: integer - window length of each model
parameters: string that specifies which model parameters to return:
BETA or R^2
Example:
========
RegressionRoll(df=df, subset = 50, dependent = 'X1', independent = ['X2'],
const = True, parameters = 'beta', win = 30)
"""
# Data subset
if subset != 0:
df = df.tail(subset)
else:
df = df
# Loopinfo
end = df.shape[0]
win = win
rng = np.arange(start = win, stop = end, step = 1)
# Subset and store dataframes
frames = {}
n = 1
for i in rng:
df_temp = df.iloc[:i].tail(win)
newname = 'df' + str(n)
frames.update({newname: df_temp})
n += 1
# Analysis on subsets
df_results = pd.DataFrame()
for frame in frames:
#print(frames[frame])
# Rolling data frames
dfr = frames[frame]
y = dependent
x = independent
if const == True:
x = sm.add_constant(dfr[x])
model = sm.OLS(dfr[y], x).fit()
else:
model = sm.OLS(dfr[y], dfr[x]).fit()
if parameters == 'beta':
theParams = model.params[0:]
coefs = theParams.to_frame()
df_temp = pd.DataFrame(coefs.T)
indx = dfr.tail(1).index[-1]
df_temp['Date'] = indx
df_temp = df_temp.set_index(['Date'])
if parameters == 'R2':
theParams = model.rsquared
df_temp = pd.DataFrame([theParams])
indx = dfr.tail(1).index[-1]
df_temp['Date'] = indx
df_temp = df_temp.set_index(['Date'])
df_temp.columns = [', '.join(independent)]
df_results = pd.concat([df_results, df_temp], axis = 0)
return(df_results)
df_rolling = RegressionRoll(df=df, subset = 50, dependent = 'X1', independent = ['X2'], const = True, parameters = 'beta',
win = 30)
输出：一个数据帧，其中包含每个 30 个数据周期窗口的 X1 上 X2 的 OLS 的 beta 估计值。
const X2
Date
2018-12-30 0.044042 0.032680
2018-12-31 0.074839 -0.023294
2019-01-01 -0.063200 0.077215
.
.
.
2019-01-16 -0.075938 -0.215108
2019-01-17 -0.143226 -0.215524
2019-01-18 -0.129202 -0.170304
