从Pandas到Statsmodels的OLS中不推荐使用的滚动窗口选项

Question

正如标题所暗示的那样，Pandas中ols命令中的滚动功能选项在statsmodels中迁移到哪里？我似乎找不到它。熊猫告诉我厄运正在起作用：

FutureWarning: The pandas.stats.ols module is deprecated and will be removed in a future version. We refer to external packages like statsmodels, see some examples here: http://statsmodels.sourceforge.net/stable/regression.html
  model = pd.ols(y=series_1, x=mmmm, window=50)

事实上，如果你做的事情如下：

import statsmodels.api as sm

model = sm.OLS(series_1, mmmm, window=50).fit()

print(model.summary())

你得到结果（窗口不会影响代码的运行）但你只得到整个时期的回归运行参数，而不是应该应该处理的每个滚动周期的一系列参数。

Answer 1

我创建了一个ols模块，旨在模仿大熊猫弃用的MovingOLS;这是here。

它有三个核心类：

• OLS：静态（单窗口）普通最小二乘回归。输出是NumPy数组
• RollingOLS：滚动（多窗口）普通最小二乘回归。输出是更高维度的NumPy数组。
• PandasRollingOLS：在pandas Series＆DataFrames中包装RollingOLS的结果。旨在模仿已弃用的pandas模块的外观。

请注意，该模块是package（我目前正在上传到PyPi的过程中）的一部分，它需要一个包间导入。

上面的前两个类完全在NumPy中实现，主要使用矩阵代数。 RollingOLS也广泛利用广播。属性很大程度上模仿statsmodels的OLS RegressionResultsWrapper。

一个例子：

import urllib.parse
import pandas as pd
from pyfinance.ols import PandasRollingOLS

# You can also do this with pandas-datareader; here's the hard way
url = "https://fred.stlouisfed.org/graph/fredgraph.csv"

syms = {
    "TWEXBMTH" : "usd", 
    "T10Y2YM" : "term_spread", 
    "GOLDAMGBD228NLBM" : "gold",
}

params = {
    "fq": "Monthly,Monthly,Monthly",
    "id": ",".join(syms.keys()),
    "cosd": "2000-01-01",
    "coed": "2019-02-01",
}

data = pd.read_csv(
    url + "?" + urllib.parse.urlencode(params, safe=","),
    na_values={"."},
    parse_dates=["DATE"],
    index_col=0
).pct_change().dropna().rename(columns=syms)
print(data.head())
#                  usd  term_spread      gold
# DATE                                       
# 2000-02-01  0.012580    -1.409091  0.057152
# 2000-03-01 -0.000113     2.000000 -0.047034
# 2000-04-01  0.005634     0.518519 -0.023520
# 2000-05-01  0.022017    -0.097561 -0.016675
# 2000-06-01 -0.010116     0.027027  0.036599

y = data.usd
x = data.drop('usd', axis=1)

window = 12  # months
model = PandasRollingOLS(y=y, x=x, window=window)

print(model.beta.head())  # Coefficients excluding the intercept
#             term_spread      gold
# DATE                             
# 2001-01-01     0.000033 -0.054261
# 2001-02-01     0.000277 -0.188556
# 2001-03-01     0.002432 -0.294865
# 2001-04-01     0.002796 -0.334880
# 2001-05-01     0.002448 -0.241902

print(model.fstat.head())
# DATE
# 2001-01-01    0.136991
# 2001-02-01    1.233794
# 2001-03-01    3.053000
# 2001-04-01    3.997486
# 2001-05-01    3.855118
# Name: fstat, dtype: float64

print(model.rsq.head())  # R-squared
# DATE
# 2001-01-01    0.029543
# 2001-02-01    0.215179
# 2001-03-01    0.404210
# 2001-04-01    0.470432
# 2001-05-01    0.461408
# Name: rsq, dtype: float64

Answer 2

用sklearn滚动测试版

import pandas as pd
from sklearn import linear_model

def rolling_beta(X, y, idx, window=255):

    assert len(X)==len(y)

    out_dates = []
    out_beta = []

    model_ols = linear_model.LinearRegression()

    for iStart in range(0, len(X)-window):        
        iEnd = iStart+window

        model_ols.fit(X[iStart:iEnd], y[iStart:iEnd])

        #store output
        out_dates.append(idx[iEnd])
        out_beta.append(model_ols.coef_[0][0])

    return pd.DataFrame({'beta':out_beta}, index=out_dates)


df_beta = rolling_beta(df_rtn_stocks['NDX'].values.reshape(-1, 1), df_rtn_stocks['CRM'].values.reshape(-1, 1), df_rtn_stocks.index.values, 255)

Answer 3

为完整性添加更快的numpy解决方案，该计算仅将计算限制为回归系数和最终估计

Numpy滚动回归函数

import numpy as np

def rolling_regression(y, x, window=60):
    """ 
    y and x must be pandas.Series
    """
# === Clean-up ============================================================
    x = x.dropna()
    y = y.dropna()
# === Trim acc to shortest ================================================
    if x.index.size > y.index.size:
        x = x[y.index]
    else:
        y = y[x.index]
# === Verify enough space =================================================
    if x.index.size < window:
        return None
    else:
    # === Add a constant if needed ========================================
        X = x.to_frame()
        X['c'] = 1
    # === Loop... this can be improved ====================================
        estimate_data = []
        for i in range(window, x.index.size+1):
            X_slice = X.values[i-window:i,:] # always index in np as opposed to pandas, much faster
            y_slice = y.values[i-window:i]
            coeff = np.dot(np.dot(np.linalg.inv(np.dot(X_slice.T, X_slice)), X_slice.T), y_slice)
            estimate_data.append(coeff[0] * x.values[window-1] + coeff[1])
    # === Assemble ========================================================
        estimate = pandas.Series(data=estimate_data, index=x.index[window-1:]) 
        return estimate             

---

笔记

在某些特定情况下，只需要回归的最终估计，x.rolling(window=60).apply(my_ols)似乎有点慢

作为提醒，回归系数可以计算为矩阵乘积，您可以在wikipedia's least squares page上阅读。通过numpy的矩阵乘法这种方法可以加速这个过程，而不是使用statsmodels中的ols。该产品以coeff = ...开头的品系表示