Python 如何在STATSOLS回归模型中包含滞后变量

在STATSOLS回归模型中是否有指定滞后自变量的方法？下面是一个示例数据帧和ols模型规范。我想在模型中加入一个滞后变量

df = pd.DataFrame({
                   "y": [2,3,7,8,1],
                   "x": [8,6,2,1,9],
                   "v": [4,3,1,3,8]
                 })

Current model:

model = sm.ols(formula = 'y ~ x + v', data=df).fit()

Desired model:

model_lag = sm.ols(formula = 'y ~ (x-1) + v', data=df).fit()

 

---

我不认为你可以在公式中称之为即时。也许用这个方法？如果这不是您需要的，请务必澄清

import statsmodels.api as sm
df['xlag'] = df['x'].shift()
df

   y  x  v  xlag
0  2  8  4   NaN
1  3  6  3   8.0
2  7  2  1   6.0
3  8  1  3   2.0
4  1  9  8   1.0

sm.formula.ols(formula = 'y ~ xlag + v', data=df).fit()

---

这已经有了一个公认的答案，但再加上我的2美分：

• 最好在换档前验证该指数（否则您的滞后可能不是您所认为的）
• 可以定义一个函数，该函数在公式中的许多地方都是可重用的

一些示例代码：

import numpy as np
import pandas as pd
import statsmodels.api as sm
import statsmodels.formula.api as smf

df = pd.DataFrame({"y": [2, 3, 7, 8, 1], "x": [8, 6, 2, 1, 9], "v": [4, 3, 1, 3, 8]})


df.index = pd.PeriodIndex(
    year=[2000, 2000, 2000, 2000, 2001], quarter=[1, 2, 3, 4, 1], freq="Q", name="period"
)


def lag(x, n, validate=True):
    """Calculates the lag of a pandas Series

    Args:
        x (pd.Series): the data to lag
        n (int): How many periods to go back (lag length)
        validate (bool, optional): Validate the series index (monotonic increasing + no gaps + no duplicates). 
                                If specified, expect the index to be a pandas PeriodIndex
                                Defaults to True.

    Returns:
        pd.Series: pd.Series.shift(n) -- lagged series
    """

    if n == 0:
        return x

    if isinstance(x, pd.Series):
        if validate:
            assert x.index.is_monotonic_increasing, (
                "\u274c" + f"x.index is not monotonic_increasing"
            )
            assert x.index.is_unique, "\u274c" + f"x.index is not unique"
            idx_full = pd.period_range(start=x.index.min(), end=x.index.max(), freq=x.index.freq)
            assert np.all(x.index == idx_full), "\u274c" + f"Gaps found in x.index"
        return x.shift(n)

    return x.shift(n)


# Manually create lag as variable:
df["x_1"] = df["x"].shift(1)
smf.ols(formula="y ~ x_1 + v", data=df).fit().summary()


# Use the defined function in the formula:
smf.ols(formula="y ~ lag(x,1) + v", data=df).fit().summary()

# ... can use in multiple places too:
smf.ols(formula="y ~ lag(x,1) + lag(v, 2)", data=df).fit().summary()