如何在statsmodel ols回归中包括滞后变量

如何解决如何在statsmodel ols回归中包括滞后变量

是否可以在statsmodel ols回归中指定滞后自变量？这是下面的示例数据框和ols模型规范。我想在模型中包含一个滞后变量。

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

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()

 

解决方法

我认为您不能在公式中即时调用它。也许使用shift方法？弄清楚这是否不是您所需要的

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()

,

这已经有一个可接受的答案，但要加上我的 2 美分：

• 在移动之前验证索引是一种很好的做法（否则您的滞后可能不是您认为的那样）
• 可以定义一个可以在公式中多处重复使用的函数

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": [4,8]})


df.index = pd.PeriodIndex(
    year=[2000,2000,2001],quarter=[1,4,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()