<p>这已经有了一个公认的答案,但再加上我的2美分:</p>
<ul>
<li>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)</li>
<li>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</li>
</ul>
<p>一些示例代码:</p>
<pre><code>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()
</code></pre>