Question

在R中，可以像

一样执行多元线性回归

temp = lm(log(volume_1[11:62])~log(price_1[11:62])+log(volume_1[10:61]))

在Python中，可以使用 R样式公式，所以我认为下面的代码也应该工作，

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

rando = lambda x: np.random.randint(low=1, high=100, size=x)

df = pd.DataFrame(data={'volume_1': rando(62), 'price_1': rando(62)})

temp = smf.ols(formula='np.log(volume_1)[11:62] ~ np.log(price_1)[11:62] + np.log(volume_1)[10:61]', 
               data=df) 
# np.log(volume_1)[10:61] express the lagged volume

但是我得到了错误

PatsyError: Number of rows mismatch between data argument and volume_1[11:62] (62 versus 51)
volume_1[11:62] ~ price_1[11:62] + volume_1[10:61]

我想不可能只对列中的部分行进行回归，因为data = df有62行，其他变量有51行。

有没有像R那样方便的回归方法？

df类型为pandas Dataframe，列名称为volume_1，price_1

Answer 1

使用patsy存储库中github question中的示例，这将是使滞后列正常工作的方法。

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

rando = lambda x: np.random.randint(low=1, high=100, size=x)

df = pd.DataFrame(data={'volume_1': rando(62), 'price_1': rando(62)})

def lag(x, n):
    if n == 0:
        return x
    if isinstance(x,pd.Series):
        return x.shift(n)

    x = x.astype('float')
    x[n:] = x[0:-n]
    x[:n] = np.nan
    return x

temp = smf.ols(formula='np.log(volume_1) ~ np.log(price_1) + np.log(lag(volume_1,1))', 
               data=df[11:62])

具有Python statsmodel的多元线性回归

1 个答案: