Question

我有一个函数，其中包含：

filter_df['returns'][i] = (
    params['const_mean_equation'] /
        (1 - params['AR_mean_equation0'] - params['MA_mean_equation0']))
filter_df['sigma_2'][i] = (
    params['const_variance_equation'] /
        (1 - params['arch0'] - params['garch0']) #unconditional variance)
filter_df['residuals'][i] = y[i] - filter_df['returns'][i]

并通过包含以下内容的字典

{'const_variance_equation': 1,
 'arch0': 2,
 'garch0': 3,
 'const_mean_equation': 4,
 'AR_mean_equation0': 5,
 'MA_mean_equation0': 6}

但是，我希望第一个变量的RHS根据提供的字典是动态的。也就是说，如果我通过这样的字典：

{'const_variance_equation': 1,
 'arch0': 2,
 'garch0': 3,
 'arch1': 2,
 'garch1': 3,
 'const_mean_equation': 4,
 'AR_mean_equation0': 5,
 'MA_mean_equation0': 6
 'AR_mean_equation1': 5,
 'MA_mean_equation1': 6}

该功能自动（动态）更改为：

filter_df['returns'][i] = (
    params['const_mean_equation'] /
        (1 - params['AR_mean_equation0'] - params['MA_mean_equation0'] -
            params['AR_mean_equation1'] - params['MA_mean_equation1']))
filter_df['sigma_2'][i] = (
    params['const_variance_equation'] /
        (1 - params['arch0'] - params['garch0'] - params['arch1'] -
            params['garch1']) #unconditional variance)
filter_df['residuals'][i] = y[i] - filter_df['returns'][i]

我不确定如何自动执行此过程。我曾想过做类似的事情：对于第一个方程(filter_df['returns'][i])，如果params[key]包含字符串'AR_mean'和'MA_mean'，请始终减去-但我不太确定如何继续该过程。使用相同的逻辑，对于sigma_2，如果所有params[key]包含字符串'arch'，则将减去所有df['date']=pd.to_datetime(df['date']) print(df.set_index('date').asfreq('D').ffill().reset_index())。有提示吗？

根据提供的字典动态更改变量的RHS

0 个答案: