Question

我正在寻找在Python中构建一个函数，该函数使用以下等式创建一个简单的OLS回归：

 Y_i - Y_i-1 = A + B(X_i - X_i-1) + E

换句话说，Y_Lag = alpha + beta（X_Lag）+错误项

当前，我有以下数据集（这是一个简短版本）

注意：Y =历史汇率

df = pd.DataFrame(np.random.randint(low=0, high=10, size=(5, 5)), columns=['Historic_Rate', 'Overnight', '1M', '3M', '6M'])

所以，我要构建的是迭代地将X变量放入一个简单的线性回归中，到目前为止，我已构建的代码如下：

#Start the iteration process for the regression to in turn fit 1 parameter

#Import required packages 

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

#Import dataset

df = pd.DataFrame(np.random.randint(low=0, high=10, size=(5, 5)), columns=['Historic_Rate', 'Overnight', '1M', '3M', '6M'])
#Y_Lag is always 1 time period only

df['Y_Lag'] = df['Historic_Rate'].shift(1)

#Begin the process with 1 lag, taking one x variable in turn

array = df[0:0]
array.drop(array.columns[[0,5]], axis=1, inplace=True)
for X in array:
    df['X_Lag'] = df['X'].shift(1)
    Model = df[df.columns[4:5]]
    Y = Model['Y_Lag']
    X = Model['X_Lag']

    Reg_model = sm.OLS(Y,X).fit()
    predictions = model.predict(X) 
    # make the predictions by the model

    # Print out the statistics
    model.summary()

因此，从本质上讲，我正在寻找创建列标题的列表，这些列表标题又将系统地遍历我的循环，每个变量将被滞后，然后针对滞后的Y变量进行回归。

我也很感谢有关如何输出模型的知识。X，其中X是数组的第X次迭代，用于动态命名变量。

Answer 1

您接近了，我想您只是在循环中将变量X与字符串'X'混淆了。我还认为您不是在计算Y_i - Y_i-1，而是将Y_i-1与X_i-1进行回归。

这是循环遍历回归的方式。我们还将使用字典来存储回归结果，并以键作为列名。

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

df = pd.DataFrame(np.random.randint(low=0, high=10, size=(5, 5)), 
                  columns=['Historic_Rate', 'Overnight', '1M', '3M', '6M'])

fit_d = {}  # This will hold all of the fit results and summaries
for col in [x for x in df.columns if x != 'Historic_Rate']:
    Y = df['Historic_Rate'] - df['Historic_Rate'].shift(1)
    # Need to remove the NaN for fit
    Y = Y[Y.notnull()]

    X = df[col] - df[col].shift(1)
    X = X[X.notnull()]

    X = sm.add_constant(X)  # Add a constant to the fit

    fit_d[col] = sm.OLS(Y,X).fit()

现在，如果您想做出一些预测，例如对于您的最后一个模型，则可以执行以下操作：

fit_d['6M'].predict(sm.add_constant(df['6M']-df['6M'].shift(1)))
#0    NaN
#1    0.5
#2   -2.0
#3   -1.0
#4   -0.5
#dtype: float64

您可以获得摘要：fit_d['6M'].summary()

                            OLS Regression Results                            
==============================================================================
Dep. Variable:          Historic_Rate   R-squared:                       0.101
Model:                            OLS   Adj. R-squared:                 -0.348
Method:                 Least Squares   F-statistic:                    0.2254
Date:                Thu, 27 Sep 2018   Prob (F-statistic):              0.682
Time:                        11:27:33   Log-Likelihood:                -9.6826
No. Observations:                   4   AIC:                             23.37
Df Residuals:                       2   BIC:                             22.14
Df Model:                           1                                         
Covariance Type:            nonrobust                                         
==============================================================================
                 coef    std err          t      P>|t|      [0.025      0.975]
------------------------------------------------------------------------------
const         -0.4332      1.931     -0.224      0.843      -8.740       7.873
6M            -0.2674      0.563     -0.475      0.682      -2.691       2.156
==============================================================================
Omnibus:                          nan   Durbin-Watson:                   2.301
Prob(Omnibus):                    nan   Jarque-Bera (JB):                0.254
Skew:                          -0.099   Prob(JB):                        0.881
Kurtosis:                       1.781   Cond. No.                         3.44
==============================================================================

Python：为简单的OLS

1 个答案: