如何摆脱Python中过多的嵌套阵列

Question

嵌套阵列

我想将上述变成以下。当我进行线性回归时，这是偶然发生的，即输出已经在1x1数组中，让我知道您是否想查看更多我的代码。看来我的beta变量是嵌套的问题。

正常阵列

获取输出

一般来说，我只是在尝试从[[array（[x x x） ]），数组（[x]），数组（[x]），阵列（[x]），array（[x]）]]

to
[[x，x，x，x，x]]

def si_model():
    dj_data = pd.read_csv("/data.tsv", sep = "\t")
    dj_data = dj_data.pct_change().dropna()
    ann_dj_data = dj_data * 252
    dj_index = ann_dj_data['^DJI']
    ann_dj_data = ann_dj_data.drop('^DJI', axis='columns')
    
      # Function to Linear Regress Each Stock onto DJ
    def model_regress(stock):
        # Fit DJ to Index Data
        DJ = np.array(dj_index).reshape(len(stock), 1)

        # Regression of each stock onto DJ
        lm = LinearRegression().fit(DJ, y=stock.to_numpy())
        resids = stock.to_numpy() - lm.predict(DJ)

        return lm.coef_, lm.intercept_, resids.std()
    
    # Run model regression on each stock
    lm_all = ann_dj_data.apply(lambda stock: model_regress(stock)).T

    # Table of the Coeffeicents
    lm_all = lm_all.rename(columns={0: 'Beta       ', 1: 'Intercept', 2: 'Rsd Std'})
        
    # Varaince of the index's returns
    dj_index_var = dj_index.std() ** 2
        
    betas = lm_all['Beta       '].to_numpy()
    resid_vars = lm_all['Rsd Std'].to_numpy() ** 2
    
    # Single index approximation of covariance matrix using identity matrix (np.eye)
    Qsi = dj_index_var * betas * betas.reshape(-1, 1) + np.eye(len(betas)) * resid_vars
    return Qsi

# Printing first five rows of approximation
Qsi = si_model()
print("Covariance Matrix")
print(Qsi[:5, :5])

Answer 1

您可以使用 squeeze（） 。

这是一个类似于您的小例子：

import numpy as np

a = np.array([17.1500691])
b = np.array([5.47690856])
c = np.array([5.47690856])
d = np.array([11.7700696])

e = list([[a,b],[c,d]])
print(e)

f = np.squeeze(np.array(e), axis=2)
print(f)

输出：

[[array([17.1500691]), array([5.47690856])], [array([5.47690856]), array([11.7700696])]]
[[17.1500691   5.47690856]
 [ 5.47690856 11.7700696 ]]