python - Pandas 成对算术类似于rolling().corr()

Question

我有一个数据框如下:

fsym                            EOS       BTC       BNB
time                                                   
2018-11-30 00:00:00+00:00 -0.051903 -0.069088 -0.058162
2018-12-01 00:00:00+00:00  0.026936  0.044739  0.040303
2018-12-02 00:00:00+00:00 -0.034843 -0.012935 -0.005900
2018-12-03 00:00:00+00:00 -0.108108 -0.070375 -0.028180
2018-12-04 00:00:00+00:00 -0.048583  0.019509  0.131986

我可以简单地计算列成对相关性:

pt = pt.rolling(3).corr()

产生:

sym                                 EOS       BTC       BNB
time                      fsym                              
2018-11-30 00:00:00+00:00 EOS        NaN       NaN       NaN
                          BTC        NaN       NaN       NaN
                          BNB        NaN       NaN       NaN
2018-12-01 00:00:00+00:00 EOS        NaN       NaN       NaN
                          BTC        NaN       NaN       NaN
                          BNB        NaN       NaN       NaN
2018-12-02 00:00:00+00:00 EOS   1.000000  0.952709  0.938688
                          BTC   0.952709  1.000000  0.999066
                          BNB   0.938688  0.999066  1.000000
2018-12-03 00:00:00+00:00 EOS   1.000000  0.998738  0.969385
                          BTC   0.998738  1.000000  0.980492
                          BNB   0.969385  0.980492  1.000000
...

如何类似地计算数据帧的成对差异？我猜这相当于使用滚动窗口 1。

编辑:正如评论中所指出的，上面的示例实际上并不是我没有注意到的列相关性。

Answer 1

如果您想要 9 列:

# test data
df = pd.DataFrame(np.arange(12).reshape(-1,3), columns=list('abc'))

s = df.values
new_cols = pd.MultiIndex.from_product([df.columns, df.columns])

pd.DataFrame((s[:,None,:] - s[:, :,  None]).reshape(len(df), -1),
             index=df.index,
             columns=new_cols)

输出:

   a        b        c      
   a  b  c  a  b  c  a  b  c
0  0  1  2 -1  0  1 -2 -1  0
1  0  1  2 -1  0  1 -2 -1  0
2  0  1  2 -1  0  1 -2 -1  0
3  0  1  2 -1  0  1 -2 -1  0