0

I've got a Series of numpy arrays:

import pandas as pd
import numpy as np
pd.Series({10: np.array([[0.72260683, 0.27739317, 0.        ],
                         [0.7187053 , 0.2812947 , 0.        ],
                         [0.71435467, 0.28564533, 1.        ],
                         [0.3268072 , 0.6731928 , 0.        ],
                         [0.31941951, 0.68058049, 1.        ],
                         [0.31260015, 0.68739985, 0.        ]]), 
           20: np.array([[0.7022099 , 0.2977901 , 0.        ],
                         [0.6983866 , 0.3016134 , 0.        ],
                         [0.69411673, 0.30588327, 1.        ],
                         [0.33857735, 0.66142265, 0.        ],
                         [0.33244109, 0.66755891, 1.        ],
                         [0.32675582, 0.67324418, 0.        ]]), 
           38: np.array([[0.68811957, 0.31188043, 0.        ],
                         [0.68425783, 0.31574217, 0.        ],
                         [0.67994496, 0.32005504, 1.        ],
                         [0.34872593, 0.65127407, 0.        ],
                         [0.34276171, 0.65723829, 1.        ],
                         [0.33722803, 0.66277197, 0.        ]])}
)

and an array of indices np.array([1, 4, 1]) indicating which rows should be filtered from consecutive arrays. The expected output would be like this:

pd.Series({10: np.array([[0.7187053 , 0.2812947 , 0.        ]]), 
           20: np.array([[0.33244109, 0.66755891, 1.        ]]), 
           38: np.array([[0.68425783, 0.31574217, 0.        ]])}
)

How can I do that? How it would be different if I would like to take out third element from each resulting array obtaining the following Series?

pd.Series({10: 0, 20: 1, 30: 0})
jakes
  • 1,964
  • 3
  • 18
  • 50

1 Answers1

3

If possible convert Series of 2d arrays to 3d array (same length)s of each 2d arrays:

a = np.array([1, 4, 1])

b = np.array(s.tolist())[np.arange(len(s)), a, 2]
print (b)
[0. 1. 0.]

c = pd.Series(b, index=s.index)
print (c)
10    0.0
20    1.0
38    0.0
dtype: float64

If want select by array of indices:

b1 = np.array(s.tolist())[np.arange(len(s)), a]
print (b1)

[[0.7187053  0.2812947  0.        ]
 [0.33244109 0.66755891 1.        ]
 [0.68425783 0.31574217 0.        ]]
jezrael
  • 822,522
  • 95
  • 1,334
  • 1,252
  • What's the difference between using `np.arange(len(s))` on first dimenstion instead of `:`? – jakes Apr 05 '20 at 13:03
  • @jakes - Difference is need `np.arrange` for select by array of indices, [like this solution, only for 1d array](https://stackoverflow.com/questions/20103779/index-2d-numpy-array-by-a-2d-array-of-indices-without-loops). – jezrael Apr 05 '20 at 13:10