大熊貓無法從numpy的數組的時間戳

Question

創建一個數據幀我有大熊貓時間戳的numpy的數組：

array([[Timestamp('2016-05-02 15:50:00+0000', tz='UTC', offset='5T'), 
     Timestamp('2016-05-02 15:50:00+0000', tz='UTC', offset='5T'), 
     Timestamp('2016-05-02 15:50:00+0000', tz='UTC', offset='5T')], 
     [Timestamp('2016-05-02 17:10:00+0000', tz='UTC', offset='5T'), 
     Timestamp('2016-05-02 17:10:00+0000', tz='UTC', offset='5T'), 
     Timestamp('2016-05-02 17:10:00+0000', tz='UTC', offset='5T')], 
     [Timestamp('2016-05-02 20:25:00+0000', tz='UTC', offset='5T'), 
     Timestamp('2016-05-02 20:25:00+0000', tz='UTC', offset='5T'), 
     Timestamp('2016-05-02 20:25:00+0000', tz='UTC', offset='5T')]], dtype=object) 

我不能創建此數組一個數據幀，作爲試圖這樣做將引發以下錯誤：

AssertionError: Number of Block dimensions (1) must equal number of axes (2) 

你可以看到數組顯然是二維的，我使用ndim進行了驗證。

爲什麼我無法創建DataFrame？

Answer 1

我認爲你可以使用list理解：

import pandas as pd 
import numpy as np 

a =np.array([[pd.Timestamp('2016-05-02 15:50:00+0000', tz='UTC', offset='5T'), 
     pd.Timestamp('2016-05-02 15:50:00+0000', tz='UTC', offset='5T'), 
     pd.Timestamp('2016-05-02 15:50:00+0000', tz='UTC', offset='5T')], 
     [pd.Timestamp('2016-05-02 17:10:00+0000', tz='UTC', offset='5T'), 
     pd.Timestamp('2016-05-02 17:10:00+0000', tz='UTC', offset='5T'), 
     pd.Timestamp('2016-05-02 17:10:00+0000', tz='UTC', offset='5T')], 
     [pd.Timestamp('2016-05-02 20:25:00+0000', tz='UTC', offset='5T'), 
     pd.Timestamp('2016-05-02 20:25:00+0000', tz='UTC', offset='5T'), 
     pd.Timestamp('2016-05-02 20:25:00+0000', tz='UTC', offset='5T')]], dtype=object) 

df = pd.DataFrame([x for x in a], columns=['a','b','c']) 
print (df) 
          a       b \ 
0 2016-05-02 15:50:00+00:00 2016-05-02 15:50:00+00:00 
1 2016-05-02 17:10:00+00:00 2016-05-02 17:10:00+00:00 
2 2016-05-02 20:25:00+00:00 2016-05-02 20:25:00+00:00 

          c 
0 2016-05-02 15:50:00+00:00 
1 2016-05-02 17:10:00+00:00 
2 2016-05-02 20:25:00+00:00 

另一種解決方案是DataFrame.from_records：

print (pd.DataFrame.from_records(a, columns=['a','b','c'])) 
          a       b \ 
0 2016-05-02 15:50:00+00:00 2016-05-02 15:50:00+00:00 
1 2016-05-02 17:10:00+00:00 2016-05-02 17:10:00+00:00 
2 2016-05-02 20:25:00+00:00 2016-05-02 20:25:00+00:00 

          c 
0 2016-05-02 15:50:00+00:00 
1 2016-05-02 17:10:00+00:00 
2 2016-05-02 20:25:00+00:00 

見alternate constructors of df。