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pandas.DataFrame - 獲取每一行的 UID?

[英]pandas.DataFrame - get the UID of each row?

使用yfinance.download()后,我得到一個pandas.DataFrame object,打印時輸出為:

Date                                                                                    
2018-01-02  18700.199219  18700.199219  18700.199219  18700.199219  18700.199219       0
2018-01-03  18953.099609  18953.099609  18953.099609  18953.099609  18953.099609       0
2018-01-04  18922.500000  18922.500000  18922.500000  18922.500000  18922.500000       0
2018-01-05  18849.800781  18849.800781  18849.800781  18849.800781  18849.800781       0
2018-01-08  18911.900391  18911.900391  18911.900391  18911.900391  18911.900391       0
...                  ...           ...           ...           ...           ...     ...
2022-03-04  23165.710938  23165.710938  23165.710938  23165.710938  23165.710938       0
2022-03-07  23148.070312  23148.070312  23148.070312  23148.070312  23148.070312       0
2022-03-08  23026.910156  23026.910156  23026.910156  23026.910156  23026.910156       0
2022-03-09  22659.669922  22659.669922  22659.669922  22659.669922  22659.669922       0
2022-03-10  22437.609375  22437.609375  22437.609375  22437.609375  22437.609375       0

[1061 rows x 6 columns]

我稱它為 UID,因為它是一個唯一標識符(即每一行都是特殊的)並且它不會被歸類為一列。

如何訪問“日期”列中的數據?

如何訪問“日期”列中的數據?

首先重置索引,並將索引轉換為常規列:

df.reset_index()['Date']

或者,將Date列保留為索引並簡單地獲取索引列:

df.index

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