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pandas.DataFrame - 获取每一行的 UID?

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

使用yfinance.download()后,我得到一个pandas.DataFrame object,打印时输出为:

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]

我称它为 UID,因为它是一个唯一标识符(即每一行都是特殊的)并且它不会被归类为一列。

如何访问“日期”列中的数据?

如何访问“日期”列中的数据?

首先重置索引,并将索引转换为常规列:

df.reset_index()['Date']

或者,将Date列保留为索引并简单地获取索引列:

df.index

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