[英]Pandas.DataFrame: How to sort rows by the largest value in each row
[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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