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[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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