pandas 从时间索引数据框中删除一行
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/16613187/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Deleting a Row from a Time Indexed Dataframe
提问by Markus W
I'm trying to delete a row in a Pandas dataframe by simply passing the date and time.
我试图通过简单地传递日期和时间来删除 Pandas 数据框中的一行。
The dataframe has the following structure:
数据框具有以下结构：
Date_Time Price1 Price2 Price3
2012-01-01 00:00:00 63.05 41.40 68.14
2012-01-01 01:00:00 68.20 42.44 59.64
2012-01-01 02:00:00 61.68 43.18 49.81
I have been trying with df = df.drop('2012-01-01 01:00:00')
我一直在尝试 df = df.drop('2012-01-01 01:00:00')
But I keep getting the following error message:
但我不断收到以下错误消息：
exceptions.ValueError: labels [2012-01-01 01:00:00] not contained in axis
Any help on either deleting the row or just deleting the values would be much appreciated.
任何有关删除行或仅删除值的帮助将不胜感激。
:-)
:-)
回答by Andy Hayden
It looks like you have to actually use the Timestamp rather than the string:
看起来您必须实际使用时间戳而不是字符串：
In [11]: df1
Out[11]:
Price1 Price2 Price3
Date_Time
2012-01-01 00:00:00 63.05 41.40 68.14
2012-01-01 01:00:00 68.20 42.44 59.64
2012-01-01 02:00:00 61.68 43.18 49.81
In [12]: df1.drop(pd.Timestamp('2012-01-01 01:00:00'))
Out[12]:
Price1 Price2 Price3
Date_Time
2012-01-01 00:00:00 63.05 41.40 68.14
2012-01-01 02:00:00 61.68 43.18 49.81
Assuming DateTime is the index, if not use
假设 DateTime 是索引，如果不使用
df1 = df.set_index('Date_Time')
回答by Rens
Alternatively, this works, too:
或者，这也有效：
df1.drop(df1.loc[df1['Date_Time'] == '2012-01-01 01:00:00'].index, inplace=True)
It's also handy when you like to drop a range of observations based on the datetime index. E.g. all observations later than 2012-01-01 01:00:00:
当您想根据日期时间索引删除一系列观察值时，它也很方便。例如所有晚于 2012-01-01 01:00:00 的观察：
df1.drop(df1.loc[df1['Date_Time'] > '2012-01-01 01:00:00'].index, inplace=True)
