pandas 将熊猫日期列转换为经过的秒数
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Converting pandas date column into seconds elapsed
提问by Fanylion
I have a pandas dataframe of multiple columns with a column of datetime64[ns] data. Time is in HH:MM:SS format. How can I convert this column of dates into a column of seconds elapsed? Like if the time said 10:00:00 in seconds that would be 36000. The seconds should be in a float64 type format.
我有一个多列的Pandas数据框，其中有一列 datetime64[ns] 数据。时间采用 HH:MM:SS 格式。如何将这一列日期转换为一列经过的秒数？就像如果时间以秒为单位表示 10:00:00 就是 36000。秒应该是 float64 类型的格式。
Example data column
示例数据列
回答by piRSquared
New Answer
Convert your text to Timedelta
新答案
将您的文本转换为Timedelta
df['Origin Time(Local)'] = pd.to_timedelta(df['Origin Time(Local)'])
df['Seconds'] = df['Origin Time(Local)'].dt.total_seconds()
Old Answer
旧答案
Consider the dataframe df
考虑数据框 df
df = pd.DataFrame(dict(Date=pd.date_range('2017-03-01', '2017-03-02', freq='2H')))
Date
0 2017-03-01 00:00:00
1 2017-03-01 02:00:00
2 2017-03-01 04:00:00
3 2017-03-01 06:00:00
4 2017-03-01 08:00:00
5 2017-03-01 10:00:00
6 2017-03-01 12:00:00
7 2017-03-01 14:00:00
8 2017-03-01 16:00:00
9 2017-03-01 18:00:00
10 2017-03-01 20:00:00
11 2017-03-01 22:00:00
12 2017-03-02 00:00:00
Subtract the most recent day from the timestamps and use total_seconds. total_secondsis an attribute of a Timedelta. We get a series of Timedeltasby taking the difference between two series of Timestamps.
从时间戳中减去最近的一天并使用total_seconds. total_seconds是 a 的一个属性Timedelta。我们Timedeltas通过取 的两个系列之间的差异得到一系列的Timestamps。
(df.Date - df.Date.dt.floor('D')).dt.total_seconds()
# equivalent to
# (df.Date - pd.to_datetime(df.Date.dt.date)).dt.total_seconds()
0 0.0
1 7200.0
2 14400.0
3 21600.0
4 28800.0
5 36000.0
6 43200.0
7 50400.0
8 57600.0
9 64800.0
10 72000.0
11 79200.0
12 0.0
Name: Date, dtype: float64
Put it in a new column
把它放在一个新的列
df.assign(seconds=(df.Date - df.Date.dt.floor('D')).dt.total_seconds())
Date seconds
0 2017-03-01 00:00:00 0.0
1 2017-03-01 02:00:00 7200.0
2 2017-03-01 04:00:00 14400.0
3 2017-03-01 06:00:00 21600.0
4 2017-03-01 08:00:00 28800.0
5 2017-03-01 10:00:00 36000.0
6 2017-03-01 12:00:00 43200.0
7 2017-03-01 14:00:00 50400.0
8 2017-03-01 16:00:00 57600.0
9 2017-03-01 18:00:00 64800.0
10 2017-03-01 20:00:00 72000.0
11 2017-03-01 22:00:00 79200.0
12 2017-03-02 00:00:00 0.0
回答by FdMon
it would work:
它会起作用：
df['time'].dt.total_seconds()
regards
问候
