Python - 将日期时间列转换为秒
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Python - Convert datetime column into seconds
提问by ScoutEU
I have a date column (called 'Time') which contains days/hours/mins etc (timedelta). I have created a new column in my dataframe and I want to convert the 'Time' column into seconds and put it in the new column for each row.
我有一个日期列（称为“时间”），其中包含天/小时/分钟等（timedelta）。我在我的数据框中创建了一个新列，我想将“时间”列转换为秒并将其放入每行的新列中。
Does anyone have any pointers? All I can find on the internet is how to convert your column, not create a new column and convert another one.
有没有人有任何指示？我在互联网上所能找到的只是如何转换您的列，而不是创建一个新列并转换另一个列。
Thank you in advance!
先感谢您！
回答by jezrael
I think you need total_seconds:
我认为你需要total_seconds：
print (df['col'].dt.total_seconds())
Sample:
样本：
df = pd.DataFrame({'date1':pd.date_range('2015-01-01', periods=3),
'date2':pd.date_range('2015-01-01 02:00:00', periods=3, freq='23H')})
print (df)
date1 date2
0 2015-01-01 2015-01-01 02:00:00
1 2015-01-02 2015-01-02 01:00:00
2 2015-01-03 2015-01-03 00:00:00
df['diff'] = df['date2'] - df['date1']
df['seconds'] = df['diff'].dt.total_seconds()
print (df)
date1 date2 diff seconds
0 2015-01-01 2015-01-01 02:00:00 02:00:00 7200.0
1 2015-01-02 2015-01-02 01:00:00 01:00:00 3600.0
2 2015-01-03 2015-01-03 00:00:00 00:00:00 0.0
df['diff'] = df['date2'] - df['date1']
df['diff'] = df['diff'].dt.total_seconds()
print (df)
date1 date2 diff
0 2015-01-01 2015-01-01 02:00:00 7200.0
1 2015-01-02 2015-01-02 01:00:00 3600.0
2 2015-01-03 2015-01-03 00:00:00 0.0
If need cast to int:
如果需要转换为int：
df['diff'] = df['date2'] - df['date1']
df['diff'] = df['diff'].dt.total_seconds().astype(int)
print (df)
date1 date2 diff
0 2015-01-01 2015-01-01 02:00:00 7200
1 2015-01-02 2015-01-02 01:00:00 3600
2 2015-01-03 2015-01-03 00:00:00 0
回答by Manuel Salgueiro
Let's assume your DataFrame's name is df.
让我们假设您的 DataFrame 的名称是df.
If you want to create a new column with the seconds you should do the following:
如果你想用秒创建一个新列，你应该执行以下操作：
df['newColumn'] = df['Time'].dt.total_seconds()
