Python-将datetime列转换为秒 [英] Python - Convert datetime column into seconds

问题描述

我有一个日期列(称为时间")，其中包含天/小时/分钟等(timedelta).我已经在数据框中创建了一个新列，我想将时间"列转换为秒，并将其放入每一行的新列中.

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.

有人有指针吗?我在互联网上能找到的就是如何转换您的列，而不是创建新列并转换另一个列.

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.

提前谢谢！

推荐答案

我认为您需要

I think you need total_seconds:

print (df['col'].dt.total_seconds())

示例:

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

如果需要强制转换为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

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