pandas 如何计算两个熊猫列之间的时间差
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How to calculate time difference between two pandas column
提问by pyd
My df looks like,
我的 df 看起来像，
start stop
0 2015-11-04 10:12:00 2015-11-06 06:38:00
1 2015-11-04 10:23:00 2015-11-05 08:30:00
2 2015-11-04 14:01:00 2015-11-17 10:34:00
4 2015-11-19 01:43:00 2015-12-21 09:04:00
print(time_df.dtypes)
start datetime64[ns]
stop datetime64[ns]
dtype: object
数据类型：对象
I am trying to find the time difference between, stop and start.
我试图找到停止和开始之间的时差。
I tried, pd.Timedelta(df_time['stop']-df_time['start'])but it gives TypeError: data type "datetime" not understood
我试过了，pd.Timedelta(df_time['stop']-df_time['start'])但它给了TypeError: data type "datetime" not understood
df_time['stop']-df_time['start']also gives same error.
df_time['stop']-df_time['start']也给出同样的错误。
My expected output,
我的预期输出，
2D,?H
1D,?H
...
...
回答by jezrael
You need omit pd.Timedelta, because difference of times return timedeltas:
您需要 omit pd.Timedelta，因为时间差异返回 timedeltas：
df_time['td'] = df_time['stop']-df_time['start']
print (df_time)
start stop td
0 2015-11-04 10:12:00 2015-11-06 06:38:00 1 days 20:26:00
1 2015-11-04 10:23:00 2015-11-05 08:30:00 0 days 22:07:00
2 2015-11-04 14:01:00 2015-11-17 10:34:00 12 days 20:33:00
EDIT: Another solution is subtract numpy arrays:
编辑：另一个解决方案是减去 numpy 数组：
df_time['td'] = df_time['stop'].values - df_time['start'].values
print (df_time)
start stop td
0 2015-11-04 10:12:00 2015-11-06 06:38:00 1 days 20:26:00
1 2015-11-04 10:23:00 2015-11-05 08:30:00 0 days 22:07:00
2 2015-11-04 14:01:00 2015-11-17 10:34:00 12 days 20:33:00
回答by Treizh
First make sure that you have dates in your column
首先确保您的列中有日期
data.loc[:, 'start'] = pd.to_datetime(data.loc[:, 'start'])
data.loc[:, 'stop'] = pd.to_datetime(data.loc[:, 'stop'])
Then substract
然后减去
data['delta'] = data['stop'] - data['start']
