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:对象
我正试图找出"停止"one_answers"开始"之间的时间差。
我试过了,pd.Timedelta(df_time['stop']-df_time['start'])
但它给出了TypeError: data type "datetime" not understood
CCD_ 3也给出了同样的错误。
我的预期输出
2D,?H
1D,?H
...
...
您需要省略pd.Timedelta
,因为时差返回时间增量:
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
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
首先确保列中有日期
data.loc[:, 'start'] = pd.to_datetime(data.loc[:, 'start'])
data.loc[:, 'stop'] = pd.to_datetime(data.loc[:, 'stop'])
然后减去
data['delta'] = data['stop'] - data['start']