Question

我有以下一列（timedelta对象），这是两个时间列之间的差异的结果：

Duration
00:12:38.260000  
00:01:00.750000  
00:19:35.260000  
00:00:29.990000

我正在尝试在此列中应用以下内容：

rolling(min_periods=3, window=5).sum()

我遇到以下错误：

No numeric types to aggregate

我应该转换我的时间吗？怎么样？

Answer 1

简短答案

用.total_seconds()转换为秒，然后求和

长答案

创建您的dataframe和duration列

import pandas as pd

dt1 = ['2019-12-01 10:00:00', '2019-12-01 10:01:00', '2019-12-01 10:00:30', '2019-12-01 10:02:30', '2019-12-01 10:05:30']
dt2 = ['2019-12-01 10:10:00', '2019-12-01 11:06:00', '2019-12-01 10:01:00', '2019-12-01 10:02:30', '2019-12-01 10:07:30']

df = pd.DataFrame({'dt1': dt1, 'dt2': dt2})
df['dt1'] = pd.to_datetime(df['dt1'])
df['dt2'] = pd.to_datetime(df['dt2'])
df['duration'] = df['dt2'] - df['dt1']
df.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 5 entries, 0 to 4
Data columns (total 3 columns):
dt1         5 non-null datetime64[ns]
dt2         5 non-null datetime64[ns]
duration    5 non-null timedelta64[ns]
dtypes: datetime64[ns](2), timedelta64[ns](1)
memory usage: 248.0 bytes

请注意，持续时间为timedelta类型。

现在用.total_seconds()转换为秒，然后求和。

df['duration_rolling_sum'] = pd.to_timedelta(df['duration'].dt.total_seconds().rolling(min_periods=3, window=5).sum(), unit='s')

df

                  dt1                 dt2 duration duration_rolling_sum
0 2019-12-01 10:00:00 2019-12-01 10:10:00 00:10:00                  NaT
1 2019-12-01 10:01:00 2019-12-01 11:06:00 01:05:00                  NaT
2 2019-12-01 10:00:30 2019-12-01 10:01:00 00:00:30             01:15:30
3 2019-12-01 10:02:30 2019-12-01 10:02:30 00:00:00             01:15:30
4 2019-12-01 10:05:30 2019-12-01 10:07:30 00:02:00             01:17:30

滚动总和和时间增量类型出现“没有要聚合的数字类型”错误

1 个答案: