`没有要聚合的数值类型`滚动求和和时间增量类型错误
`没有要聚合的数值类型`滚动求和和时间增量类型错误
提问于 2019-12-24 11:59:26
我有以下列(timedelta对象)，它是两个时间列之间差异的结果：
Duration
00:12:38.260000
00:01:00.750000
00:19:35.260000
00:00:29.990000 我尝试在本专栏中应用以下内容：
rolling(min_periods=3, window=5).sum()我有以下错误：
No numeric types to aggregate
我应该转换我的工期吗？多么?
回答 1
Stack Overflow用户
发布于 2019-12-24 14:55:14
short answer
使用.total_seconds()转换为秒，然后使用sum
long answer
创建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，持续时间的类型为timedelta。
现在使用.total_seconds()转换为秒，然后使用sum。
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/59463672
复制相关文章
相似问题
腾讯云开发者
