我有以下的DF
lst = [[1548828606206000000, 1548840373139000000],
[1548841285708000000, 1548841458405000000],
[1548842198276000000, 1548843109519000000],
[1548844022821000000, 1548844934207000000],
[1548845431090000000, 1548845539219000000],
[1548845555332000000, 1548845846621000000],
[1548847176147000000, 1548851020030000000],
[1548851704053000000, 1548852256143000000],
[1548852436514000000, 1548855900767000000],
[1548856817770000000, 1548857162183000000],
[1548858736931000000, 1548858979032000000]]
df = pd.DataFrame(lst,columns =['start','end'])
df['start'] = pd.to_datetime(df['start'])
df['end'] = pd.to_datetime(df['end'])
在我的虚拟DF然后6小时应为60分钟(每小时最大) - 0时10分06秒=○点49分54秒。对于第七和第八应该是每个1:00:00作为结束时间是9时26分十三秒。对于9号应该是零点26分13秒加上所有的在下面.rows与第九小时重叠09:44间隔 - 09:41 = 3mins和60分钟-00:56 = 4分钟。因此总为第九应26+ 3 + 4〜= 00:32:28
我最初的apporach是合并的开始和结束,加上虚拟点每3排,上采样到1S,得到行之间的区别,概括起来只有实际行。必须有这样做的更pythonic的方法。任何暗示将是巨大的。
IIUC,是这样的:
df.apply(lambda x: pd.to_timedelta(pd.Series(1, index=pd.date_range(x.start, x.end, freq='S'))
.groupby(pd.Grouper(freq='H')).count(), unit='S'), axis=1).sum()
输出:
2019-01-30 06:00:00 00:49:54
2019-01-30 07:00:00 01:00:00
2019-01-30 08:00:00 01:00:00
2019-01-30 09:00:00 00:32:28
2019-01-30 10:00:00 00:33:43
2019-01-30 11:00:00 00:40:24
2019-01-30 12:00:00 00:45:37
2019-01-30 13:00:00 00:45:01
2019-01-30 14:00:00 00:09:48
Freq: H, dtype: timedelta64[ns]
或者把它降到小时,尝试:
df.apply(lambda r: pd.to_timedelta(pd.Series(1, index=pd.date_range(r.start, r.end, freq='S'))
.pipe(lambda x: x.groupby(x.index.hour).count()), unit='S'), axis=1)\
.sum()
输出:
6 00:49:54
7 01:00:00
8 01:00:00
9 00:32:28
10 00:33:43
11 00:40:24
12 00:45:37
13 00:45:01
14 00:09:48
dtype: timedelta64[ns]