我每小时观察几个表现出每日季节性的变量。我想在24小时之前用相应变量的值填充所有缺少的值。
理想情况下，我的函数将从最旧到最新填充缺失的值。因此，如果有25个连续的缺失值，则用与第一个缺失值相同的值填充第25个缺失值。在这种情况下，使用Series.map（）失败。
value desired_output
hour
2019-08-17 00:00:00 58.712986 58.712986
2019-08-17 01:00:00 28.904234 28.904234
2019-08-17 02:00:00 14.275149 14.275149
2019-08-17 03:00:00 58.777087 58.777087
2019-08-17 04:00:00 95.964955 95.964955
2019-08-17 05:00:00 64.971372 64.971372
2019-08-17 06:00:00 95.759469 95.759469
2019-08-17 07:00:00 98.675457 98.675457
2019-08-17 08:00:00 77.510319 77.510319
2019-08-17 09:00:00 56.492446 56.492446
2019-08-17 10:00:00 90.968924 90.968924
2019-08-17 11:00:00 66.647501 66.647501
2019-08-17 12:00:00 7.756725 7.756725
2019-08-17 13:00:00 49.328135 49.328135
2019-08-17 14:00:00 28.634033 28.634033
2019-08-17 15:00:00 65.157161 65.157161
2019-08-17 16:00:00 93.127539 93.127539
2019-08-17 17:00:00 98.806335 98.806335
2019-08-17 18:00:00 94.789761 94.789761
2019-08-17 19:00:00 63.518037 63.518037
2019-08-17 20:00:00 89.524433 89.524433
2019-08-17 21:00:00 48.076081 48.076081
2019-08-17 22:00:00 5.027928 5.027928
2019-08-17 23:00:00 0.417763 0.417763
2019-08-18 00:00:00 29.933627 29.933627
2019-08-18 01:00:00 61.726948 61.726948
2019-08-18 02:00:00 NaN 14.275149
2019-08-18 03:00:00 NaN 58.777087
2019-08-18 04:00:00 NaN 95.964955
2019-08-18 05:00:00 NaN 64.971372
2019-08-18 06:00:00 NaN 95.759469
2019-08-18 07:00:00 NaN 98.675457
2019-08-18 08:00:00 NaN 77.510319
2019-08-18 09:00:00 NaN 56.492446
2019-08-18 10:00:00 NaN 90.968924
2019-08-18 11:00:00 NaN 66.647501
2019-08-18 12:00:00 NaN 7.756725
2019-08-18 13:00:00 NaN 49.328135
2019-08-18 14:00:00 NaN 28.634033
2019-08-18 15:00:00 NaN 65.157161
2019-08-18 16:00:00 NaN 93.127539
2019-08-18 17:00:00 NaN 98.806335
2019-08-18 18:00:00 NaN 94.789761
2019-08-18 19:00:00 NaN 63.518037
2019-08-18 20:00:00 NaN 89.524433
2019-08-18 21:00:00 NaN 48.076081
2019-08-18 22:00:00 NaN 5.027928
2019-08-18 23:00:00 NaN 0.417763
2019-08-19 00:00:00 NaN 29.933627
2019-08-19 01:00:00 NaN 61.726948
2019-08-19 02:00:00 NaN 14.275149
2019-08-19 03:00:00 NaN 58.777087
2019-08-19 04:00:00 NaN 95.964955
2019-08-19 05:00:00 NaN 64.971372
2019-08-19 06:00:00 NaN 95.759469
2019-08-19 07:00:00 NaN 98.675457
2019-08-19 08:00:00 NaN 77.510319
2019-08-19 09:00:00 NaN 56.492446
2019-08-19 10:00:00 NaN 90.968924
2019-08-19 11:00:00 NaN 66.647501
2019-08-19 12:00:00 NaN 7.756725
2019-08-19 13:00:00 61.457913 61.457913
2019-08-19 14:00:00 52.429383 52.429383
2019-08-19 15:00:00 79.016485 79.016485
2019-08-19 16:00:00 77.724758 77.724758
2019-08-19 17:00:00 62.205810 62.205810
2019-08-19 18:00:00 15.841707 15.841707
2019-08-19 19:00:00 72.196028 72.196028
2019-08-19 20:00:00 5.497441 5.497441
2019-08-19 21:00:00 30.737596 30.737596
2019-08-19 22:00:00 65.550690 65.550690
2019-08-19 23:00:00 3.543332 3.543332
import pandas as pd
from dateutil.relativedelta import relativedelta as rel_delta
df['isna'] = df['value'].isna()
df['value'] = df.index.map(lambda t: df.at[t - rel_delta(hours=24), 'value'] if df.at[t,'isna'] and t - rel_delta(hours=24) >= df.index.min() else df.at[t, 'value'])
完成这种幼稚的前向填充的最有效方法是什么？
答案 0 :(得分：3)
IIUC，只需groupby时间和ffill()
df['resuts'] = df.groupby(df.hour.dt.time).value.ffill()
如果hour是您的索引，只需执行df.index.time。
检查：
>>> (df['results'] == df['desired_output']).all()
True
答案 1 :(得分：0)
这行不通吗？
df['value'] = df['value'].fillna(df.index.hour)
答案 2 :(得分：0)
将日期和时间分成两列作为字符串。称为df。
Date Time Value
0 2019-08-17 00:00:00 58.712986
1 2019-08-17 01:00:00 28.904234
2 2019-08-17 02:00:00 14.275149
3 2019-08-17 03:00:00 58.777087
4 2019-08-17 04:00:00 95.964955
然后进行数据重塑，将“时间”（Time）设置为列标题，然后每小时进行一次填充。
（df重塑）
Date 00:00:00 01:00:00 02:00:00 03:00:00 04:00:00
2019-08-17 58.712986 28.904234 14.275149 58.777087 95.964955
2019-08-18 29.933627 61.726948 NaN NaN NaN
2019-08-19 NaN NaN NaN NaN NaN
（df填充）
Date 00:00:00 01:00:00 02:00:00 03:00:00 04:00:00
2019-08-17 58.712986 28.904234 14.275149 58.777087 95.964955
2019-08-18 29.933627 61.726948 14.275149 58.777087 95.964955
2019-08-19 29.933627 61.726948 14.275149 58.777087 95.964955
（代码）
(df.set_index(['Date','Time')['Value']
.unstack()
.ffill()
.stack()
.reset_index(name='Value')