熊猫分组小时并计算缺货时间
我的时间序列如下：
| datetime_create | quantity_old | quantity_new | quantity_diff | is_stockout |
| 2018-02-15 08:12:54.289 | 16 | 15 | -1 | False |
| 2018-02-15 08:14:10.619 | 15 | 13 | -2 | False |
| 2018-02-15 08:49:15.962 | 13 | 9 | -4 | False |
| 2018-02-15 08:51:04.740 | 9 | 8 | -1 | False |
| 2018-02-15 08:56:37.086 | 8 | 7 | -1 | False |
| 2018-02-15 09:23:22.858 | 7 | 5 | -2 | False |
| 2018-02-15 10:16:50.324 | 5 | 4 | -1 | False |
| 2018-02-15 10:19:25.071 | 4 | 3 | -1 | False |
| 2018-02-15 10:33:22.788 | 3 | 2 | -1 | False |
| 2018-02-15 10:33:34.125 | 2 | 0 | -2 | True |
| 2018-02-15 16:45:24.747 | 0 | 1 | 1 | False |
| 2018-02-15 16:48:29.996 | 1 | 0 | -1 | True |
| 2018-02-17 10:42:58.325 | 0 | 42 | 42 | False |
| 2018-02-17 10:47:07.380 | 42 | 41 | -1 | False |
| 2018-02-17 11:42:31.008 | 41 | 40 | -1 | False |
| 2018-02-17 11:48:31.070 | 40 | 39 | -1 | False |
| 2018-02-17 12:39:13.681 | 39 | 38 | -1 | False |
| 2018-02-17 12:48:00.286 | 38 | 37 | -1 | False |
| 2018-02-17 12:56:59.203 | 37 | 36 | -1 | False |
| 2018-02-17 13:18:12.285 | 36 | 35 | -1 | False |
| 2018-02-17 13:29:53.465 | 35 | 34 | -1 | False |
| 2018-02-17 14:54:55.810 | 34 | 33 | -1 | False |
| 2018-02-17 15:53:38.816 | 33 | 32 | -1 | False |
| 2018-02-17 16:28:08.076 | 32 | 31 | -1 | False |
| 2018-02-17 16:45:18.965 | 31 | 30 | -1 | False |
| 2018-02-17 16:59:11.111 | 30 | 29 | -1 | False |
| 2018-02-17 17:18:53.646 | 29 | 27 | -2 | False |
| 2018-02-17 17:44:43.508 | 27 | 26 | -1 | False |
| 2018-02-17 19:34:49.701 | 26 | 25 | -1 | False |
| 2018-02-17 20:49:00.205 | 25 | 24 | -1 | False |
| 2018-02-18 07:14:22.207 | 24 | 22 | -2 | False |
| 2018-02-18 08:35:41.560 | 22 | 20 | -2 | False |
| 2018-02-18 10:22:18.825 | 20 | 19 | -1 | False |
| 2018-02-18 10:28:33.909 | 19 | 18 | -1 | False |
| 2018-02-18 10:37:30.427 | 18 | 17 | -1 | False |
| 2018-02-18 10:50:55.265 | 17 | 16 | -1 | False |
| 2018-02-18 11:17:53.359 | 16 | 15 | -1 | False |
| 2018-02-18 11:42:29.214 | 0 | 30 | 30 | False |
| 2018-02-18 11:58:19.113 | 15 | 14 | -1 | False |
| 2018-02-18 11:58:56.432 | 14 | 13 | -1 | False |
| 2018-02-18 12:06:48.438 | 13 | 12 | -1 | False |
| 2018-02-18 12:21:43.634 | 12 | 11 | -1 | False |
| 2018-02-18 12:44:46.288 | 11 | 9 | -2 | False |
| 2018-02-18 13:26:01.952 | 9 | 8 | -1 | False |
| 2018-02-18 13:26:40.940 | 8 | 9 | 1 | False |
| 2018-02-18 13:27:34.090 | 9 | 8 | -1 | False |
| 2018-02-18 13:27:52.443 | 8 | 9 | 1 | False |
| 2018-02-18 13:28:58.832 | 9 | 8 | -1 | False |
| 2018-02-18 14:56:49.105 | 8 | 7 | -1 | False |
| 2018-02-18 16:00:32.212 | 7 | 6 | -1 | False |
| 2018-02-18 16:28:20.175 | 6 | 5 | -1 | False |
| 2018-02-18 16:31:48.741 | 5 | 3 | -2 | False |
| 2018-02-18 16:40:33.922 | 3 | 2 | -1 | False |
| 2018-02-18 16:56:17.864 | 2 | 1 | -1 | False |
| 2018-02-18 17:15:01.065 | 1 | 2 | 1 | False |
| 2018-02-18 17:40:43.062 | 2 | 1 | -1 | False |
| 2018-02-18 17:55:50.520 | 1 | 0 | -1 | True |
| 2018-02-18 18:20:21.664 | 30 | 29 | -1 | False |
| 2018-02-18 21:38:10.645 | 29 | 28 | -1 | False |
| 2018-02-19 06:36:04.564 | 28 | 27 | -1 | False |
| 2018-02-19 08:49:23.080 | 27 | 26 | -1 | False |
我想计算一天中每小时的总缺货时间，例如
| date | 0 | 1 | 2 | 3 | ... | 23 |
| ---------- | --- | --- | --- | --- | --- | --- |
| 2018-02-15 | 10 | 0 | 0 | 10 | ... | 13 |
| 2018-02-16 | 6 | 0 | 7 | 10 | ... | 20 |
| 2018-02-17 | 6 | 0 | 0 | 10 | ... | 20 |
规则：
- 按小时分组
- 我可以在一小时内访问所有行。
-
计算两次之间的时间
- 起点：
is_stockout从False到True - 终点：
is_stockout从True到False
一个小时内。 可能有很多
start point和end point - 起点：
- 将索引更改为天，将列更改为24小时。
看起来有点像new-syntax-to-window-and-resample-operations
我认为我需要使用
df.resample('H').apply(caluclate_time_in_hour)
但这似乎还不够：
-
df.resample('H')结果索引是小时，而不是列 -
如何编写正确的
caluclate_time_in_hour？我认为apply无法做到这一点。我写了一个伪代码：
def caluclate_time_in_hour(item): # note: item here is stockcount . not just True or False global last_time global is_stockout global data cur_time = item.name # I need pandas return every row even that hour doesn't have data # so that no need to check the how many hours elasped. if item is np.nan: if is_stockout: data[cur_time.hour] = 60*60 else: data[cur_time.hour] = 0 if is_stockout: if item > 0: data[cur_time.hour] += cur_time - last_time else: is_stockout = False else: if item = 0: is_stockout = True last_time = item.name return data.copy()如何知道此项是本小时的最后一项，以便我可以退回
data？这是apply问题。也许我需要熊猫按小时归还所有行才能申请。
我只是想知道我是否可以通过pandas内置函数来完成上述任务，而无需循环所有行来构造新的DataFrame。
例如，2018-02-15 ~ 2018-02-16具有以下两条记录：
| datetime_create | quantity_old | quantity_new | quantity_diff | is_stockout |
| 2018-02-14 00:45:00 | 40 | 10 | -30 | False |
| 2018-02-15 12:45:00 | 10 | 2 | -8 | False |
| 2018-02-15 13:45:00 | 2 | 1 | -1 | False |
| 2018-02-15 16:45:00 | 1 | 0 | -1 | True |
| 2018-02-16 10:42:00 | 0 | 42 | 42 | False |
| 2018-02-16 13:42:00 | 42 | 40 | -2 | False |
| 2018-02-16 19:42:00 | 40 | 38 | -2 | False |
| 2018-02-17 20:42:00 | 38 | 40 | 2 | False |
# duplicate above
| 2018-02-18 00:45:00 | 40 | 10 | -30 | False |
| 2018-02-19 12:45:00 | 10 | 2 | -8 | False |
| 2018-02-19 13:45:00 | 2 | 1 | -1 | False |
| 2018-02-19 16:45:00 | 1 | 0 | -1 | True |
| 2018-02-20 10:42:00 | 0 | 42 | 42 | False |
| 2018-02-20 13:42:00 | 42 | 40 | -2 | False |
| 2018-02-20 19:42:00 | 40 | 38 | -2 | False |
| 2018-02-21 20:42:00 | 38 | 40 | 2 | False |
csv：
datetime_create,quantity_old,quantity_new,quantity_diff,is_stockout
2018-02-14 00:45:00,40,10,-30,False
2018-02-15 12:45:00,10,2,-8,False
2018-02-15 13:45:00,2,1,-1,False
2018-02-15 16:45:00,1,0,-1,True
2018-02-16 10:42:00,0,42,42,False
2018-02-16 13:42:00,42,40,-2,False
2018-02-16 19:42:00,40,38,-2,False
2018-02-17 20:42:00,38,40,2,False
2018-02-18 00:45:00,40,10,-30,False
2018-02-19 12:45:00,10,2,-8,False
2018-02-19 13:45:00,2,1,-1,False
2018-02-19 16:45:00,1,0,-1,True
2018-02-20 10:42:00,0,42,42,False
2018-02-20 13:42:00,42,40,-2,False
2018-02-20 19:42:00,40,38,-2,False
2018-02-21 20:42:00,38,40,2,False
会有结果（这里的时间单位是分钟，代表美丽）：
date,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23
2018-02-14,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0
2018-02-15,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,15.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0
2018-02-16,60.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0,42.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0
2018-02-17,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0
2018-02-18,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0
2018-02-19,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,15.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0
2018-02-20,60.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0,60.0,42.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0
2018-02-21,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0,0.0
1 个答案:
答案 0 :(得分：1)
我认为需要先resample分钟，然后向前填充NaN s，转换为inetger s，并为Series添加DataFrame.squeeze。
然后通过date和hour与sum进行聚合，最后通过unstack进行整形：
s = df[['is_stockout']].resample('T').ffill().astype(int).squeeze()
df1 = s.groupby([s.index.date, s.index.hour]).sum().unstack(fill_value=0)
print (df1)
datetime_create 0 1 2 3 4 5 6 7 8 9 ... 14 15 16 17 \
2018-02-14 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0
2018-02-15 0 0 0 0 0 0 0 0 0 0 ... 0 0 15 60
2018-02-16 60 60 60 60 60 60 60 60 60 60 ... 0 0 0 0
2018-02-17 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0
2018-02-18 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0
2018-02-19 0 0 0 0 0 0 0 0 0 0 ... 0 0 15 60
2018-02-20 60 60 60 60 60 60 60 60 60 60 ... 0 0 0 0
2018-02-21 0 0 0 0 0 0 0 0 0 0 ... 0 0 0 0
datetime_create 18 19 20 21 22 23
2018-02-14 0 0 0 0 0 0
2018-02-15 60 60 60 60 60 60
2018-02-16 0 0 0 0 0 0
2018-02-17 0 0 0 0 0 0
2018-02-18 0 0 0 0 0 0
2018-02-19 60 60 60 60 60 60
2018-02-20 0 0 0 0 0 0
2018-02-21 0 0 0 0 0 0
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