如何找出python pandas dataframe列（日期格式）中的空白？

Question

name,year
AAA,2015-11-02 22:00:00
AAA,2015-11-02 23:00:00
AAA,2015-11-03 00:00:00
AAA,2015-11-03 01:00:00
AAA,2015-11-03 02:00:00
AAA,2015-11-03 05:00:00
ZZZ,2015-09-01 00:00:00
ZZZ,2015-11-01 01:00:00
ZZZ,2015-11-01 07:00:00
ZZZ,2015-11-01 08:00:00
ZZZ,2015-11-01 09:00:00
ZZZ,2015-11-01 12:00:00

null

name,year
AAA,2015-11-02 22:00:00,0
AAA,2015-11-02 23:00:00,0
AAA,2015-11-03 00:00:00,0
AAA,2015-11-03 01:00:00,0
AAA,2015-11-03 02:00:00,2
AAA,2015-11-03 05:00:00,0
ZZZ,2015-09-01 00:00:00,0
ZZZ,2015-11-01 01:00:00,5
ZZZ,2015-11-01 07:00:00,0
ZZZ,2015-11-01 08:00:00,0
ZZZ,2015-11-01 09:00:00,2
ZZZ,2015-11-01 12:00:00,0

CSV-2:

name,prev_year,next_year,gaps
AAA,2015-11-03 02:00:00,2015-11-03 05:00:00,2015-11-03 03:00:00
AAA,2015-11-03 02:00:00,2015-11-03 05:00:00,2015-11-03 04:00:00
ZZZ,2015-11-01 01:00:00,2015-11-01 07:00:00,2015-11-01 02:00:00
ZZZ,2015-11-01 01:00:00,2015-11-01 07:00:00,2015-11-01 03:00:00
ZZZ,2015-11-01 01:00:00,2015-11-01 07:00:00,2015-11-01 04:00:00
ZZZ,2015-11-01 01:00:00,2015-11-01 07:00:00,2015-11-01 05:00:00
ZZZ,2015-11-01 01:00:00,2015-11-01 07:00:00,2015-11-01 06:00:00
ZZZ,2015-11-01 09:00:00,2015-11-01 12:00:00,2015-11-01 10:00:00
ZZZ,2015-11-01 09:00:00,2015-11-01 12:00:00,2015-11-01 11:00:00

我试了如下：

df['year'] = pd.to_datetime(df['year'], format='%Y-%m-%d %H:%M:%S')
mask = df.groupby("name").year.diff() > pd.Timedelta('0 days 01:00:00')

Answer 1

要将gap添加到数据帧中，需要重新分配生成的掩码。要得到总小时数，您可以简单地除以1小时：

df['year'] = pd.to_datetime(df['year'], format='%Y-%m-%d %H:%M:%S')
df['Gap'] = (df.groupby("name").year.diff() / pd.to_timedelta('1 hour')).fillna(0)

这给出了以下数据帧：

   name                year     Gap
0   AAA 2015-11-02 22:00:00     0.0
1   AAA 2015-11-02 23:00:00     1.0
2   AAA 2015-11-03 00:00:00     1.0
3   AAA 2015-11-03 01:00:00     1.0
4   AAA 2015-11-03 02:00:00     1.0
5   AAA 2015-11-03 05:00:00     3.0
6   ZZZ 2015-09-01 00:00:00     0.0
7   ZZZ 2015-11-01 07:00:00     6.0
8   ZZZ 2015-11-01 08:00:00     1.0
9   ZZZ 2015-11-01 09:00:00     1.0
10  ZZZ 2015-11-01 12:00:00     3.0

为了得到其开始时间旁边的间隙，并与您希望的“CSV-1”的方式保持一致，我们只需将其上移一行，然后在填充na值之前减去1：

df['Gap'] = ((df.groupby("name").year.diff() / pd.to_timedelta('1 hour')).shift(-1) - 1).fillna(0)

   name                year  Gap
0   AAA 2015-11-02 22:00:00  0.0
1   AAA 2015-11-02 23:00:00  0.0
2   AAA 2015-11-03 00:00:00  0.0
3   AAA 2015-11-03 01:00:00  0.0
4   AAA 2015-11-03 02:00:00  2.0
5   AAA 2015-11-03 05:00:00  0.0
6   ZZZ 2015-11-01 01:00:00  5.0
7   ZZZ 2015-11-01 07:00:00  0.0
8   ZZZ 2015-11-01 08:00:00  0.0
9   ZZZ 2015-11-01 09:00:00  2.0
10  ZZZ 2015-11-01 12:00:00  0.0

df['prev_year'] = df['year']
df['next_year'] = df.groupby('name')['year'].shift(-1)

df.set_index('year', inplace=True)
df = df.groupby('name', as_index=False)\
       .resample(rule='1H')\
       .ffill()\
       .reset_index()

gaps = df[df['year'] != df['prev_year']][['name', 'prev_year', 'next_year', 'year']]

gaps.rename({'year': 'gaps'}, index='columns', inplace=True)

   name           prev_year           next_year                year
5   AAA 2015-11-03 02:00:00 2015-11-03 05:00:00 2015-11-03 03:00:00
6   AAA 2015-11-03 02:00:00 2015-11-03 05:00:00 2015-11-03 04:00:00
9   ZZZ 2015-11-01 01:00:00 2015-11-01 07:00:00 2015-11-01 02:00:00
10  ZZZ 2015-11-01 01:00:00 2015-11-01 07:00:00 2015-11-01 03:00:00
11  ZZZ 2015-11-01 01:00:00 2015-11-01 07:00:00 2015-11-01 04:00:00
12  ZZZ 2015-11-01 01:00:00 2015-11-01 07:00:00 2015-11-01 05:00:00
13  ZZZ 2015-11-01 01:00:00 2015-11-01 07:00:00 2015-11-01 06:00:00
17  ZZZ 2015-11-01 09:00:00 2015-11-01 12:00:00 2015-11-01 10:00:00
18  ZZZ 2015-11-01 09:00:00 2015-11-01 12:00:00 2015-11-01 11:00:00

df['year'] = pd.to_datetime(df['year'], format='%Y-%m-%d %H:%M:%S')
df['Gap'] = ((df.groupby("name").year.diff() / pd.to_timedelta('1 hour')).shift(-1) - 1).fillna(0)

df.to_csv('csv-1.csv', index=False)

df['prev_year'] = df['year']
df['next_year'] = df.groupby('name')['year'].shift(-1)

df.set_index('year', inplace=True)
df = df.groupby('name', as_index=False)\
       .resample(rule='1H')\
       .ffill()\
       .reset_index()

gaps = df[df['year'] != df['prev_year']][['name', 'prev_year', 'next_year', 'year']]

gaps.rename({'year': 'gaps'}, index='columns', inplace=True)

gaps.to_csv('csv-2.csv', index=False)