2

我有一个如下所示的熊猫数据框：

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

我想找出数据框的年份列中与特定名称相关的空白。例如，

  1. AAA 名称在“2015-11-03 02:00:00”日期前有 2 小时的差距。
  2. ZZZ 名称与“2015-11-01 01:00:00”日期有 5 小时的差距。
  3. ZZZ 名称在“2015-11-01 09:00:00”日期前有 2 小时的差距。

我想生成两个包含内容的 csv 文件：

CSV-1：

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

我尝试如下：

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')
4

1 回答 1

3

要使您的数据框空白，您需要重新分配mask您生成的数据框。要获得总小时数，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)

这为我们提供了以下数据框：

   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

为了在其开始时间旁边获得间隙并与您希望它为“csv-1”的方式保持一致，我们只需将其向上移动一行并在填充 na 值之前减去一：

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

这得到：

   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

为了获得您的第二个 csv，我们可以执行以下操作：

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)

首先，我们设置“之前”和“之后”列。然后通过将索引更改为'year'，我们可以使用该.resample()方法来填充我们所有缺失的时间。通过使用ffill()当我们重新采样时，我们将最后一条可用记录复制到我们添加的所有新行中。我们知道，当 时'prev_year' != 'year'，我们位于框架中以前不存在的行上，因此是间隙之一，因此我们过滤到那些行，选择我们需要的列并重命名它们。这给出了：

   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

总而言之，您的脚本可能如下所示：

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)
于 2018-06-20T07:25:15.777 回答