删除采样频率不同的连续重复项-Python

数据帧如下所示：

0, 3710.968017578125, 2012-01-07T03:13:43.859Z
1, 3710.968017578125, 2012-01-07T03:13:48.890Z
2, 3712.472900390625, 2012-01-07T03:13:53.906Z
3, 3712.472900390625, 2012-01-07T03:13:58.921Z
4, 3713.110107421875, 2012-01-07T03:14:03.900Z
5, 3713.110107421875, 2012-01-07T03:14:03.937Z
6, 3713.89892578125, 2012-01-07T03:14:13.900Z
7, 3713.89892578125, 2012-01-07T03:14:13.968Z
8, 3713.89892578125, 2012-01-07T03:14:19.000Z
9, 3714.64990234375, 2012-01-07T03:14:24.000Z
10, 3714.64990234375, 2012-01-07T03:14:24.015Z
11, 3714.64990234375, 2012-01-07T03:14:29.000Z
12, 3714.64990234375, 2012-01-07T03:14:29.031Z

在某些行中，有具有毫秒不同时间戳的行，我想删除它们，只保留具有毫秒不同时间戳的行。有些行的毫秒和秒值相同，而第9行到第12行则不同，因此，我不能使用

a.loc[a.shift（）！=a]

所需的输出将是：

0, 3710.968017578125, 2012-01-07T03:13:43.859Z
1, 3710.968017578125, 2012-01-07T03:13:48.890Z
2, 3712.472900390625, 2012-01-07T03:13:53.906Z
3, 3712.472900390625, 2012-01-07T03:13:58.921Z
4, 3713.110107421875, 2012-01-07T03:14:03.900Z
6, 3713.89892578125, 2012-01-07T03:14:13.900Z
8, 3713.89892578125, 2012-01-07T03:14:19.000Z
9, 3714.64990234375, 2012-01-07T03:14:24.000Z
11, 3714.64990234375, 2012-01-07T03:14:29.000Z

尝试：

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我希望它能自我解释。

您可以使用下面的脚本。我没有得到你的数据框列名，所以我在下面发明了列['x'，'date\u time']

df = pd.DataFrame([
(3710.968017578125, pd.to_datetime('2012-01-07T03:13:43.859Z')),
(3710.968017578125, pd.to_datetime('2012-01-07T03:13:48.890Z')),
(3712.472900390625, pd.to_datetime('2012-01-07T03:13:53.906Z')),
(3712.472900390625, pd.to_datetime('2012-01-07T03:13:58.921Z')),
(3713.110107421875, pd.to_datetime('2012-01-07T03:14:03.900Z')),
(3713.110107421875, pd.to_datetime('2012-01-07T03:14:03.937Z')),
(3713.89892578125, pd.to_datetime('2012-01-07T03:14:13.900Z')),
(3713.89892578125, pd.to_datetime('2012-01-07T03:14:13.968Z')),
(3713.89892578125, pd.to_datetime('2012-01-07T03:14:19.000Z')),
(3714.64990234375, pd.to_datetime('2012-01-07T03:14:24.000Z')),
(3714.64990234375, pd.to_datetime('2012-01-07T03:14:24.015Z')),
(3714.64990234375, pd.to_datetime('2012-01-07T03:14:29.000Z')),
(3714.64990234375, pd.to_datetime('2012-01-07T03:14:29.031Z'))], 
    columns=['x', 'date_time'])

• 创建“time_diff”列以获取 当前行和下一行的日期时间
• 也只能得到这些差异 没有或超过1秒
• 下降温度柱时间差

df = pd.DataFrame([
(3710.968017578125, pd.to_datetime('2012-01-07T03:13:43.859Z')),
(3710.968017578125, pd.to_datetime('2012-01-07T03:13:48.890Z')),
(3712.472900390625, pd.to_datetime('2012-01-07T03:13:53.906Z')),
(3712.472900390625, pd.to_datetime('2012-01-07T03:13:58.921Z')),
(3713.110107421875, pd.to_datetime('2012-01-07T03:14:03.900Z')),
(3713.110107421875, pd.to_datetime('2012-01-07T03:14:03.937Z')),
(3713.89892578125, pd.to_datetime('2012-01-07T03:14:13.900Z')),
(3713.89892578125, pd.to_datetime('2012-01-07T03:14:13.968Z')),
(3713.89892578125, pd.to_datetime('2012-01-07T03:14:19.000Z')),
(3714.64990234375, pd.to_datetime('2012-01-07T03:14:24.000Z')),
(3714.64990234375, pd.to_datetime('2012-01-07T03:14:24.015Z')),
(3714.64990234375, pd.to_datetime('2012-01-07T03:14:29.000Z')),
(3714.64990234375, pd.to_datetime('2012-01-07T03:14:29.031Z'))], 
    columns=['x', 'date_time'])

df['time_diff'] = df.groupby('x')['date_time'].diff()
df = df[(df['time_diff'].isnull()) | (df['time_diff'].map(lambda x: x.seconds > 1))]
df = df.drop(['time_diff'], axis=1)
df