我一直在寻找有关如何在熊猫数据框中有选择地删除连续重复项的所有问题/答案，仍然无法弄清楚以下情况：

import pandas as pd
import numpy as np

def random_dates(start, end, n, freq, seed=None):
    if seed is not None:
        np.random.seed(seed)

    dr = pd.date_range(start, end, freq=freq)
    return pd.to_datetime(np.sort(np.random.choice(dr, n, replace=False)))

date = random_dates('2018-01-01', '2018-01-12', 20, 'H', seed=[3, 1415])

data = {'Timestamp': date, 
        'Message': ['Message received.','Sending...', 'Sending...', 'Sending...', 'Work in progress...', 'Work in progress...', 
                    'Message received.','Sending...', 'Sending...','Work in progress...',
                    'Message received.','Sending...', 'Sending...', 'Sending...','Work in progress...', 'Work in progress...', 'Work in progress...',
                    'Message received.','Sending...', 'Sending...']}

df = pd.DataFrame(data, columns = ['Timestamp', 'Message'])

我有以下数据框：

             Timestamp              Message
0  2018-01-02 03:00:00    Message received.
1  2018-01-02 11:00:00           Sending...
2  2018-01-03 04:00:00           Sending...
3  2018-01-04 11:00:00           Sending...
4  2018-01-04 16:00:00  Work in progress...
5  2018-01-04 17:00:00  Work in progress...
6  2018-01-05 05:00:00    Message received.
7  2018-01-05 11:00:00           Sending...
8  2018-01-05 17:00:00           Sending...
9  2018-01-06 02:00:00  Work in progress...
10 2018-01-06 14:00:00    Message received.
11 2018-01-07 07:00:00           Sending...
12 2018-01-07 20:00:00           Sending...
13 2018-01-08 01:00:00           Sending...
14 2018-01-08 02:00:00  Work in progress...
15 2018-01-08 15:00:00  Work in progress...
16 2018-01-09 00:00:00  Work in progress...
17 2018-01-10 03:00:00    Message received.
18 2018-01-10 09:00:00           Sending...
19 2018-01-10 14:00:00           Sending...

我想仅在“消息”为“工作进行中…”时才将连续重复项放在df [‘消息’]列中，并保留第一个实例（例如，这里需要删除索引5、15和16）我想得到：

             Timestamp              Message
0  2018-01-02 03:00:00    Message received.
1  2018-01-02 11:00:00           Sending...
2  2018-01-03 04:00:00           Sending...
3  2018-01-04 11:00:00           Sending...
4  2018-01-04 16:00:00  Work in progress...
6  2018-01-05 05:00:00    Message received.
7  2018-01-05 11:00:00           Sending...
8  2018-01-05 17:00:00           Sending...
9  2018-01-06 02:00:00  Work in progress...
10 2018-01-06 14:00:00    Message received.
11 2018-01-07 07:00:00           Sending...
12 2018-01-07 20:00:00           Sending...
13 2018-01-08 01:00:00           Sending...
14 2018-01-08 02:00:00  Work in progress...
17 2018-01-10 03:00:00    Message received.
18 2018-01-10 09:00:00           Sending...
19 2018-01-10 14:00:00           Sending...

我尝试过类似帖子中提供的解决方案，例如：

df['Message'].loc[df['Message'].shift(-1) != df['Message']]

我还计算了消息的长度：

df['length'] = df['Message'].apply(lambda x: len(x))

并写了一个条件丢弃为：

df.loc[(df['length'] ==17) | (df['length'] ==10) | ~df['Message'].duplicated(keep='first')]

看起来更好，但索引14、15和16仍被完全删除，因此行为不佳，请参阅：

             Timestamp              Message  length
0  2018-01-02 03:00:00    Message received.      17
1  2018-01-02 11:00:00           Sending...      10
2  2018-01-03 04:00:00           Sending...      10
3  2018-01-04 11:00:00           Sending...      10
4  2018-01-04 16:00:00  Work in progress...      19
6  2018-01-05 05:00:00    Message received.      17
7  2018-01-05 11:00:00           Sending...      10
8  2018-01-05 17:00:00           Sending...      10
10 2018-01-06 14:00:00    Message received.      17
11 2018-01-07 07:00:00           Sending...      10
12 2018-01-07 20:00:00           Sending...      10
13 2018-01-08 01:00:00           Sending...      10
17 2018-01-10 03:00:00    Message received.      17
18 2018-01-10 09:00:00           Sending...      10
19 2018-01-10 14:00:00           Sending...      10

感谢您的时间和帮助！