我一直在查看有关如何在 Pandas 数据框中有选择地删除连续重复项的所有问题/答案，但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...
我想删除 df['Message'] 列中的连续重复项，仅当 'Message' 为 'Work in progress...' 并保留第一个实例(此处例如需要删除索引 5、15 和 16)，理想情况下我想得到:
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))
并写了一个有条件的丢弃:
df.loc[(df['length'] ==17) | (df['length'] ==10) | ~df['Message'].duplicated(keep='first')]
它看起来更好，但仍然完全删除了索引 14、15 和 16，因此行为不端，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!
最佳答案
首先过滤第一个连续值，比较 Series.shift 和链式掩码，过滤所有没有 Work in progress... 的行值(value)观:
df = df[(df['Message'].shift() != df['Message']) | (df['Message'] != 'Work in progress...')]
print (df)
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...
关于python - Pandas 有选择地删除连续的重复项，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/59874217/