pandas drop consecutive duplicates selectively

六眼飞鱼酱① 提交于 2020-02-14 10:47:51

问题


I have been looking at all questions/answers about to how drop consecutive duplicates selectively in a pandas dataframe, still cannot figure out the following scenario:

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'])

I have the following dataframe:

             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...

I want to drop the consecutive duplicates in df['Message'] column ONLY when 'Message' is 'Work in progress...' and keep the first instance (here e.g. Index 5, 15 and 16 need to be dropped), ideally I would like to get:

             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...

I have tried solutions offered in similar posts like:

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

I also calculated the length of the Messages:

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

and wrote a conditional drop as:

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

It looks better but still Index 14, 15, and 16 are dropped altogether, thus it is ill-behaved, see:

             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

Your time and help is appreciated!


回答1:


First filter first consecutive values with compare by Series.shift and chain mask with filter all rows with no Work in progress... values:

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...



回答2:


You can first get all Messages with 'Work in Progress' and compare them with the previous element and then filter:

condition = (df['Message'] == 'Work in progress...') & (df['Message']==df['Message'].shift(1))

df[~condition]

     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...


来源:https://stackoverflow.com/questions/59874217/pandas-drop-consecutive-duplicates-selectively

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