根据列值对有序pandas数据帧中的行进行分组

Question

I have a question about grouping only certain rows together in a pandas dataframe (that is ordered by timestamp), depending on their column values.

So here is an example:

df=pd.DataFrame({"text":["Hello.",
                    "I had a question.", 
                    "Hi!",
                    "Yes how can I help?",
                    "Do you ship to the UK?"
                    ],
            "timestamp":[
                        pd.Timestamp('20131213 11:50:00'),
                        pd.Timestamp('20131213 11:51:00'),
                        pd.Timestamp('20131213 11:52:00'),
                        pd.Timestamp('20131213 11:53:00'),
                        pd.Timestamp('20131213 11:54:00')
                        ],
            "direction":["In","In","Out","Out","In"]})

This is what the dataframe looks like:

This dataframe is ordered by timestamp and could be (for example) a chat thread where direction "In" could be one person talking and "Out" is another person talking.

What I would like to get is something like this:

In the final dataframe, the text of the rows are grouped together into one row if they are the same direction, but rows are only grouped together until you reach a row with a different direction. AND the order of the messages is retained.

Does anyone have any ideas?

Answer 1

Setup

operations = {
    'text': ' '.join,
    'direction': 'first',
}

---

Using agg and a common trick to group by consecutive values:

df.groupby(df.direction.ne(df.direction.shift()).cumsum()).agg(operations)

                               text direction
direction
1          Hello. I had a question.        In
2           Hi! Yes how can I help?       Out
3            Do you ship to the UK?        In

Answer 2

What about doing something like this:

# indicate direction changes
df['dir'] = df.direction.shift(1).bfill()
df['dir_change'] = df.apply(lambda x: 1 if x.direction != x.dir else 0, axis=1)

# create new groups
df['new_group'] = df.dir_change.cumsum()

# group on new groups and aggregate the text
agg_df = df.groupby('new_group').agg({'text':lambda x: ' '.join(list(x)), 'timestamp':'first'})