Find list of indices before specific value in column for each unique id

Question

I have dataframe like this:

id |    date   | status
________________________
...     ...      ...
1  |2020-01-01 | reserve
1  |2020-01-02 | sold
2  |2020-01-01 | free
3  |2020-01-03 | reserve
3  |2020-01-25 | signed
3  |2020-01-30 | sold
...     ...      ...
10 |2020-01-02 | signed
10 |2020-02-15 | sold 
...     ...      ....

I want to find indices of all rows with status sold and then for all rows back 29 days of these (rows with status sold) assign 1 and 0 in else case.

The desired dataframe is this

id |    date   | status  | label
_________________________________
...     ...      ...        ...
1  |2020-01-01 | reserve | 1 
1  |2020-01-02 | sold    | 1
2  |2019-12-02 | free    | 0    # no sold status for 2
3  |2020-01-03 | reserve | 1
3  |2020-01-25 | signed  | 1
3  |2020-01-30 | sold    | 1
...     ...      ...        ...
10 |2020-01-02 | signed  | 0
10 |2020-02-15 | sold    | 1    # more than 29 days from 2020-02-15
...     ...      ....       ...

I attemped to use apply(), but I found out I can’t call function like that

def make_labels(df):    
    
    def get_indices(df):
        return list(df[df['date'] >= df.iloc[-1]['date'] - timedelta(days=29)].index)

    df.sort_values(['id', 'date'], inplace=True)
    zero_labels = pd.Series(0, index = df.index, name='sold_labels')    
    one_lables = df.groupby('id')['status'].apply(lambda s: get_indices if s.iloc[-1] == 'sold').sum()
    zero_labels.loc[one_lables] = 1
    
    return zero_labels

df['label'] = make_labels(df)

dataframe constructor of the input:

d = {'id': [1, 1, 2, 3, 3, 3, 10, 10], 
     'date': ['2020-01-01', '2020-01-02', '2020-01-01', '2020-01-03', '2020-01-25', '2020-01-30', '2020-01-02', '2020-02-15'],
     'status': ['reserve', 'sold', 'free', 'reserve', 'signed', 'sold', 'signed', 'sold']
    }
df = pd.DataFrame(data=d)

Asked By: Michael

||

Source

Answer 1

You can use groupby.transform to get the sold date per group, then compare the difference to 29 days:

df['date'] = pd.to_datetime(df['date'])

ref = (df['date']
 .where(df['status'].eq('sold'))
 .groupby(df['id'])
 .transform('first')
 )

df['label'] = (df['date'].rsub(ref)
               .le('29days')
               .astype(int)
               )

Output:

   id       date   status  label
0   1 2020-01-01  reserve      1
1   1 2020-01-02     sold      1
2   2 2020-01-01     free      0
3   3 2020-01-03  reserve      1
4   3 2020-01-25   signed      1
5   3 2020-01-30     sold      1
6  10 2020-01-02   signed      0
7  10 2020-02-15     sold      1

Answered By: mozway

Find list of indices before specific value in column for each unique id

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