问题
I have a data-frame as follows:
df_raw_dates = pd.DataFrame({"id": [102, 102, 102, 103, 103, 103, 104], "val": [9,2,4,7,6,3,2], "dates": [pd.Timestamp(2002, 1, 1), pd.Timestamp(2002, 3, 3), pd.Timestamp(2003, 4, 4), pd.Timestamp(2003, 8, 9), pd.Timestamp(2005, 2, 3), pd.Timestamp(2005, 2, 8), pd.Timestamp(2005, 2, 3)]})
id val dates
0 102 9 2002-01-01
1 102 2 2002-03-03
2 102 4 2003-04-04
3 103 7 2003-08-09
4 103 6 2005-02-03
5 103 3 2005-02-08
6 104 2 2005-02-03
What I want to achieve is instead of having the dates column to have a column diff_dates that will represent the difference between consecutive dates per id where the first entry for each id in the diff_dates column will be 0. With that said, the resulting data frame should be:
df_processed_dates = pd.DataFrame({"id": [102, 102, 102, 103, 103, 103, 104], "val": [9,2,4,7,6,3,2], "diff_dates": [0, 61, 397, 0, 544, 5, 0]})
id val diff_dates
0 102 9 0
1 102 2 61
2 102 4 397
3 103 7 0
4 103 6 544
5 103 3 5
6 104 2 0
Looking forward to your answers!
回答1:
Use groupby + diff. You can then extract the days component and fill missing entries with fillna.
df_raw_dates.groupby('id').dates.diff().dt.days.fillna(0, downcast='infer')
0 0
1 61
2 397
3 0
4 544
5 5
6 0
Name: dates, dtype: int64
To assign this back as a new column, do
df_raw_dates['date_diff'] = (
df_raw_dates
.pop('dates')
.groupby(df_raw_dates['id'])
.diff()
.dt.days
.fillna(0, downcast='infer'))
df_raw_dates
id val date_diff
0 102 9 0
1 102 2 61
2 102 4 397
3 103 7 0
4 103 6 544
5 103 3 5
6 104 2 0
来源：https://stackoverflow.com/questions/53814805/difference-between-consecutive-dates-in-pandas-groupby