How could I remove duplicates if duplicates mean less than 30days? - sql

Question

(using sql or pandas)
I want to delete records if the Date difference between two records is less than 30 days.
But first record of ID must be remained.
#example
ROW ID DATE
1 A 2020-01-01 -- first
2 A 2020-01-03
3 A 2020-01-31
4 A 2020-02-05
5 A 2020-02-28
6 A 2020-03-09
7 B 2020-03-06 -- first
8 B 2020-05-07
9 B 2020-06-02
#expected results
ROW ID DATE
1 A 2020-01-01
4 A 2020-02-05
6 A 2020-03-09
7 B 2020-03-06
8 B 2020-05-07
ROW 2,3 are within 30 days from ROW 1
ROW 5 is within 30 days from ROW 4
ROW 9 is within 30 days from ROW 8

Answer 1

To cope with your task it is not possible to call any
vectorized methods.
The cause is that after a row is recognized as a duplicate, then
this row "does not count" when you check further rows.
E.g. after rows 2020-01-03 and 2020-01-31 were deleted (as
"too close" to the previous row) then 2020-02-05 row should be
left, because now the distance to the previous row (2020-01-01)
is big enough.
So I came up with a solution based on a "function with memory":
def isDupl(elem):
if isDupl.prev is None:
isDupl.prev = elem
return False
dDiff = (elem - isDupl.prev).days
rv = dDiff <= 30
if not rv:
isDupl.prev = elem
return rv
This function should be invoked for each DATE in the
current group (with same ID) but before that isDupl.prev
must be set to None.
So the function to apply to each group of rows is:
def isDuplGrp(grp):
isDupl.prev = None
return grp.DATE.apply(isDupl)
And to get the expected result, run:
df[~(df.groupby('ID').apply(isDuplGrp).reset_index(level=0, drop=True))]
(you may save it back to df).
The result is:
ROW ID DATE
0 1 A 2020-01-01
3 4 A 2020-02-05
5 6 A 2020-03-09
6 7 B 2020-03-06
7 8 B 2020-05-07
And finally, a remark about the other solution:
It contains rows:
3 4 A 2020-02-05
4 5 A 2020-02-28
which are only 23 days apart, so this solution is wrong.
The same pertains to rows:
5 A 2020-02-28
6 A 2020-03-09
which are also too close in time.

Answer 2

You can try this:
Convert date to datetime64
Get the first date from each group df.groupby('ID')['DATE'].transform('first')
Add a filter to keep only dates greater than 30 days
Append the first date of each group to the dataframe
Code:
df['DATE'] = pd.to_datetime(df['DATE'])
df1 = df[(df['DATE'] - df.groupby('ID')['DATE'].transform('first')) >= pd.Timedelta(30, unit='D')]
df1 = df1.append(df.groupby('ID', as_index=False).agg('first')).sort_values(by=['ID', 'DATE'])
print(df1)
ROW ID DATE
0 1 A 2020-01-01
2 3 A 2020-01-31
3 4 A 2020-02-05
4 5 A 2020-02-28
5 6 A 2020-03-09
1 7 B 2020-03-06
7 8 B 2020-05-07
8 9 B 2020-06-02