熊猫在df中找到与上一行具有相同值的最后一行

Question

I have a df ,

acct_no    code    date           id
100        10      01/04/2019     22
100        10      01/03/2019     22
100        10      01/05/2019     22
200        20      01/06/2019     33
200        20      01/05/2019     33
200        20      01/07/2019     33

I want to first sort the df in ascending order for date when acct_no and code are the same,

df.sort_values(['acct_no', 'code', 'date'], inplace=True)

then I am wondering what the way to find the last row whose acct_no , code are the same as the previous row, the result need to look like,

  acct_no    code    date           id
  100        10      01/05/2019     22
  200        20      01/07/2019     33

Answer 1

You can also try with groupby.last() :

df.groupby(['acct_no', 'code'],as_index=False).last()

---

   acct_no  code        date  id
0      100    10  01/05/2019  22
1      200    20  01/07/2019  33

Answer 2

Use DataFrame.drop_duplicates , but first convert column to datetimes:

#if dates are first use dayfirst=True
df['date'] = pd.to_datetime(df['date'], dayfirst=True)
#if months are first
#df['date'] = pd.to_datetime(df['date'])
df1 = (df.sort_values(['acct_no', 'code', 'date'])
         .drop_duplicates(['acct_no', 'code'], keep='last'))
print (df1)
   acct_no  code       date  id
2      100    10 2019-05-01  22
5      200    20 2019-07-01  33