如何根据序列中的下一个值删除行?(熊猫)

Question

我有以下数据框架:

id  date       outcome
3   03/05/2019  no
3   29/05/2019  no
3   04/09/2019  no
3   30/10/2019  yes
3   03/05/2020  no
5   03/12/2019  no
5   26/12/2019  no
5   27/01/2020  yes
5   03/06/2020  yes
6   04/05/2019  no
6   27/10/2019  no
6   26/11/2019  yes
6   28/11/2019  yes
6   29/11/2019  yes
6   20/12/2019  yes
6   27/12/2019  yes
6   29/12/2019  yes
6   03/01/2020  yes
6   14/01/2020  yes
6   11/02/2020  yes
6   13/02/2020  yes
6   18/02/2020  yes
6   13/04/2020  yes
6   14/04/2020  yes
6   24/04/2020  yes
6   30/04/2020  yes
6   05/05/2020  no

，其中根据id分组，日期按升序排列。

我想删除一行，如果它之后的行有相同的结果。这是id 3的期望结果:

id  date       outcome
3   04/09/2019  no
3   30/10/2019  yes
3   03/05/2020  no

现在我已经创建了一个像这样的蒙版:

m1 = (df['alerts'] == df['alerts'].shift(-1))

但不确定下一步需要做什么来应用条件并删除指定的行。我假设我将做一个groupby与id列…

Answer 1

让我们做

m1 = (df['outcome'] !=
df['outcome'].shift()).cumsum()
out = df.groupby([df['id'],m1]).head(1)
id        date outcome
0    3  03/05/2019      no
3    3  30/10/2019     yes
4    3  03/05/2020      no
5    5  03/12/2019      no
7    5  27/01/2020     yes
9    6  04/05/2019      no
11   6  26/11/2019     yes
26   6  05/05/2020      no

Answer 2

这解决你的问题了吗?

new_df = pd.DataFrame()
current = "yes"
for i, row in df.iterrows():
if row['outcome'] != current:
current = row['outcome']
new_df = new_df.append(row)