Question

我想找到列日期（pd.to_datetime格式）的最小值，而不是“ 1777-07-07”，这基本上是离群值。输入数据帧如图所示

    col2        date
    b1a2  1777-07-07
    b1a2  2012-09-14
    b1a2  1777-07-07
    b1a2  1777-07-07
    b1a2  2017-09-14
    b1a2  2019-09-24
    b1a2  2012-09-14
    b1a2  2012-09-14
    b1a2  2012-09-28
    a1l2  1777-07-07
    a1l2  2012-09-24
    a1l2  2012-09-24
    a1l2  2002-09-28
    a1l2  2012-09-24
    a1l2  2008-09-14
    a1l2  2012-09-24

所以当我尝试以下内容

    df = df.join(df.groupby(['col2'])['date'].agg({'earliest':'min'}),on=['disability_case_id'])
    df = df.join(df.groupby(['col2'])['date'].agg({'latest':'max'}),on=['disability_case_id'])

这给了我最大和最小的值，如图所示

    col2        date earliset   latest
    b1a2  1777-07-07 1777-07-07 2019-09-24
    b1a2  2012-09-14 1777-07-07 2019-09-24
    b1a2  2017-09-14 1777-07-07 2019-09-24
    b1a2  2019-09-24 1777-07-07 2019-09-24
    b1a2  2012-09-14 1777-07-07 2019-09-24
    b1a2  2012-09-14 1777-07-07 2019-09-24
    b1a2  2012-09-28 1777-07-07 2019-09-24
    a1l2  1777-07-07 1777-07-07 2012-09-28
    a1l2  2012-09-24 1777-07-07 2012-09-28
    a1l2  2012-09-28 1777-07-07 2012-09-28
    a1l2  2002-09-28 1777-07-07 2012-09-28
    a1l2  2012-09-24 1777-07-07 2012-09-28
    a1l2  2008-09-14 1777-07-07 2012-09-28
    a1l2  2012-09-24 1777-07-07 2012-09-28

但是我想避免离群值，我的预期输出是

b1a2  1777-07-07 2012-09-14 2019-09-24
b1a2  2012-09-14 2012-09-14 2019-09-24
b1a2  2017-09-14 2012-09-14 2019-09-24
b1a2  2019-09-24 2012-09-14 2019-09-24
b1a2  2012-09-14 2012-09-14 2019-09-24
b1a2  2012-09-14 2012-09-14 2019-09-24
b1a2  2012-09-28 2012-09-14 2019-09-24
a1l2  1777-07-07 2002-09-28 2012-09-28
a1l2  2012-09-24 2002-09-28 2012-09-28
a1l2  2012-09-28 2002-09-28 2012-09-28
a1l2  2002-09-28 2002-09-28 2012-09-28
a1l2  2012-09-24 2002-09-28 2012-09-28
a1l2  2008-09-14 2002-09-28 2012-09-28
a1l2  2012-09-24 2002-09-28 2012-09-28

Answer 1

在恒定离群值的情况下，在分组依据之前遮罩。使用transform广播回原始DataFrame。

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

s = df.date.where(df.date.ne('1777-07-07')).groupby(df.col2)
df['earliest'] = s.transform('min')
df['latest'] = s.transform('max')

输出：

    col2       date   earliest     latest
0   b1a2 1777-07-07 2012-09-14 2019-09-24
1   b1a2 2012-09-14 2012-09-14 2019-09-24
2   b1a2 1777-07-07 2012-09-14 2019-09-24
3   b1a2 1777-07-07 2012-09-14 2019-09-24
4   b1a2 2017-09-14 2012-09-14 2019-09-24
5   b1a2 2019-09-24 2012-09-14 2019-09-24
6   b1a2 2012-09-14 2012-09-14 2019-09-24
7   b1a2 2012-09-14 2012-09-14 2019-09-24
8   b1a2 2012-09-28 2012-09-14 2019-09-24
9   a1l2 1777-07-07 2002-09-28 2012-09-24
10  a1l2 2012-09-24 2002-09-28 2012-09-24
11  a1l2 2012-09-24 2002-09-28 2012-09-24
12  a1l2 2002-09-28 2002-09-28 2012-09-24
13  a1l2 2012-09-24 2002-09-28 2012-09-24
14  a1l2 2008-09-14 2002-09-28 2012-09-24
15  a1l2 2012-09-24 2002-09-28 2012-09-24

Answer 2

屏蔽无效值，然后像以前一样继续操作。

u = df['date'].mask(df['date'].eq('1777-07-07')).groupby(df['col2']).agg(['min', 'max'])

df.merge(u, left_on='col2', right_index=True)

    col2       date        min        max
0   b1a2 1777-07-07 2012-09-14 2019-09-24
1   b1a2 2012-09-14 2012-09-14 2019-09-24
2   b1a2 1777-07-07 2012-09-14 2019-09-24
3   b1a2 1777-07-07 2012-09-14 2019-09-24
4   b1a2 2017-09-14 2012-09-14 2019-09-24
5   b1a2 2019-09-24 2012-09-14 2019-09-24
6   b1a2 2012-09-14 2012-09-14 2019-09-24
7   b1a2 2012-09-14 2012-09-14 2019-09-24
8   b1a2 2012-09-28 2012-09-14 2019-09-24
9   a1l2 1777-07-07 2002-09-28 2012-09-24
10  a1l2 2012-09-24 2002-09-28 2012-09-24
11  a1l2 2012-09-24 2002-09-28 2012-09-24
12  a1l2 2002-09-28 2002-09-28 2012-09-24
13  a1l2 2012-09-24 2002-09-28 2012-09-24
14  a1l2 2008-09-14 2002-09-28 2012-09-24
15  a1l2 2012-09-24 2002-09-28 2012-09-24

groupby执行列后，将第二个最小值和最大值输入新列

2 个答案:

输出：