Question

df1 = pd.DataFrame({'id':[44,44,44,88,88,90,95],
               'Old Status': ['Draft','Submit','Return','Submit','Accept',
               'Draft','Draft'],
               'New Status' : ['Submit','Return','Reject','Accept','Develop',
                              'Submit','Reject'],
                              'Datetime': ['2018-10-24 08:12:02',
                              '2018-10-24 18:12:02', '2018-11-24 08:56:02',
                              '2018-10-24 10:12:02','2018-10-29 13:17:02',
                              '2018-12-30 08:43:12', '2019-01-24 06:12:02']
                              }, columns = ['id','Old Status', 'New Status', 'Datetime'])
df1['Datetime'] = pd.to_datetime(df1['Datetime'])                              
df1
   id Old Status New Status            Datetime
0  44      Draft     Submit 2018-10-24 08:12:02
1  44     Sumbit     Return 2018-10-24 18:12:02
2  44     Return     Reject 2018-11-24 08:56:02
3  88     Submit     Accept 2018-10-24 10:12:02
4  88     Accept    Develop 2018-10-29 13:17:02
5  90      Draft     Submit 2018-12-30 08:43:12
6  95      Draft     Reject 2019-01-24 06:12:02

我具有上述格式的数据框，但是在可视化数据时我需要使事情变得更容易，因此我需要两列，即“状态输入”和“状态输出”日期。对于任何Datetime.loc[n]，“状态输入”将等于Status Out列，Datetime.loc[n+1]将等于id。

下一行具有新的id时，可以假设New Status是当前状态，因此Status Out将是null。

我一直在研究，但似乎找不到与此相关的任何问题。因此，我开始使用循环，但是感觉很丑陋，我知道必须有更多的“熊猫”方式来做到这一点。

到目前为止，我有以下内容。然后，我计划添加条件来处理id的更改，然后转换为数据框，但是感觉很错：

df['Status In'] = df['Datetime']
s_out = [0]*(df['Status In'].count()-1)
for el in range(0,df['Status In'].count()-1):
    s_out[el] = df['Status In'].iloc[el+1]

最终结果类似于：

   id Old Status New Status           Status In           Status Out
0  44      Draft     Submit 2018-10-24 08:12:02  2018-10-24 18:12:02
1  44     Sumbit     Return 2018-10-24 18:12:02  2018-11-24 08:56:02
2  44     Return     Reject 2018-11-24 08:56:02                  NaN
3  88     Submit     Accept 2018-10-24 10:12:02  2018-10-29 13:17:02
4  88     Accept    Develop 2018-10-29 13:17:02                  NaN
5  90      Draft     Submit 2018-12-30 08:43:12                  NaN
6  95      Draft     Reject 2019-01-24 06:12:02                  NaN

在Python / Pandas中是否有更好，更清洁的方法而不使用for循环和if语句？

Answer 1

首先使用shift，然后使用Series.where遮罩eq：

shifted = df1.groupby('id')['Datetime','Old Status'].shift(-1)
print (shifted)
             Datetime Old Status
0 2018-10-24 18:12:02     Submit
1 2018-11-24 08:56:02     Return
2                 NaT        NaN
3 2018-10-29 13:17:02     Accept
4                 NaT        NaN
5                 NaT        NaN
6                 NaT        NaN

df1['Status Out'] = shifted['Datetime'].where(df1['New Status'].eq(shifted['Old Status']))
print (df1)
   id Old Status New Status            Datetime          Status Out
0  44      Draft     Submit 2018-10-24 08:12:02 2018-10-24 18:12:02
1  44     Submit     Return 2018-10-24 18:12:02 2018-11-24 08:56:02
2  44     Return     Reject 2018-11-24 08:56:02                 NaT
3  88     Submit     Accept 2018-10-24 10:12:02 2018-10-29 13:17:02
4  88     Accept    Develop 2018-10-29 13:17:02                 NaT
5  90      Draft     Submit 2018-12-30 08:43:12                 NaT
6  95      Draft     Reject 2019-01-24 06:12:02                 NaT

熊猫：创建一列，其中行等于另一列中的下一行

1 个答案: