熊猫分组多条件和日期差计算

Question

我坚持了解使用的方法。我有以下数据框：

test2

我需要：

1. 按相同的“ CODE”分组，
2. 检查“ DESC”是否不同
3. 检查“ TYPE”是否相同
4. 计算满足前两个命令的日期之间的月份差异

预期输出如下：

Answer 1

以下代码使用.drop_duplicates()和.duplicated()从数据框中保留或丢弃具有重复值的行。

您如何计算一个月的差额？一个月可以是28、30或31天。您可以将最终结果除以30，并获得月数差异的指示。所以我暂时保留了几天。

import pandas as pd

df = {'CODE': ['BBLGLC70M','BBLGLC70M','ZZTNRD77', 'ZZTNRD77', 'AACCBD', 'AACCBD', 'BCCDN', 'BCCDN', 'BCCDN'],
      'DATE': ['16/05/2019','25/09/2019', '16/03/2020', '27/02/2020', '16/07/2020', '21/07/2020', '13/02/2020', '23/07/2020', '27/02/2020'],
      'TYPE': ['PRI', 'PRI', 'PRI', 'PRI', 'PUB', 'PUB', 'PUB', 'PRI', 'PUB'],
      'DESC' : ['KO', 'OK', 'KO', 'KO', 'KO', 'OK', 'KO', 'OK', 'OK']
       }

df = pd.DataFrame(df)
df['DATE'] = pd.to_datetime(df['DATE'], format = '%d/%m/%Y')

# only keep rows that have the same code and type 
df = df[df.duplicated(subset=['CODE', 'TYPE'], keep=False)]

# throw out rows that have the same code and desc
df = df.drop_duplicates(subset=['CODE', 'DESC'], keep=False)

# find previous date
df = df.sort_values(by=['CODE', 'DATE'])
df['previous_date'] = df.groupby('CODE')['DATE'].transform('shift')

# drop rows that don't have a previous date
df = df.dropna()

# calculate the difference between current date and previous date
df['difference_in_dates'] = (df['DATE'] - df['previous_date'])

这将导致以下df：

CODE        DATE        TYPE    DESC    previous_date   difference_in_dates
AACCBD      2020-07-21  PUB     OK      2020-07-16      5 days
BBLGLC70M   2019-09-25  PRI     OK      2019-05-16      132 days
BCCDN       2020-02-27  PUB     OK      2020-02-13      14 days