熊猫分组多条件和日期差计算

编程问答 2022-06-04

问题描述

我坚持了解使用的方法。我有以下数据框：

test2

enter image description here

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：

enter image description here

解决方法

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','AACCBD','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','PUB','PUB'],'DESC' : ['KO','OK','KO','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