Pandas groupby多条件及日期差计算
Pandas groupby multi conditions and date difference calculation
我无法理解要使用的方法。我有以下数据框:
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')
df
我需要:
- groupby相同'CODE',
- 检查 'DESC' 是否不一样
- 检查 'TYPE' 是否相同
- 计算满足前两个命令的日期之间的月差
预期输出如下:
以下代码使用 .drop_duplicates() and .duplicated() 保留或丢弃数据框中具有重复值的行。
你如何计算一个月的差异?一个月可以是 28、30 或 31 天。您可以将最终结果除以 30,得到月数差异的指示。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
我无法理解要使用的方法。我有以下数据框:
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')
df
我需要:
- groupby相同'CODE',
- 检查 'DESC' 是否不一样
- 检查 'TYPE' 是否相同
- 计算满足前两个命令的日期之间的月差
预期输出如下:
以下代码使用 .drop_duplicates() and .duplicated() 保留或丢弃数据框中具有重复值的行。
你如何计算一个月的差异?一个月可以是 28、30 或 31 天。您可以将最终结果除以 30,得到月数差异的指示。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