Python 更改日期计算之间的数据帧周数
我有一个像这样的数据框Python 更改日期计算之间的数据帧周数,python,pandas,dataframe,Python,Pandas,Dataframe,我有一个像这样的数据框 from pandas import Timestamp df = pd.DataFrame({'inventory_created_date': [Timestamp('2016-08-17 00:00:00'), Timestamp('2016-08-17 00:00:00'), Times
from pandas import Timestamp
df = pd.DataFrame({'inventory_created_date': [Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00')],
'rma_processed_date': [Timestamp('2017-09-25 00:00:00'),
Timestamp('2018-01-08 00:00:00'),
Timestamp('2018-04-21 00:00:00'),
Timestamp('2018-08-10 00:00:00'),
Timestamp('2018-10-17 00:00:00'),
Timestamp('2018-11-08 00:00:00'),
Timestamp('2019-07-18 00:00:00'),
Timestamp('2020-01-30 00:00:00'),
Timestamp('2020-04-20 00:00:00'),
Timestamp('2020-06-09 00:00:00')],
'uniqueid':['9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959'],
'rma_created_date':[Timestamp('2017-07-31 00:00:00'),
Timestamp('2017-12-12 00:00:00'),
Timestamp('2018-04-03 00:00:00'),
Timestamp('2018-07-23 00:00:00'),
Timestamp('2018-09-28 00:00:00'),
Timestamp('2018-10-24 00:00:00'),
Timestamp('2019-06-21 00:00:00'),
Timestamp('2019-12-03 00:00:00'),
Timestamp('2020-04-03 00:00:00'),
Timestamp('2020-05-18 00:00:00')],
'time_in_weeks':[50, 69, 85, 101, 110, 114, 148, 172, 189, 196],
'failure_status':[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]})
时间（以周为单位）rma\u创建日期rma\u处理日期2017-12-12rma\u创建日期2017-09-251169112018-04-03的rma\u创建日期
，在2018-01-08的第二行中有rma\u处理日期
。因此，这两个日期之间的周数为12
。因此，第三行中的85
应成为12
这就是我到目前为止所做的
def clean_df(df):
'''
This function will fix the time_in_weeks column to calculate the correct number of weeks
when there is multiple failured for an item.
'''
# Sort by rma_created_date
df = df.sort_values(by=['rma_created_date'])
# Convert date columns into datetime
df['inventory_created_date'] = pd.to_datetime(df['inventory_created_date'], errors='coerce')
df['rma_processed_date'] = pd.to_datetime(df['rma_processed_date'], errors='coerce')
df['rma_created_date'] = pd.to_datetime(df['rma_created_date'], errors='coerce')
# If we have rma_processed_dates that are of 1/1/1900 then just drop that row
df = df[~(df['rma_processed_date'] == '1900-01-01')]
# Correct the time_in_weeks column
df['time_in_weeks']=np.where(df.uniqueid.duplicated(keep='first'),df.rma_processed_date.dt.isocalendar().week.sub(df.rma_processed_date.dt.isocalendar().week.shift(1)),df.time_in_weeks)
return df
df = clean_df(df)
当我把这个函数应用到这个例子中时，我得到的就是这个
df = pd.DataFrame({'inventory_created_date': [Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00'),
Timestamp('2016-08-17 00:00:00')],
'rma_processed_date': [Timestamp('2017-09-25 00:00:00'),
Timestamp('2018-01-08 00:00:00'),
Timestamp('2018-04-21 00:00:00'),
Timestamp('2018-08-10 00:00:00'),
Timestamp('2018-10-17 00:00:00'),
Timestamp('2018-11-08 00:00:00'),
Timestamp('2019-07-18 00:00:00'),
Timestamp('2020-01-30 00:00:00'),
Timestamp('2020-04-20 00:00:00'),
Timestamp('2020-06-09 00:00:00')],
'uniqueid':['9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959',
'9907937959'],
'rma_created_date':[Timestamp('2017-07-31 00:00:00'),
Timestamp('2017-12-12 00:00:00'),
Timestamp('2018-04-03 00:00:00'),
Timestamp('2018-07-23 00:00:00'),
Timestamp('2018-09-28 00:00:00'),
Timestamp('2018-10-24 00:00:00'),
Timestamp('2019-06-21 00:00:00'),
Timestamp('2019-12-03 00:00:00'),
Timestamp('2020-04-03 00:00:00'),
Timestamp('2020-05-18 00:00:00')],
'time_in_weeks':[50, 4294967259, 14, 16, 10, 3, 4294967280, 4294967272, 12, 7],
'failure_status':[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]})
显然，计算是不正确的，这让我相信这一定是出了什么问题
df['time_in_weeks']=np.where(df.uniqueid.duplicated(keep='first'),df.rma_processed_date.dt.isocalendar().week.sub(df.rma_processed_date.dt.isocalendar().week.shift(1)),df.time_in_weeks)
如果有人有任何建议，我将不胜感激
以周为单位的时间列预计为[50,11,12,13,7,1,32,20,9,4]
让我们移动rma\u处理的日期然后从rma\u创建的日期中减去它
最后使用.dt.days
得到天数，然后除以7
得到周数，最后，使用update
更新时间（以周为单位）
列：
weeks = df['rma_created_date'].sub(df['rma_processed_date'].shift()).dt.days.div(7).round()
df['time_in_weeks'].update(weeks)
结果:
inventory_created_date rma_processed_date uniqueid rma_created_date time_in_weeks failure_status
0 2016-08-17 2017-09-25 9907937959 2017-07-31 50 1
1 2016-08-17 2018-01-08 9907937959 2017-12-12 11 1
2 2016-08-17 2018-04-21 9907937959 2018-04-03 12 1
3 2016-08-17 2018-08-10 9907937959 2018-07-23 13 1
4 2016-08-17 2018-10-17 9907937959 2018-09-28 7 1
5 2016-08-17 2018-11-08 9907937959 2018-10-24 1 1
6 2016-08-17 2019-07-18 9907937959 2019-06-21 32 1
7 2016-08-17 2020-01-30 9907937959 2019-12-03 20 1
8 2016-08-17 2020-04-20 9907937959 2020-04-03 9 1
9 2016-08-17 2020-06-09 9907937959 2020-05-18 4 1
请不要张贴图片。禁止在StackOverflow上使用图像。请花点时间阅读@ShubhamSharma调整和changed@ShubhamSharma谢谢你把它添加到Posts weeks的列表中了？那你是怎么得到结果的？我们不应该只更新现有的dftime\u in-weeks
列吗？@justanewbweeks
是一个pandas.Series
。我们可以使用update
方法用新的weeks
系列更新现有列time\u in\u weeks
。