如何通过查看一个数据框中的日期落在另一个数据框中的日期范围内来合并Pandas数据框？

Question

我有两个具有员工数据的数据框，如下所示。 一个数据文件包含员工数据，其中包括员工生病的日期，而另一个数据文件包含员工工作的日期（即，显示为日期范围）。 我想通过查看特定雇员的“病假”在“工作范围”内的位置来合并这两个文件（希望是在熊猫中）。 例如，在下面的图像/数据中，员工1在11/25 / 2015、12 / 23/2015和10/12/2015患病。 这些分别属于“工作范围”，分别为11/21/2015-11/29 / 2015、12 / 21/2015-12/29/2015和10/9/2015-10/17/2015。

员工工作日期数据：

 ╔══════════╦════════════╦════════════╗ ║ Employee ║ datein ║ dateout ║ ╠══════════╬════════════╬════════════╣ ║ 1 ║ 11/21/2015 ║ 11/29/2015 ║ ║ 2 ║ 12/9/2015 ║ 12/14/2015 ║ ║ 3 ║ 11/10/2015 ║ 11/19/2015 ║ ║ 4 ║ 11/11/2015 ║ 11/17/2015 ║ ║ 5 ║ 11/30/2015 ║ 12/8/2015 ║ ║ 1 ║ 12/21/2015 ║ 12/29/2015 ║ ║ 2 ║ 1/7/2016 ║ 1/12/2016 ║ ║ 3 ║ 12/10/2015 ║ 12/19/2015 ║ ║ 4 ║ 12/10/2015 ║ 12/16/2015 ║ ║ 5 ║ 12/30/2015 ║ 1/7/2016 ║ ║ 1 ║ 10/9/2015 ║ 10/17/2015 ║ ║ 2 ║ 10/27/2015 ║ 11/1/2015 ║ ║ 3 ║ 9/28/2015 ║ 10/7/2015 ║ ║ 4 ║ 9/29/2015 ║ 10/5/2015 ║ ╚══════════╩════════════╩════════════╝ 

员工病假日期数据：

 ╔══════════╦════════════╦═══════════╗ ║ Employee ║ sickDate ║ sickness ║ ╠══════════╬════════════╬═══════════╣ ║ 1 ║ 11/25/2015 ║ flu ║ ║ 10 ║ 11/21/2015 ║ hd ║ ║ 21 ║ 9/20/2015 ║ other ║ ║ 1 ║ 12/23/2015 ║ other ║ ║ 4 ║ 12/13/2015 ║ vacationx ║ ║ 7 ║ 7/21/2015 ║ cough ║ ║ 3 ║ 10/1/2015 ║ rash ║ ║ 4 ║ 10/5/2015 ║ other ║ ║ 5 ║ 1/7/2016 ║ eyex ║ ║ 2 ║ 12/12/2015 ║ tanx ║ ║ 1 ║ 10/12/2015 ║ fatiguex ║ ╚══════════╩════════════╩═══════════╝ 

合并数据：

 ╔══════════╦════════════╦════════════╦════════════╦═══════════╗ ║ Employee ║ datein ║ dateout ║ sickDate ║ sickness ║ ╠══════════╬════════════╬════════════╬════════════╬═══════════╣ ║ 1 ║ 11/21/2015 ║ 11/29/2015 ║ 11/25/2015 ║ flu ║ ║ 2 ║ 12/9/2015 ║ 12/14/2015 ║ 12/12/2015 ║ tanx ║ ║ 3 ║ 11/10/2015 ║ 11/19/2015 ║ ║ ║ ║ 4 ║ 11/11/2015 ║ 11/17/2015 ║ ║ ║ ║ 5 ║ 11/30/2015 ║ 12/8/2015 ║ ║ ║ ║ 1 ║ 12/21/2015 ║ 12/29/2015 ║ 12/23/2015 ║ other ║ ║ 2 ║ 1/7/2016 ║ 1/12/2016 ║ ║ ║ ║ 3 ║ 12/10/2015 ║ 12/19/2015 ║ ║ ║ ║ 4 ║ 12/10/2015 ║ 12/16/2015 ║ 12/13/2015 ║ vacationx ║ ║ 5 ║ 12/30/2015 ║ 1/7/2016 ║ 1/7/2016 ║ eyex ║ ║ 1 ║ 10/9/2015 ║ 10/17/2015 ║ 10/12/2015 ║ fatiguex ║ ║ 2 ║ 10/27/2015 ║ 11/1/2015 ║ ║ ║ ║ 3 ║ 9/28/2015 ║ 10/7/2015 ║ 10/1/2015 ║ rash ║ ║ 4 ║ 9/29/2015 ║ 10/5/2015 ║ 10/5/2015 ║ other ║ ╚══════════╩════════════╩════════════╩════════════╩═══════════╝ 

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如何在Pandas或python中做到这一点？ （谢谢您的帮助！）

Answer 1

您需要将此数据作为df1和set_index('Employee')放入pd.DataFrame( ... ) set_index('Employee')

 ╔══════════╦════════════╦════════════╗ ║ Employee ║ datein ║ dateout ║ ╠══════════╬════════════╬════════════╣ ║ 1 ║ 11/21/2015 ║ 11/29/2015 ║ ║ 2 ║ 12/9/2015 ║ 12/14/2015 ║ ║ 3 ║ 11/10/2015 ║ 11/19/2015 ║ ║ 4 ║ 11/11/2015 ║ 11/17/2015 ║ ║ 5 ║ 11/30/2015 ║ 12/8/2015 ║ ║ 1 ║ 12/21/2015 ║ 12/29/2015 ║ ║ 2 ║ 1/7/2016 ║ 1/12/2016 ║ ║ 3 ║ 12/10/2015 ║ 12/19/2015 ║ ║ 4 ║ 12/10/2015 ║ 12/16/2015 ║ ║ 5 ║ 12/30/2015 ║ 1/7/2016 ║ ║ 1 ║ 10/9/2015 ║ 10/17/2015 ║ ║ 2 ║ 10/27/2015 ║ 11/1/2015 ║ ║ 3 ║ 9/28/2015 ║ 10/7/2015 ║ ║ 4 ║ 9/29/2015 ║ 10/5/2015 ║ ╚══════════╩════════════╩════════════╝ 

然后将此数据作为df2和set_index('Employee')放入pd.DataFrame( ... ) set_index('Employee')

 ╔══════════╦════════════╦═══════════╗ ║ Employee ║ sickDate ║ sickness ║ ╠══════════╬════════════╬═══════════╣ ║ 1 ║ 11/25/2015 ║ flu ║ ║ 10 ║ 11/21/2015 ║ hd ║ ║ 21 ║ 9/20/2015 ║ other ║ ║ 1 ║ 12/23/2015 ║ other ║ ║ 4 ║ 12/13/2015 ║ vacationx ║ ║ 7 ║ 7/21/2015 ║ cough ║ ║ 3 ║ 10/1/2015 ║ rash ║ ║ 4 ║ 10/5/2015 ║ other ║ ║ 5 ║ 1/7/2016 ║ eyex ║ ║ 2 ║ 12/12/2015 ║ tanx ║ ║ 1 ║ 10/12/2015 ║ fatiguex ║ ╚══════════╩════════════╩═══════════╝ 

最后， df = df1.join(df2).reset_index()

Answer 2

考虑内部和外部大熊猫合并方法。 下面假设日期为datetime格式，可能需要从字符串对象进行转换：

workdf['datein'] = pd.to_datetime(workdf['datein'])
workdf['dateout'] = pd.to_datetime(workdf['dateout'])
sickdf['sickDate'] = pd.to_datetime(sickdf['sickDate'])

# INNER MERGE ON BOTH DFs WHERE SICK DAYS REPEAT FOR MATCHING EMPLOYEE ROW IN WORK DAYS
mergedf = pd.merge(workdf, sickdf, on='Employee', how="inner")

# OUTER MERGE TO KEEP ALL WORK DAY RECORDS WITH FILTERED SICK DAYS DATA SET
finaldf = pd.merge(mergedf[(mergedf['sickDate'] - mergedf['datein'] >= 0) &
                           (mergedf['dateout'] - mergedf['sickDate'] >= 0)],
                   workdf, on=['Employee', 'datein', 'dateout'], how="outer")

finaldf = finaldf.sort(['Employee','datein','dateout']).reset_index(drop=True)

结果

#    Employee     datein      dateout     sickDate   sickness
#0          1 2015-10-09   2015-10-17   2015-10-12   fatiguex
#1          1 2015-11-21   2015-11-29   2015-11-25        flu
#2          1 2015-12-21   2015-12-29   2015-12-23      other
#3          2 2015-10-27   2015-11-01          NaT        NaN
#4          2 2015-12-09   2015-12-14   2015-12-12       tanx
#5          2 2016-01-07   2016-01-12          NaT        NaN
#6          3 2015-09-28   2015-10-07   2015-10-01       rash
#7          3 2015-11-10   2015-11-19          NaT        NaN
#8          3 2015-12-10   2015-12-19          NaT        NaN
#9          4 2015-09-29   2015-10-05   2015-10-05      other
#10         4 2015-11-11   2015-11-17          NaT        NaN
#11         4 2015-12-10   2015-12-16   2015-12-13  vacationx
#12         5 2015-11-30   2015-12-08          NaT        NaN
#13         5 2015-12-30   2016-01-07   2016-01-07       eyex