如何通過查看一個數據框中的日期落在另一個數據框中的日期范圍內來合並Pandas數據框？

Question

我有兩個具有員工數據的數據框，如下所示。 一個數據文件包含員工數據，其中包括員工生病的日期，而另一個數據文件包含員工工作的日期（即，顯示為日期范圍）。 我想通過查看特定雇員的“病假”在“工作范圍”內的位置來合並這兩個文件（希望是在熊貓中）。 例如，在下面的圖像/數據中，員工1在11/25 / 2015、12 / 23/2015和10/12/2015患病。 這些分別屬於“工作范圍”，分別為11/21/2015-11/29 / 2015、12 / 21/2015-12/29/2015和10/9/2015-10/17/2015。

員工工作日期數據：

 ╔══════════╦════════════╦════════════╗ ║ 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 ║ ╚══════════╩════════════╩════════════╝ 

員工病假日期數據：

 ╔══════════╦════════════╦═══════════╗ ║ 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 ║ ╚══════════╩════════════╩═══════════╝ 

合並數據：

 ╔══════════╦════════════╦════════════╦════════════╦═══════════╗ ║ 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中做到這一點？ （謝謝您的幫助！）

Answer 1

您需要將此數據作為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 ║ ╚══════════╩════════════╩════════════╝ 

然后將此數據作為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

考慮內部和外部大熊貓合並方法。 下面假設日期為datetime格式，可能需要從字符串對象進行轉換：

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)

結果

#    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