python - Pandas 中两个特定日期时间范围之间出现的数字

Question

我有 2 个 CSV 文件，如下所示。

1. 我想要一个新列差异，其中...
   • 如果手机号码出现在 Book_date...App_date 的日期范围内:差异 = 差异App_date > 和 Occur_date
   • 如果未在该日期范围内发生，则返回 NaN。
2. 我还想根据唯一类别和手机号码对其进行过滤

csv_1

Mobile_Number    Book_Date       App_Date

503477334    2018-10-12       2018-10-18
506002884    2018-10-12       2018-10-19
501022162    2018-10-12       2018-10-16
503487338    2018-10-13       2018-10-13
506012887    2018-10-13       2018-10-21
503427339    2018-10-14       2018-10-17

csv_2

Mobile_Number    Occur_Date    

503477334        2018-10-16
506002884        2018-10-21
501022162        2018-10-15
503487338        2018-10-13
501428449        2018-10-18
506012887        2018-10-14

我想要在 csv_1 中添加一个新列，其中如果手机号码出现在 csv_2 中的 Book_date 和 App_date 的日期范围内，则 App_date 与 Occur_date 之间的差值或 NaN(如果该手机号码未出现在该日期范围内)。输出应该是

输出

Mobile_Number    Book_Date       App_Date   Difference

503477334    2018-10-12       2018-10-18       2
506002884    2018-10-12       2018-10-19      -2
501022162    2018-10-12       2018-10-16       1
503487338    2018-10-13       2018-10-13       0
506012887    2018-10-13       2018-10-21       7 
503427339    2018-10-14       2018-10-17       NaN

编辑:

如果我想根据上述两个 csv 文件的唯一类别和 mobile_number 对其进行过滤。如何做同样的事情？

csv_1

Category     Mobile_Number   Book_Date       App_Date

A              503477334    2018-10-12       2018-10-18
B              503477334    2018-10-07       2018-10-16
C              501022162    2018-10-12       2018-10-16
A              503487338    2018-10-13       2018-10-13
C              506012887    2018-10-13       2018-10-21
E              503427339    2018-10-14       2018-10-17

csv_2

Category     Mobile_Number    Occur_Date    

A              503477334        2018-10-16
B              503477334        2018-10-13
A              501022162        2018-10-15
A              503487338        2018-10-13
F              501428449        2018-10-18
C              506012887        2018-10-14

我希望根据 Mobile_Number 和类别过滤输出

输出

Category     Mobile_Number    Book_Date       App_Date   Difference

A              503477334    2018-10-12       2018-10-18       2
B              503477334    2018-10-07       2018-10-16       3
C              501022162    2018-10-12       2018-10-16       NaN
A              503487338    2018-10-13       2018-10-13       0
C              506012887    2018-10-13       2018-10-21       7 
E              503427339    2018-10-14       2018-10-17       NaN

Answer 1

使用Series.map对于与 Mobile_Number 匹配的新 Series 以及列之间的测试值，请使用 Series.between ，然后通过掩码使用 numpy.where 赋值:

df1['Book_Date'] = pd.to_datetime(df1['Book_Date'])
df1['App_Date'] = pd.to_datetime(df1['App_Date'])
df2['Occur_Date'] = pd.to_datetime(df2['Occur_Date'])

s1 = df2.drop_duplicates('Mobile_Number').set_index('Mobile_Number')['Occur_Date']
s2 = df1['Mobile_Number'].map(s1)

m = s2.between(df1['Book_Date'], df1['App_Date'])

#solution with no mask
df1['Difference1'] = df1['App_Date'].sub(s2).dt.days
#solution with test between
df1['Difference2'] = np.where(m, df1['App_Date'].sub(s2).dt.days, np.nan)
print (df1)
   Mobile_Number  Book_Date   App_Date Difference  Difference1  Difference2
0      503477334 2018-10-12 2018-10-18 2018-10-16          2.0          2.0
1      506002884 2018-10-12 2018-10-19 2018-10-21         -2.0          NaN
2      501022162 2018-10-12 2018-10-16 2018-10-15          1.0          1.0
3      503487338 2018-10-13 2018-10-13 2018-10-13          0.0          0.0
4      506012887 2018-10-13 2018-10-21 2018-10-14          7.0          7.0
5      503427339 2018-10-14 2018-10-17        NaT          NaN          NaN

编辑:

您可以使用 merge 代替 map 来连接 2 列:

df1['Book_Date'] = pd.to_datetime(df1['Book_Date'])
df1['App_Date'] = pd.to_datetime(df1['App_Date'])
df2['Occur_Date'] = pd.to_datetime(df2['Occur_Date'])

df3 = df1.merge(df2, on=['Category','Mobile_Number'], how='left')
print (df3)
  Category  Mobile_Number  Book_Date   App_Date Occur_Date
0        A      503477334 2018-10-12 2018-10-18 2018-10-16
1        B      503477334 2018-10-07 2018-10-16 2018-10-13
2        C      501022162 2018-10-12 2018-10-16        NaT
3        A      503487338 2018-10-13 2018-10-13 2018-10-13
4        C      506012887 2018-10-13 2018-10-21 2018-10-14
5        E      503427339 2018-10-14 2018-10-17        NaT

m = df3['Occur_Date'].between(df3['Book_Date'], df3['App_Date'])
#print (m)

df3['Difference2'] = np.where(m, df3['App_Date'].sub(df3['Occur_Date']).dt.days, np.nan)
print (df3)
  Category  Mobile_Number  Book_Date   App_Date Occur_Date  Difference2
0        A      503477334 2018-10-12 2018-10-18 2018-10-16          2.0
1        B      503477334 2018-10-07 2018-10-16 2018-10-13          3.0
2        C      501022162 2018-10-12 2018-10-16        NaT          NaN
3        A      503487338 2018-10-13 2018-10-13 2018-10-13          0.0
4        C      506012887 2018-10-13 2018-10-21 2018-10-14          7.0
5        E      503427339 2018-10-14 2018-10-17        NaT          NaN