Question

我有一个如下数据框：

name   country    Join Date      End date 
Wrt     IND        1-2-2016      8-9-2017
Grt     China      3-2-2015     12-6-2018
frt     France     8-3-2017     continuing 
srt     Scottland   9-4-2018     continuing
crt     china       9-7-2016     7-8-2018

我正在尝试查找加入日期和结束日期之间的差异。我尝试使用f9['Num of days'] = f9['End date '] - f9['Join Date']，但收到以下错误：

TypeError: unsupported operand type(s) for -: 'DatetimeIndex' and 'float'

我的预期输出应该是：

   name   country    Join Date      End date   diff 
   Wrt     IND        1-2-2016      8-9-2017   395
   Grt     China      3-2-2017      12-6-2018  160
   frt     France     8-3-2017     continuing  continuing
   srt     Scottland   9-4-2018     continuing  continuing
   crt     china       9-7-2017     7-8-2018     280

Answer 1

首先使用参数errors='coerce'将两列都转换为日期时间，以获取缺少值的值，如果日期错误，例如字符串continuing，并且在必要时还添加参数dayfirst=True，然后减去值，用{{ 3}}从时间增量开始，最后在必要时用Series.dt.days替换误码值：

f9['Join Date'] = pd.to_datetime(f9['Join Date'], errors='coerce', dayfirst=True)
f9['End date'] = pd.to_datetime(f9['End date'], errors='coerce', dayfirst=True)

f9['Num of days'] = (f9['End date'] - f9['Join Date']).dt.days.fillna('continuing')
print (f9)
  name    country  Join Date   End date Num of days
0  Wrt        IND 2016-02-01 2017-09-08         585
1  Grt      China 2015-02-03 2018-06-12        1225
2  frt     France 2017-03-08        NaT  continuing
3  srt  Scottland 2018-04-09        NaT  continuing
4  crt      china 2016-07-09 2018-08-07         759

或者：

f9['Join Date'] = pd.to_datetime(f9['Join Date'], errors='coerce')
f9['End date'] = pd.to_datetime(f9['End date'], errors='coerce')

f9['Num of days'] = (f9['End date'] - f9['Join Date']).dt.days.fillna('continuing')
print (f9)
  name    country  Join Date   End date Num of days
0  Wrt        IND 2016-01-02 2017-08-09         585
1  Grt      China 2015-03-02 2018-12-06        1375
2  frt     France 2017-08-03        NaT  continuing
3  srt  Scottland 2018-09-04        NaT  continuing
4  crt      china 2016-09-07 2018-07-08         669

最后一步应该是替换丢失的值，但丢失datetime的列，获取与datetimes混合的字符串，因此以后类似datetime的函数失败：

f9['End date'] = f9['End date'].fillna('continuing')
print (f9)
  name    country  Join Date             End date Num of days
0  Wrt        IND 2016-01-02  2017-08-09 00:00:00         585
1  Grt      China 2015-03-02  2018-12-06 00:00:00        1375
2  frt     France 2017-08-03           continuing  continuing
3  srt  Scottland 2018-09-04           continuing  continuing
4  crt      china 2016-09-07  2018-07-08 00:00:00         669

编辑：

您可以从顶部或底部添加多个条件，也可以使用Series.fillna函数：

f9['Join Date'] = pd.to_datetime(f9['Join Date'], errors='coerce')
f9['End date'] = pd.to_datetime(f9['End date'], errors='coerce')

f9['Num of days'] = (f9['End date'] - f9['Join Date']).dt.days

m1 = f9['Num of days'] > 730
m2 = f9['Num of days'].between(365, 730)
m3 = f9['Num of days'] < 365 
m4 = f9['Num of days'].isna()

f9['Status'] = np.select([m1, m2, m3,m4], ['U','L', 'N','EOL']) 

f9[['End date','Num of days']] = f9[['End date','Num of days']].fillna('continuing')
print (f9)

  name    country  Join Date             End date Num of days Status
0  Wrt        IND 2016-01-02  2017-08-09 00:00:00         585      L
1  Grt      China 2015-03-02  2018-12-06 00:00:00        1375      U
2  frt     France 2017-08-03           continuing  continuing    EOL
3  srt  Scottland 2018-09-04           continuing  continuing    EOL
4  crt      china 2016-09-07  2018-07-08 00:00:00         669      L

另一个想法是使用Series.between进行装箱：

f9['Join Date'] = pd.to_datetime(f9['Join Date'], errors='coerce')
f9['End date'] = pd.to_datetime(f9['End date'], errors='coerce')

f9['Num of days'] = (f9['End date'] - f9['Join Date']).dt.days

f9['Status']=pd.cut(f9['Num of days'],bins=[-np.inf, 365, 730, np.inf],labels=['U','L', 'N'])
f9['Status'] = f9['Status'].cat.add_categories(['EOL']).fillna('EOL')
f9[['End date','Num of days']] = f9[['End date','Num of days']].fillna('continuing')
print (f9)
  name    country  Join Date             End date Num of days Status
0  Wrt        IND 2016-01-02  2017-08-09 00:00:00         585      L
1  Grt      China 2015-03-02  2018-12-06 00:00:00        1375      N
2  frt     France 2017-08-03           continuing  continuing    EOL
3  srt  Scottland 2018-09-04           continuing  continuing    EOL
4  crt      china 2016-09-07  2018-07-08 00:00:00         669      L

Answer 2

首先使用to_datetime转换日期中的两列
然后使用.dt.date

减去并获得天数

df = pd.DataFrame(data={'name':['wrt','grt','frt'],
                   'country':['ind','china','france'],
                   'join_date':['1-2-2016','3-2-2015','8-3-2017'],
                   'end_date':['8-9-2017','12-6-2018','continuing']})

df['join_date'] = pd.to_datetime(df['join_date'],errors='coerce').dt.date
df['end_date'] = pd.to_datetime(df['end_date'],errors='coerce').dt.date

df['diff'] = (df['end_date'] - df['join_date']).dt.days
df = df[['join_date','end_date','diff']].fillna('continuing')
print(df)

Answer 3

在这里您可以做的是将“加入日期”和“结束日期”系列转换为numpy数组，并为此dtype = np.datetime64进行比较，然后将差值数组存储到数据帧中。还要用您要填写的任何日期的当前数据时间填写“连续”单元格。（取决于您的情况）

Answer 4

这是可以在jupyter笔记本中运行的可行解决方案。

试图获取两个日期列之间的差异

4 个答案: