问题描述
import numpy as np
import pandas as pd
mydata = {'ID1': [1,1,2,3,4],'ID2': [1,4,5,6],'Date1': ['2011-04-23','2011-05-13','2012-04-23','2012-05-13','2011-08-23','2011-08-26'],'Date2': ['2011-04-25','2011-05-23','2012-04-1','2011-05-18','2011-08-24','2011-08-29']
}
mydata = pd.DataFrame(mydata)
我想创建一个新列,例如天,如下所示:如果ID1是唯一的,则为-1;如果ID1不是唯一的,则计算Date1(带滞后)和Date2之间的差。下面的代码以某种方式起作用;对于唯一的ID1,它不会产生-1。这也有点奇怪。感谢您提供任何其他解决方案的帮助。
mydata['Date1'] = pd.to_datetime(mydata['Date1'])
mydata['Date2'] = pd.to_datetime(mydata['Date2'])
mydata = mydata.sort_values(['ID1','Date1'],ascending=[True,True])
diff_time = mydata['Date2'].rsub(mydata['Date1'].shift(-1),axis=0)
mydata['days'] = np.where(mydata['ID1']==mydata['ID1'].shift(-1),(diff_time.dt.days*24+diff_time.astype(str).str.split('[ :]').str[2].astype(float))/24,0)
输出:
ID1 ID2 Date1 Date2 days
0 1 1 2011-04-23 2011-04-25 18.0
1 1 2 2011-05-13 2011-05-23 0.0
2 2 3 2012-04-23 2012-04-01 0.0 # 0.0 here should be -1 as ID1 is unique
4 3 5 2011-08-23 2011-08-24 263.0
3 3 4 2012-05-13 2011-05-18 0.0
5 4 6 2011-08-26 2011-08-29 0.0 # 0.0 here should be -1 as ID1 is unique
解决方法
您可以将DataFrameGroupBy.shift
和Series.duplicated
中的numpy.where
与{{3}}中的string
重复使用ID
:
-1