如何使用.loc标记groupby pandas的新列

时间:2018-03-07 03:48:41

标签: python pandas

我有一个由'Key'分组的df。我想标记允许日期与另一个允许日期匹配的组内的任何行。

df =  pd.DataFrame({'Key': ['10003', '10003', '10003', '10003', '10003','10003','10034', '10034'], 
           'Num1': [12,13,13,13,13,13,16,13],
           'Num2': [121,122,122,124,125,126,127,128],
          'admit': [20120506, 20120508, 20121010,20121010,20121010,20121110,20120516,20120520],
      'discharge': [20120508, 20120508, 20121012,20121016,20121023,20121111,20120518,20120522]})
df['admit'] = pd.to_datetime(df['admit'], format='%Y%m%d')
df['discharge'] = pd.to_datetime(df['discharge'], format='%Y%m%d')

初始df

    Key     Num1    Num2    admit       discharge
0   10003   12      121     2012-05-06  2012-05-08
1   10003   13      122     2012-05-08  2012-05-08
2   10003   13      122     2012-10-10  2012-10-12
3   10003   13      124     2012-10-10  2012-10-16
4   10003   13      125     2012-10-10  2012-10-23
5   10003   13      126     2012-11-10  2012-11-11
6   10034   16      127     2012-05-16  2012-05-18
7   10034   13      128     2012-05-20  2012-05-22

最终df

    Key     Num1    Num2    admit       discharge   flag
0   10003   12      121     2012-05-06  2012-05-08  0
1   10003   13      122     2012-05-08  2012-05-08  0
2   10003   13      122     2012-10-10  2012-10-12  1
3   10003   13      124     2012-10-10  2012-10-16  1
4   10003   13      125     2012-10-10  2012-10-23  1
5   10003   13      126     2012-11-10  2012-11-11  0
6   10034   16      127     2012-05-16  2012-05-18  0
7   10034   13      128     2012-05-20  2012-05-22  0

df维度是1.5亿乘400,所以我在考虑使用.loc而不是.apply可能更适合这个。但我愿意接受建议。

我的代码:

df.loc[df.groupby('Key').duplicated(subset='admit'),'flag'] = 1

然而,这是一个我应该使用的错误。

AttributeError: Cannot access callable attribute 'duplicated' of 'DataFrameGroupBy' objects, try using the 'apply' method

1 个答案:

答案 0 :(得分:1)

您需要申请

df.loc[df.groupby('Key').apply(lambda x : x.duplicated(subset='admit',keep=False)).values,'flag']=1
df
Out[300]: 
     Key  Num1  Num2      admit  discharge  flag
0  10003    12   121 2012-05-06 2012-05-08   NaN
1  10003    13   122 2012-05-08 2012-05-08   NaN
2  10003    13   122 2012-10-10 2012-10-12   1.0
3  10003    13   124 2012-10-10 2012-10-16   1.0
4  10003    13   125 2012-10-10 2012-10-23   1.0
5  10003    13   126 2012-11-10 2012-11-11   NaN
6  10034    16   127 2012-05-16 2012-05-18   NaN
7  10034    13   128 2012-05-20 2012-05-22   NaN
相关问题