我有一个数据框df
,每个组的最后一行(groupby STK_ID
)是NaN:
>>> print df
sales opr_pft net_pft
STK_ID RPT_Date
002138 20130331 2.0703 0.3373 0.2829
20130630 NaN NaN NaN
20130930 7.4993 1.2248 1.1630
20140122 NaN NaN NaN
600004 20130331 11.8429 3.0816 2.1637
20130630 24.6232 6.2152 4.5135
20130930 37.9673 9.2088 6.6463
20140122 NaN NaN NaN
600809 20130331 27.9517 9.9426 7.5182
20130630 40.6460 13.9414 9.8572
20130930 53.0501 16.8081 11.8605
20140122 NaN NaN NaN
现在我希望fillna是每个组的最后一行及其前一行,结果应该是这样的:
sales opr_pft net_pft
STK_ID RPT_Date
002138 20130331 2.0703 0.3373 0.2829
20130630 NaN NaN NaN **(Not fillna this row)**
20130930 7.4993 1.2248 1.1630
20140122 7.4993 1.2248 1.1630
600004 20130331 11.8429 3.0816 2.1637
20130630 24.6232 6.2152 4.5135
20130930 37.9673 9.2088 6.6463
20140122 37.9673 9.2088 6.6463
600809 20130331 27.9517 9.9426 7.5182
20130630 40.6460 13.9414 9.8572
20130930 53.0501 16.8081 11.8605
20140122 53.0501 16.8081 11.8605
我几乎完成了它:df.groupby(level=0).apply(lambda grp: grp.fillna(method='ffill'))
,它生成如下:
sales opr_pft net_pft
STK_ID RPT_Date
002138 20130331 2.0703 0.3373 0.2829
20130630 2.0703 0.3373 0.2829
20130930 7.4993 1.2248 1.1630
20140122 7.4993 1.2248 1.1630
600004 20130331 11.8429 3.0816 2.1637
20130630 24.6232 6.2152 4.5135
20130930 37.9673 9.2088 6.6463
20140122 37.9673 9.2088 6.6463
600809 20130331 27.9517 9.9426 7.5182
20130630 40.6460 13.9414 9.8572
20130930 53.0501 16.8081 11.8605
20140122 53.0501 16.8081 11.8605
这不是我想要的,它通过组内的行填充。那么如何填写Pandas中每组的最后一行?
答案 0 :(得分:5)
您可以在groupby中使用其他功能:
def f(g):
last = len(g.values)-1
g.iloc[last,:] = g.iloc[last-1,:]
return g
print df.groupby(level=0).apply(f)
输出:
sales opr_pft net_pft
STK_ID RPT_Date
2138 20130331 2.0703 0.3373 0.2829
20130630 NaN NaN NaN
20130930 7.4993 1.2248 1.1630
20140122 7.4993 1.2248 1.1630
600004 20130331 11.8429 3.0816 2.1637
20130630 24.6232 6.2152 4.5135
20130930 37.9673 9.2088 6.6463
20140122 37.9673 9.2088 6.6463
600809 20130331 27.9517 9.9426 7.5182
20130630 40.6460 13.9414 9.8572
20130930 53.0501 16.8081 11.8605
20140122 53.0501 16.8081 11.8605