我有4个投资组合a,b,c,d,可以在一段时间内采用“不”或“自己”的值。 (下面包含代码以方便复制)
ano=('a','no',datetime(2018,1,1), datetime(2018,1,2))
aown=('a','own',datetime(2018,1,3), datetime(2018,1,4))
bno=('b','no',datetime(2018,1,1), datetime(2018,1,5))
bown=('b','own',datetime(2018,1,6), datetime(2018,1,7))
cown=('c','own',datetime(2018,1,9), datetime(2018,1,10))
down=('d','own',datetime(2018,1,9), datetime(2018,1,9))
sch=pd.DataFrame([ano,aown,bno,bown,cown,down],columns=['portf','base','st','end'])
进度摘要:
portf base st end
0 a no 2018-01-01 2018-01-02
1 a own 2018-01-03 2018-01-04
2 b no 2018-01-01 2018-01-05
3 b own 2018-01-06 2018-01-07
4 c own 2018-01-09 2018-01-10
5 d own 2018-01-09 2018-01-09
我尝试过的方法:创建一个保持数据框并根据计划表填写值。不幸的是,第一个投资组合“ a”被覆盖了
df=pd.DataFrame(index=pd.date_range(min(sch.st),max(sch.end)),columns=['portf','base'])
for row in range(len(sch)):
df.loc[sch['st'][row]:sch['end'][row],['portf','base']]= sch.loc[row,['portf','base']].values
portf base
2018-01-01 b no
2018-01-02 b no
2018-01-03 b no
2018-01-04 b no
2018-01-05 b no
2018-01-06 b own
2018-01-07 b own
2018-01-08 NaN NaN
2018-01-09 d own
2018-01-10 c own
所需的输出:
2018-01-01 (('a','no'), ('b','no'))
2018-01-02 (('a','no'), ('b','no'))
2018-01-03 (('a','own'), ('b','no'))
2018-01-04 (('a','own'), ('b','no'))
2018-01-05 ('b','no')
...
我敢肯定有一种更简单的方法可以实现这一目标,但是可能这是我以前从未遇到过的例子。提前非常感谢!
答案 0 :(得分:0)
我将以不同的方式组织数据,索引是日期,portf的列,值是基数。
首先,我们需要重塑数据并重新采样到每日字段。这是一个简单的枢轴。
cols = ['portf', 'base']
s = (df.reset_index()
.melt(cols+['index'], value_name='date')
.set_index('date')
.groupby(cols+['index'], group_keys=False)
.resample('D').ffill()
.drop(columns=['variable', 'index'])
.reset_index())
res = s.pivot(index='date', columns='portf')
res = res.resample('D').first() # Recover missing dates between
res
base
portf a b c d
2018-01-01 no no NaN NaN
2018-01-02 no no NaN NaN
2018-01-03 own no NaN NaN
2018-01-04 own no NaN NaN
2018-01-05 NaN no NaN NaN
2018-01-06 NaN own NaN NaN
2018-01-07 NaN own NaN NaN
2018-01-08 NaN NaN NaN NaN
2018-01-09 NaN NaN own own
2018-01-10 NaN NaN own NaN
如果您需要其他输出,我们可以通过一些不理想的Series.apply
电话到达那里。这对于大型DataFrame来说将是非常糟糕的。我会认真考虑保留以上内容。
s.set_index('date').apply(tuple, axis=1).groupby('date').apply(tuple)
date
2018-01-01 ((a, no), (b, no))
2018-01-02 ((a, no), (b, no))
2018-01-03 ((a, own), (b, no))
2018-01-04 ((a, own), (b, no))
2018-01-05 ((b, no),)
2018-01-06 ((b, own),)
2018-01-07 ((b, own),)
2018-01-09 ((c, own), (d, own))
2018-01-10 ((c, own),)
dtype: object