第二列表示重置的熊猫仔

时间:2019-04-12 16:35:47

标签: python pandas dataframe

我需要计算以Wgt列中的新值指示的某个频率重置的累积乘积。

例如,在由以下对象产生的DataFrame中:

df = pd.DataFrame(np.random.lognormal(0, 0.01, 27), pd.date_range('2019-01-06', '2019-02-01'), columns=['Chg'])
df['Wgt'] = df['Chg'].asfreq('W')
df.loc[df.Wgt > 0, 'Wgt'] = np.random.uniform(0.5, 1, df.Wgt.count())

                 Chg       Wgt
2019-01-06  1.014571  0.861546
2019-01-07  1.018993       NaN
2019-01-08  1.017461       NaN
2019-01-09  1.003788       NaN
2019-01-10  1.014106       NaN
2019-01-11  0.995758       NaN
2019-01-12  0.989058       NaN
2019-01-13  0.995897  0.602225
2019-01-14  1.007336       NaN
2019-01-15  1.004143       NaN
...

我想计算一个新列Agg,其值为:

  1. 如果df.Wgt != np.nan,则df.Agg = df.Wgt
  2. 其他df.Agg = df.Agg.shift() * df.Chg

即,在此示例中,Agg为:

                 Chg    Wgt         Agg
1/6/2019    1.014571    0.861546    0.861546
1/7/2019    1.018993    NaN         0.877909343
1/8/2019    1.017461    NaN         0.893238518
1/9/2019    1.003788    NaN         0.896622106
1/10/2019   1.014106    NaN         0.909269857
1/11/2019   0.995758    NaN         0.905412734
1/12/2019   0.989058    NaN         0.895505708
1/13/2019   0.995897    0.602225    0.602225
1/14/2019   1.007336    NaN         0.606642923
1/15/2019   1.004143    NaN         0.609156244
...

有什么令人讨厌的方法?

1 个答案:

答案 0 :(得分:2)

np.wherecumprod一起使用

s=df.loc[df.Wgt.isnull(),'Chg'].groupby(df.Wgt.notna().cumsum()).cumprod()
np.where(df.Wgt.notna(),df.Wgt,s*df.Wgt.ffill())
Out[531]: 
array([0.861546  , 0.87790934, 0.89323852, 0.89662211, 0.90926986,
       0.90541273, 0.89550571, 0.602225  , 0.60664292, 0.60915624])