熊猫：在第二列中两次出现之间的obs计数填充一列

Question

说我有下面的DataFrame，它具有0/1项，具体取决于在某个月内是否发生了什么。

Y = [0,0,1,1,0,0,0,0,1,1,1]
X = pd.date_range(start = "2010", freq = "MS", periods = len(Y))

df = pd.DataFrame({'R': Y},index = X)



            R
2010-01-01  0
2010-02-01  0
2010-03-01  1
2010-04-01  1
2010-05-01  0
2010-06-01  0
2010-07-01  0
2010-08-01  0
2010-09-01  1
2010-10-01  1
2010-11-01  1

我想要创建一个第二列，该列列出直到下一次出现1为止的月数。

也就是说，我需要：

            R  F
2010-01-01  0  2
2010-02-01  0  1
2010-03-01  1  0
2010-04-01  1  0
2010-05-01  0  4
2010-06-01  0  3
2010-07-01  0  2
2010-08-01  0  1
2010-09-01  1  0
2010-10-01  1  0
2010-11-01  1  0

我已经尝试过的：我还没走很远，但是我能够填补第一部分

A = list(df.index)
T = df[df['R']==1]

a = df.index[0]
b = T.index[0]
c = A.index(b) - A.index(a)

df.loc[a:b, 'F'] = np.linspace(c,0,c+1)

            R    F
2010-01-01  0  2.0
2010-02-01  0  1.0
2010-03-01  1  0.0
2010-04-01  1  NaN
2010-05-01  0  NaN
2010-06-01  0  NaN
2010-07-01  0  NaN
2010-08-01  0  NaN
2010-09-01  1  NaN
2010-10-01  1  NaN
2010-11-01  1  NaN

编辑可能最好提供一个跨多年的原始示例。

Y = [0,0,1,1,0,0,0,0,1,1,1,0,0,1,1,1,0,1,1,1]
X = pd.date_range(start = "2010", freq = "MS", periods = len(Y))

df = pd.DataFrame({'R': Y},index = X)

Answer 1

这是我的方式

s=df.R.cumsum()
df.loc[df.R==0,'F']=s.groupby(s).cumcount(ascending=False)+1
df.F.fillna(0,inplace=True)

df
Out[12]: 
            R    F
2010-01-01  0  2.0
2010-02-01  0  1.0
2010-03-01  1  0.0
2010-04-01  1  0.0
2010-05-01  0  4.0
2010-06-01  0  3.0
2010-07-01  0  2.0
2010-08-01  0  1.0
2010-09-01  1  0.0
2010-10-01  1  0.0
2010-11-01  1  0.0