Pandas: fill one column with count of # of obs between occurrences in a 2nd column

Here is my way

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

Create a series containing your dates, mask this series when your R series is not equal to 1, bfill, and subtract!

---

u = df.index.to_series()

ii = u.where(df.R.eq(1)).bfill()

12 * (ii.dt.year - u.dt.year) + (ii.dt.month - u.dt.month)

2010-01-01    2
2010-02-01    1
2010-03-01    0
2010-04-01    0
2010-05-01    4
2010-06-01    3
2010-07-01    2
2010-08-01    1
2010-09-01    0
2010-10-01    0
2010-11-01    0
Freq: MS, dtype: int64

Here is a way that worked for me, not as elegant as @user3483203 but it does the job.

df['F'] = 0 
for i in df.index: 
     j = i 
     while df.loc[j, 'R'] == 0: 
         df.loc[i, 'F'] =df.loc[i, 'F'] + 1 
         j=j+1                                                                                                                      
df  

################
Out[39]: 
        index  R  F
0  2010-01-01  0  2
1  2010-02-01  0  1
2  2010-03-01  1  0
3  2010-04-01  1  0
4  2010-05-01  0  4
5  2010-06-01  0  3
6  2010-07-01  0  2
7  2010-08-01  0  1
8  2010-09-01  1  0
9  2010-10-01  1  0
10 2010-11-01  1  0

In [40]: