Splitting a dataframe based on column values

Question

I have a dataframe like such

 EndDate
2007-10-31              0
2007-11-30    -0.03384464
2007-12-31     -0.0336299
2008-01-31   -0.009448923
2008-02-29              


        

Answer 1

Alexander's solution didn't work. There is a small error. The code should be:

d = {n: df2.iloc[rows] 
 for n, rows in df2.groupby('group_no').groups.items()}

Answer 2

First, you can create group numbers by comparing the value column to zero and then taking a cumulative sum of these boolean values.

df['group_no'] = (df.val == 0).cumsum()
>>> df.head(6)
      EndDate       val  group_no
0  2007-10-31  0.000000         1
1  2007-11-30 -0.033845         1
2  2007-12-31 -0.033630         1
3  2008-01-31 -0.009449         1
4  2008-02-29  0.000000         2
5  2008-03-31 -0.057450         2

Next, you can use a dictionary comprehension together with loc to select the relevant group_no dataframe. To get the last group number, I get the last value using iat for location based indexing.

d = {i: df.loc[df.group_no == i, ['EndDate', 'val']] 
     for i in range(1, df.group_no.iat[-1])}

>>> d
{1:       EndDate       val
 0  2007-10-31  0.000000
 1  2007-11-30 -0.033845
 2  2007-12-31 -0.033630
 3  2008-01-31 -0.009449, 
 2:       EndDate       val
 4  2008-02-29  0.000000
 5  2008-03-31 -0.057450
 6  2008-04-30 -0.038694, 
 3:       EndDate       val
 7  2008-05-31  0.000000
 8  2008-06-30 -0.036245
 9  2008-07-31 -0.005286}

EDIT As suggested by @DSM, using groupby appears to be about 6x faster based on a sample dataframe with 15k rows.

d = {n: df2.ix[rows] 
     for n, rows in enumerate(df2.groupby('group_no').groups)}