Splitting data in Pandas/Python

You can use groupby here, using the compare-cumsum-groupby pattern (here let's say that the column with the Es is called "letter"):

>>> grouped = df.groupby((df["letter"] == "E").cumsum())
>>> frames = [g for k,g in grouped]
>>> for frame in frames:
...     print(frame)
...     print("--")
...     
  letter
0      E
1      T
2      U
--
  letter
3      E
4      M
--
  letter
5      E
--
  letter
6      E
--

This works because we can compare everything to E, creating a Series of booleans:

>>> df["letter"] == "E"
0     True
1    False
2    False
3     True
4    False
5     True
6     True
Name: letter, dtype: bool

and then if we take the cumulative sum of that we get

>>> (df["letter"] == "E").cumsum()
0    1
1    1
2    1
3    2
4    2
5    3
6    4
Name: letter, dtype: int32

where each new group has its own number. Reading the split-apply-combine section of the documentation is probably a good idea-- you might not even need to break everything up into subframes if the operation you want to perform on the groups is already supported.

You can first find indexes of rows with that value, then you can iterate through them and get slices of the dataframe:

def break_df_by(df, column, value):  
    ret = []
    indexes = df[df.ix[:, column] == value].index #[2, 5, 10] in the example below
    for n,i in enumerate(indexes):
        if n == 0: #if we want values before first 'E'
            ret.append(df[:i])
        elif n == len(indexes) - 1: #if we want values after last 'E'
            ret.append(df[i:])
        elif n > 0:
            ret.append(df[indexes[n-1]:i])

    return ret

An example, dataframe is:

    1
0   T
1   V
2   E
3   T
4   V
5   E
6   T
7   U
8   N
9   V
10  E
11  T
12  U
13  N

Result:

break_df_by(df, 1, 'E')

[   1
 0  T
 1  V,    1
 2  E
 3  T
 4  V,    1
 5  E
 6  T
 7  U
 8  N
 9  V,     1
 10  E
 11  T
 12  U
 13  N]