### Question :

I have a pandas.DataFrame with measurements taken at consecutive points in time. Along with each measurement the system under observation had a distinct state at each point in time. Hence, the DataFrame also contains a column with the state of the system at each measurement. State changes are much slower than the measurement interval. As a result, the column indicating the states might look like this (index: state):

```
1: 3
2: 3
3: 3
4: 3
5: 4
6: 4
7: 4
8: 4
9: 1
10: 1
11: 1
12: 1
13: 1
```

Is there an easy way to retrieve the indices of each segment of consecutively equal states. That means I would like to get something like this:

```
[[1,2,3,4], [5,6,7,8], [9,10,11,12,13]]
```

The result might also be in something different than plain lists.

The only solution I could think of so far is manually iterating over the rows, finding segment change points and reconstructing the indices from these change points, but I have the hope that there is an easier solution.

##
Answer #1:

One-liner:

```
df.reset_index().groupby('A')['index'].apply(np.array)
```

Code for example:

```
In [1]: import numpy as np
In [2]: from pandas import *
In [3]: df = DataFrame([3]*4+[4]*4+[1]*4, columns=['A'])
In [4]: df
Out[4]:
A
0 3
1 3
2 3
3 3
4 4
5 4
6 4
7 4
8 1
9 1
10 1
11 1
In [5]: df.reset_index().groupby('A')['index'].apply(np.array)
Out[5]:
A
1 [8, 9, 10, 11]
3 [0, 1, 2, 3]
4 [4, 5, 6, 7]
```

You can also directly access the information from the groupby object:

```
In [1]: grp = df.groupby('A')
In [2]: grp.indices
Out[2]:
{1L: array([ 8, 9, 10, 11], dtype=int64),
3L: array([0, 1, 2, 3], dtype=int64),
4L: array([4, 5, 6, 7], dtype=int64)}
In [3]: grp.indices[3]
Out[3]: array([0, 1, 2, 3], dtype=int64)
```

To address the situation that DSM mentioned you could do something like:

```
In [1]: df['block'] = (df.A.shift(1) != df.A).astype(int).cumsum()
In [2]: df
Out[2]:
A block
0 3 1
1 3 1
2 3 1
3 3 1
4 4 2
5 4 2
6 4 2
7 4 2
8 1 3
9 1 3
10 1 3
11 1 3
12 3 4
13 3 4
14 3 4
15 3 4
```

Now groupby both columns and apply the lambda function:

```
In [77]: df.reset_index().groupby(['A','block'])['index'].apply(np.array)
Out[77]:
A block
1 3 [8, 9, 10, 11]
3 1 [0, 1, 2, 3]
4 [12, 13, 14, 15]
4 2 [4, 5, 6, 7]
```

##
Answer #2:

You could use np.diff() to test where a segment starts/ends and iterate over those results. Its a very simple solution, so probably not the most performent one.

```
a = np.array([3,3,3,3,3,4,4,4,4,4,1,1,1,1,4,4,12,12,12])
prev = 0
splits = np.append(np.where(np.diff(a) != 0)[0],len(a)+1)+1
for split in splits:
print np.arange(1,a.size+1,1)[prev:split]
prev = split
```

Results in:

```
[1 2 3 4 5]
[ 6 7 8 9 10]
[11 12 13 14]
[15 16]
[17 18 19]
```