**EDIT 2: To answer the OP new requirement**

```
ranges = []
for key, group in groupby(enumerate(data), lambda (index, item): index - item):
group = map(itemgetter(1), group)
if len(group) > 1:
ranges.append(xrange(group[0], group[-1]))
else:
ranges.append(group[0])
```

Output:

```
[xrange(2, 5), xrange(12, 17), 20]
```

You can replace xrange with range or any other custom class.

Python docs have a very neat recipe for this:

```
from operator import itemgetter
from itertools import groupby
data = [2, 3, 4, 5, 12, 13, 14, 15, 16, 17]
for k, g in groupby(enumerate(data), lambda (i,x):i-x):
print(map(itemgetter(1), g))
```

Output:

```
[2, 3, 4, 5]
[12, 13, 14, 15, 16, 17]
```

If you want to get the exact same output, you can do this:

```
ranges = []
for k, g in groupby(enumerate(data), lambda (i,x):i-x):
group = map(itemgetter(1), g)
ranges.append((group[0], group[-1]))
```

output:

```
[(2, 5), (12, 17)]
```

**EDIT:** The example is already explained in the documentation but maybe I should explain it more:

The key to the solution is

differencing with a range so that

consecutive numbers all appear in same

group.

If the data was: `[2, 3, 4, 5, 12, 13, 14, 15, 16, 17]`

Then `groupby(enumerate(data), lambda (i,x):i-x)`

is equivalent of the following:

```
groupby(
[(0, 2), (1, 3), (2, 4), (3, 5), (4, 12),
(5, 13), (6, 14), (7, 15), (8, 16), (9, 17)],
lambda (i,x):i-x
)
```

The lambda function subtracts the element index from the element value. So when you apply the lambda on each item. You’ll get the following keys for groupby:

```
[-2, -2, -2, -2, -8, -8, -8, -8, -8, -8]
```

groupby groups elements by equal key value, so the first 4 elements will be grouped together and so forth.

I hope this makes it more readable.

`python 3`

version may be helpful for beginners

import the libraries required first

```
from itertools import groupby
from operator import itemgetter
ranges =[]
for k,g in groupby(enumerate(data),lambda x:x[0]-x[1]):
group = (map(itemgetter(1),g))
group = list(map(int,group))
ranges.append((group[0],group[-1]))
```