API experiment:  lambda grouper based on sliding window

related #3066, #414, looking for an expressive, flexible 
way of splitting frames into multiple frames based on 
predicate functions.

Not optimized, just trying it out.

- The number of args in the predicate function determines the width
of the sliding window 
- The df.groupby() `offset` kwd determines window offset
relative to current element.
- The df.groupby() `fill` kwd determines the value provided
to the predicate instead of "phantom" locations when the window
slides past the edge of of array.

The predicate function also gets a `_` variable injected into it's
context, which intially 0, and is updated after each call with the value
just returned. That is, `_`  holds the "group label of the previous item.
By updating the value as we go along we can always compute the
label for the "next" group. Somewhat like `reduce`

#3066 was prompted by this [SO question](http://stackoverflow.com/questions/13353233/best-way-to-split-a-dataframe-given-an-edge/15449992#15449992)
Note how concise doing that is, using this idiom (much more complex 
splitting is easy as well):

```python
In [1]: def prnt_dfs(g):
   ...:    for d in zip(*list(g))[-1]:
   ...:        print d 
   ...:        print ""
   ...: 

In [23]: ix = ['A','A','A','B','A','A','B','A','A']
    ...: df = pd.DataFrame(np.random.randn(9),index=ix)
    ...: prnt_dfs(df.groupby(lambda p,c: _+int(p=='B'),offset=-1))
          0
A -0.214422
A  1.373191
A -1.209503
B -0.258765

          0
A  1.014395
A -0.381113
B  0.707706

          0
A -2.038331
A  1.719545

```


```python
In [2]: df = pd.DataFrame(range(4),index=['X0','X1','B','X3'])
   ...: 

# start new df if current is B
In [3]: g=(df.groupby(lambda c,n: _+int(c == 'B'))) 
   ...: prnt_dfs(g)
   ...: 
    0
X0  0
X1  1

    0
B   2
X3  3

# start new df if next is B
In [4]: g=(df.groupby(lambda c,n: _+int(n == 'B'))) 
   ...: prnt_dfs(g)
   ...: 
    0
X0  0

    0
X1  1
B   2
X3  3

# start new df if prev was B
In [5]: g=(df.groupby(lambda p,c: _+int(p == 'B'),offset=-1)) 
   ...: prnt_dfs(g)
   ...: 
    0
X0  0
X1  1
B   2

    0
X3  3

# start new df if prevprev was B
In [6]: g=(df.groupby(lambda p,pp,c: _+int(pp == 'B'),offset=-1)) 
   ...: prnt_dfs(g)
   ...: 
    0
X0  0
X1  1
B   2

    0
X3  3


In [7]: # start new df if value 2 ahead >5 or (current + previous) <= 3
   ...: df = pd.DataFrame(range(10),index=[1,2,5,3,7,8,1,1,3,1])
   ...: g=(df.groupby(lambda p,c,n,nn: _+int(c+p<=3 or nn == 'C'),offset=-1,fill=0)) 
   ...: prnt_dfs(g)
   0
1  0

   0
2  1

   0
5  2

   0
3  3
7  4
8  5
1  6
1  7
3  8

   0
1  9



```