PERF: transform speedup
http://stackoverflow.com/questions/22072943/faster-way-to-transform-group-with-mean-value-in-pandas/22073449#22073449

I think this actually will work in a general case. but prob only really makes sense when you have a cythonized work function (or a ufunc) that is much faster than iteratively calling the groups.

```
np.random.seed(0)

N = 120000
N_TRANSITIONS = 1400

# generate groups
transition_points = np.random.permutation(np.arange(N))[:N_TRANSITIONS]
transition_points.sort()
transitions = np.zeros((N,), dtype=np.bool)
transitions[transition_points] = True
g = transitions.cumsum()

df = pd.DataFrame({ "signal" : np.random.rand(N)})
```

```
In [44]: grp = df["signal"].groupby(g)

In [45]: result2 = df["signal"].groupby(g).transform(np.mean)

In [47]: %timeit df["signal"].groupby(g).transform(np.mean)
1 loops, best of 3: 535 ms per loop
```

Using broadcasting

```
 In [43]: result = pd.concat([ Series([r]*len(grp.groups[i])) for i, r in enumerate(grp.mean().values) ],ignore_index=True)

In [42]: %timeit pd.concat([ Series([r]*len(grp.groups[i])) for i, r in enumerate(grp.mean().values) ],ignore_index=True)
10 loops, best of 3: 119 ms per loop

In [46]: result.equals(result2)
Out[46]: True
```

I think you might need to set the index of the returned on the broadcast result (it happens to work here because its a default index

```
result = pd.concat([ Series([r]*len(grp.groups[i])) for i, r in enumerate(grp.mean().values) ],ignore_index=True)
result.index = df.index
```

Final Result is best

```
pd.Series(np.repeat(grp.mean().values, grp.count().values))
```
