issue when shifting with Timedelta in a groupby
Hello the awesome Pandas team!

Consider the example below

```
data = pd.DataFrame({'mydate' : [pd.to_datetime('2016-06-06'),
                                 pd.to_datetime('2016-06-08'),
                                 pd.to_datetime('2016-06-09'),
                                 pd.to_datetime('2016-06-10'),
                                 pd.to_datetime('2016-06-12'),
                                 pd.to_datetime('2016-06-13')],
                     'myvalue' : [1, 2, 3, 4, 5, 6],
                     'group' : ['A', 'A', 'A', 'B', 'B', 'B']})

data.set_index('mydate', inplace = True)
Out[58]: 
           group  myvalue
mydate                   
2016-06-06     A        1
2016-06-08     A        2
2016-06-09     A        3
2016-06-10     B        4
2016-06-12     B        5
2016-06-13     B        6

```
Now I need to compute the difference between the current value of `myvalue` and its lagged value, where by lagged I actually mean **lagged by 1 day** (if possible).

So this returns a result, but its not what I need

```
data['delta_one'] = data.groupby('group').myvalue.transform(lambda x: x - x.shift(1))

data
Out[56]: 
           group  myvalue  delta_one
mydate                              
2016-06-06     A        1        nan
2016-06-08     A        2     1.0000
2016-06-09     A        3     1.0000
2016-06-10     B        4        nan
2016-06-12     B        5     1.0000
2016-06-13     B        6     1.0000
```


This is what I need, but it does not work

```
data['delta_two'] = data.groupby('group').myvalue.transform(lambda x: x - x.shift(1, pd.Timedelta('1 days')))

  File "C:\Users\john\AppData\Local\Continuum\anaconda3\lib\site-packages\pandas\core\internals.py", line 120, in __init__
    len(self.mgr_locs)))

ValueError: Wrong number of items passed 4, placement implies 3
```

Any ideas? Is this a bug? I think I am using the correct pandonic syntax here.

Thanks!
