issue when shifting with Timedelta in a groupby #20492
Description
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!