Pandas: Combine resampling with groupby and calculate time differences - group-by

Question

I am doing data analysis with trading data. I would like to use Pandas in order to examine the times when the traders are active.
In particular, I try to extract the difference in minutes between the dates of every first trade of every trader for each day and cumulate it to a monthly basis
The data looks like this:
Timestamp (Datetime) | Buyer | Volume
--------------------------------------
2012-01-01 09:00:00 | John | 10
2012-01-01 10:00:00 | Mark | 10
2012-01-01 16:00:00 | Mark | 10
2012-01-01 11:00:00 | Kevin | 10
2012-02-01 10:00:00 | Mark | 10
2012-02-01 09:00:00 | John | 10
2012-02-01 17:00:00 | Mark | 10
Right now I use resampling to retrieve the first trade on a daily basis. However, I want to group also by the buyer to calculate the differences in their trading dates. Like this
Timestamp (Datetime) | Buyer | Volume
--------------------------------------
2012-01-01 09:00:00 | John | 10
2012-01-01 10:00:00 | Mark | 10
2012-01-01 11:00:00 | Kevin | 10
2012-01-02 10:00:00 | Mark | 10
2012-01-02 09:00:00 | John | 10
Overall I am looking to calculate the differences in minutes between the first trades on a daily basis for each trader.
Update
For example in the case of John on the 2012-01-01: Dist = 60 (Diff John-Mark) + 120 (Diff John-Kevin) = 180
I would highly appreciate if anyone has an idea how to do this.
Thank you

Answer 1

Your original frame (the resampled one)
In [71]: df_orig
Out[71]:
buyer date volume
0 John 2012-01-01 09:00:00 10
1 Mark 2012-01-01 10:00:00 10
2 Kevin 2012-01-01 11:00:00 10
3 Mark 2012-01-02 10:00:00 10
4 John 2012-01-02 09:00:00 10
Set the index to the date column, keeping the date column in place
In [75]: df = df_orig.set_index('date',drop=False)
Create this aggregation function
def f(frame):
frame.sort('date',inplace=True)
frame['start'] = frame.date.iloc[0]
return frame
Groupby the single date
In [74]: x = df.groupby(pd.TimeGrouper('1d')).apply(f)
Create the differential in minutes
In [86]: x['diff'] = (x.date-x.start).apply(lambda x: float(x.item().total_seconds())/60)
In [87]: x
Out[87]:
buyer date volume start diff
date
2012-01-01 2012-01-01 09:00:00 John 2012-01-01 09:00:00 10 2012-01-01 09:00:00 0
2012-01-01 10:00:00 Mark 2012-01-01 10:00:00 10 2012-01-01 09:00:00 60
2012-01-01 11:00:00 Kevin 2012-01-01 11:00:00 10 2012-01-01 09:00:00 120
2012-01-02 2012-01-02 09:00:00 John 2012-01-02 09:00:00 10 2012-01-02 09:00:00 0
2012-01-02 10:00:00 Mark 2012-01-02 10:00:00 10 2012-01-02 09:00:00 60
Here's the explanation. We use the TimeGrouper to have the grouping by date, where a frame is passed to the function f. This function, then uses the first date of the day (the sort is necessary here). You subtract this from the date on the entry to get a timedelta64, which is then massaged to minutes (this is a bit hacky right now because of some numpy issues, should be more natural in 0.12)
Thanks for you update, I originally thought you wanted the diff per buyer, not from the first buyer, but that's just a minor tweak.
Update:
To track the buyer name as well (which corresponds to the start date), just include
it in the function f
def f(frame):
frame.sort('date',inplace=True)
frame['start'] = frame.date.iloc[0]
frame['start_buyer'] = frame.buyer.iloc[0]
return frame
Then can groupby on this at the end:
In [14]: x.groupby(['start_buyer']).sum()
Out[14]:
diff
start_buyer
John 240