I have a DataFrame Like following.
df = pd.DataFrame({'id' : [1,1,2,3,2],
'value' : ["a","b","a","a","c"], 'Time' : ['6/Nov/2012 23:59:59 -0600','6/Nov/2012 00:00:05 -0600','7/Nov/2012 00:00:09 -0600','27/Nov/2012 00:00:13 -0600','27/Nov/2012 00:00:17 -0600']})
I need to get an output like following.
combined_id | enter time | exit time | time difference
combined_id should be created by grouping 'id' and 'value'
g = df.groupby(['id', 'value'])
Following doesn’t work with grouping by two columns. (How to use first() and last() here as enter and exit times?)
df['enter'] = g.apply(lambda x: x.first())
To get difference would following work?
df['delta'] = (df['exit']-df['enter'].shift()).fillna(0)
First ensure you're column is a proper datetime column:
In [11]: df['Time'] = pd.to_datetime(df['Time'])
Now, you can do the groupby and use agg with the first and last groupby methods:
In [12]: g = df.groupby(['id', 'value'])
In [13]: res = g['Time'].agg({'first': 'first', 'last': 'last'})
In [14]: res = g['Time'].agg({'enter': 'first', 'exit': 'last'})
In [15]: res['time_diff'] = res['exit'] - res['enter']
In [16]: res
Out[16]:
exit enter time_diff
id value
1 a 2012-11-06 23:59:59 2012-11-06 23:59:59 0 days
b 2012-11-06 00:00:05 2012-11-06 00:00:05 0 days
2 a 2012-11-07 00:00:09 2012-11-07 00:00:09 0 days
c 2012-11-27 00:00:17 2012-11-27 00:00:17 0 days
3 a 2012-11-27 00:00:13 2012-11-27 00:00:13 0 days
Note: this is a bit of a boring example since there is only one item in each group...
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