Python Pandas Dataframe GroupBy Size based on condition

Question

I have a dataframe 'df' that looks like this:

id  date1   date2
1   11/1/2016   11/1/2016
1   11/1/2016   11/2/2016
1   11/1/2016   11/1/2016
1   11/1/2016   11/2/2016
1   11/2/2016   11/2/2016
2   11/1/2016   11/1/2016
2   11/1/2016   11/2/2016
2   11/1/2016   11/1/2016
2   11/2/2016   11/2/2016
2   11/2/2016   11/2/2016

What I would like to do is to groupby the id, then get the size for each id where date1=date2. The result should look like:

id  samedate    count
1   11/1/2016    2 
1   11/2/2016    1 
2   11/1/2016    2 
2   11/2/2016    2 

I have tried this:

gb=df.groupby(id').apply(lambda x: x[x.date1== x.date2]['date1'].size())

And get this error:

TypeError: 'int' object is not callable

You could certainly flag each instance where the date1 and date2 are equal, then count those flags for each id by each samedate, but I have to believe there is a groupby option for this.

Answer 1

You can use boolean indexing first and then aggregate size:

df.date1 = pd.to_datetime(df.date1)
df.date2 = pd.to_datetime(df.date2)

df = df[df.date1 == df.date2]
gb=df.groupby(['id', 'date1']).size().reset_index(name='count')
print (gb)
   id      date1  count
0   1 2016-11-01      2
1   1 2016-11-02      1
2   2 2016-11-01      2
3   2 2016-11-02      2

Timings:

In [79]: %timeit (df[df.date1 == df.date2].groupby(['id', 'date1']).size().reset_index(name='count'))
100 loops, best of 3: 3.84 ms per loop

In [80]: %timeit (df.groupby(['id', 'date1']).apply(lambda x: (x['date1'] == x['date2']).sum()).reset_index())
100 loops, best of 3: 7.57 ms per loop

Code for timings:

#len df = 10k
df = pd.concat([df]*1000).reset_index(drop=True)
#print (df)

df.date1 = pd.to_datetime(df.date1)
df.date2 = pd.to_datetime(df.date2)