频率计数唯一值熊猫

Question

I have a Pandas Series as follow :

2014-05-24 23:59:49     1.3
2014-05-24 23:59:50    2.17
2014-05-24 23:59:50    1.28
2014-05-24 23:59:51    1.30
2014-05-24 23:59:51    2.17
2014-05-24 23:59:53    2.17
2014-05-24 23:59:58    2.17
Name: api_id, Length: 483677

I'm trying to count for each id the frequency per day. For now I'm doing this :

count = {}
for x in apis.unique():
    count[x] = apis[apis == x].resample('D','count')
count_df = pd.DataFrame(count)

That gives me what I want which is :

            ...    2.13   2.17   2.4  2.6  2.7  3.5(user)  3.9  4.2   5.1  5.6  
timestamp   ...                                                                 
2014-05-22  ...     391  49962  3727  161    2        444  113   90  1398   90  
2014-05-23  ...     450  49918  3861  187    1        450  170   90   629   90  
2014-05-24  ...     396  46359  3603  172    3        513  171   89   622   90  

But is there a way to do so without the for loop ?

Answer 1

You can use the value_counts function for this ( docs ), applying this after a groupby (which is similar to the resample('D') you did, but resample is expecting an aggregated output so we have to use the more general groupby in this case). With a small example:

In [16]: s = pd.Series([1,1,2,2,1,2,5,6,2,5,4,1], index=pd.date_range('2012-01-01', periods=12, freq='8H'))

In [17]: counts = s.groupby(pd.Grouper(freq='D')).value_counts()

In [18]: counts
Out[18]: 
2012-01-01  1    2
            2    1
2012-01-02  2    2
            1    1
2012-01-03  2    1
            6    1
            5    1
2012-01-04  1    1
            5    1
            4    1
dtype: int64

To get this in the desired format, you can just unstack this (move the second level row indices to the columns):

In [19]: counts.unstack()
Out[19]: 
             1   2   4   5   6
2012-01-01   2   1 NaN NaN NaN
2012-01-02   1   2 NaN NaN NaN
2012-01-03 NaN   1 NaN   1   1
2012-01-04   1 NaN   1   1 NaN

Note: for the use of groupby(pd.Grouper(freq='D')) you need pandas 0.14. If you have al older version, you can use groupby(pd.TimeGrouper(freq='D')) to obtain exactly the same. This is also similar to doing groupby(s.index.date) (with the difference you have then datetime.date objects in the index).