python - Pandas Efficient Count 列中的唯一值然后找到该计数的最大值

Question

数据框 (df) 如下所示:

    Date                Caller  Called
0   2011-01-01 00:00:00 Sarah   Claire
1   2011-01-01 00:00:00 Sarah   Ryan
2   2011-01-01 00:00:00 Sarah   Alex
3   2011-01-02 00:00:00 Sarah   Max
4   2011-01-02 00:00:00 Sarah   Phoebe


number_date =  df.groupby(['Caller',pd.DatetimeIndex(df['Date']).normalize()])
number_date['Called'].unique().apply(lambda x: len(x))

这符合我的预期。她在第 1 天打了 3 个电话，在第 2 天打了 2 个电话:

Caller            
Sarah   2011-01-01    3
        2011-01-02    2

如何修改为:

Caller   MaxCallsOneDay
Sarah    3

Answer 1

你可以试试nunique和 nlargest :

print df
        Date  Caller  Called
0 2011-01-01   Sarah  Claire
1 2011-01-01   Sarah    Ryan
2 2011-01-01  Sarah1    Ryan
3 2011-01-01  Sarah1   Ryan1
4 2011-01-01  Sarah1    Ryan
5 2011-01-01   Sarah    Alex
6 2011-01-02   Sarah     Max

number_date=df.groupby(['Caller',pd.DatetimeIndex(df['Date']).normalize()])['Called'].nunique()
              .groupby(level=0).nlargest(1).reset_index(level=1, drop=True).reset_index()

number_date.columns = ['Caller','Date', 'MaxCallsOneDay']
print number_date
   Caller       Date  MaxCallsOneDay
0   Sarah 2011-01-01               3
1  Sarah1 2011-01-01               2

如果您不需要 Date 列，请使用 max :

df = df.groupby(['Caller',pd.DatetimeIndex(df['Date']).normalize()])['Called'].nunique()
       .groupby(level=0).max().reset_index(name='MaxCallsOneDay')
print df
   Caller  MaxCallsOneDay
0   Sarah               3
1  Sarah1               2

时间 - len(df) = 7k:

In [531]: %timeit df.groupby(['Caller',pd.DatetimeIndex(df['Date']).normalize()])['Called'].unique().apply(lambda x: len(x)).groupby(level=0).nlargest(1).reset_index(level=1, drop=True).reset_index()
The slowest run took 4.80 times longer than the fastest. This could mean that an intermediate result is being cached 
10 loops, best of 3: 8.58 ms per loop

In [532]: %timeit df.groupby(['Caller',pd.DatetimeIndex(df['Date']).normalize()])['Called'].nunique().groupby(level=0).nlargest(1).reset_index(level=1, drop=True).reset_index()
100 loops, best of 3: 7.07 ms per loop

In [547]: %timeit df.groupby(['Caller',pd.DatetimeIndex(df['Date']).normalize()])['Called'].nunique().groupby(level=0).max().reset_index(name='MaxCallsOneDay')
100 loops, best of 3: 3.52 ms per loop