基于熊猫的多种条件的Grouby和计数总和

给定一个数据框，我如下：

import pandas as pd
import datetime

df = pd.DataFrame([[2, 3],[2, 1],[2, 1],[3, 4],[3, 1],[3, 1],[3, 1],[3, 1],[4, 2],[4, 1],[4, 1],[4, 1]], columns=['id', 'count'])
df['date'] = [datetime.datetime.strptime(x,'%Y-%m-%d %H:%M:%S') for x in 
              ['2016-12-28 15:17:00','2016-12-28 15:29:00','2017-01-05 09:32:00','2016-12-03 18:10:00','2016-12-10 11:31:00',
                '2016-12-14 09:32:00','2016-12-18 09:31:00','2016-12-22 09:32:00','2016-11-28 15:31:00','2016-12-01 16:11:00',
               '2016-12-10 09:31:00','2016-12-13 12:06:00']]

我想根据条件使用grouby：对于数据相同的数据id，如果它们的日期差小于4天，则将它们视为相同的组，否则创建一个新列new_id，然后我将对grouby进行计数并基于计算总和new_id。

我可以通过以下代码得到结果，但是它太慢了，我怎样才能使其更有效？

df.sort_values(by=['id', 'date'], ascending = [True, False], inplace = True)
df['id'] = df['id'].astype(str)
df['id_up'] = df['id'].shift(-1)
df['id_down'] = df['id'].shift(1)
df['date_up'] = df['date'].shift(-1)       
df['date_diff'] = df.apply(lambda df: (df['date'] - df['date_up'])/datetime.timedelta(days=1) if df['id'] == df['id_up'] else 0, axis=1)
df = df.reset_index()
df = df.drop(['index','id_up','id_down','date_up'],axis=1)
df['new'] = ''
for i in range(df.shape[0]):   
    if i == 0:
        df.loc[i,'new'] = 1
    else:
        if df.loc[i,'id'] != df.loc[i-1,'id']:
            df.loc[i,'new'] = 1
        else:
            if df.loc[i-1,'date_diff'] <= 4:
                df.loc[i,'new'] = df.loc[i-1,'new']
            else:
                df.loc[i,'new'] = df.loc[i-1,'new'] + 1
df['new'] = df['id'].astype(str) + '-' +  df['new'].astype(str)
df1 = df.groupby('new')['date'].min()
df1 = df1.reset_index()
df1.rename(columns={"date": "first_date"}, inplace=True)
df = pd.merge(df, df1, on='new')
df1 = df.groupby('new')['date'].max()
df1 = df1.reset_index()
df1.rename(columns={"date": "last_date"}, inplace=True)
df = pd.merge(df, df1, on='new')
df1 = df.groupby('new')['count'].sum()
df1 = df1.reset_index()
df1.rename(columns={"count": "count_sum"}, inplace=True)
df = pd.merge(df, df1, on='new')
print(df)

出：

    id  count         date      date_diff   new       first_date       last_date    count_sum
0   2   1   2017-01-05 09:32:00 7.752083    2-1 2017-01-05 09:32:00 2017-01-05 09:32:00 1
1   2   1   2016-12-28 15:29:00 0.008333    2-2 2016-12-28 15:17:00 2016-12-28 15:29:00 4
2   2   3   2016-12-28 15:17:00 0.000000    2-2 2016-12-28 15:17:00 2016-12-28 15:29:00 4
3   3   1   2016-12-22 09:32:00 4.000694    3-1 2016-12-22 09:32:00 2016-12-22 09:32:00 1
4   3   1   2016-12-18 09:31:00 3.999306    3-2 2016-12-10 11:31:00 2016-12-18 09:31:00 3
5   3   1   2016-12-14 09:32:00 3.917361    3-2 2016-12-10 11:31:00 2016-12-18 09:31:00 3
6   3   1   2016-12-10 11:31:00 6.722917    3-2 2016-12-10 11:31:00 2016-12-18 09:31:00 3
7   3   4   2016-12-03 18:10:00 0.000000    3-3 2016-12-03 18:10:00 2016-12-03 18:10:00 4
8   4   1   2016-12-13 12:06:00 3.107639    4-1 2016-12-10 09:31:00 2016-12-13 12:06:00 2
9   4   1   2016-12-10 09:31:00 8.722222    4-1 2016-12-10 09:31:00 2016-12-13 12:06:00 2
10  4   1   2016-12-01 16:11:00 3.027778    4-2 2016-11-28 15:31:00 2016-12-01 16:11:00 3
11  4   2   2016-11-28 15:31:00 0.000000    4-2 2016-11-28 15:31:00 2016-12-01 16:11:00 3