Question

我有以下函数，其中df是一个159538行x 3列的pandas数据帧：

dfs = []
for i in df['email_address']:
    data = df[df['email_address'] == i]
    data['difference'] = data['ts_placed'].diff().astype('timedelta64[D]')
    repeat = []
    for a in data['difference']:
        if a > 10:
            repeat.append(0)
        elif a <= 10:
            repeat.append(1)
        else:
            repeat.append(0)
    data['repeat'] = repeat
    dfs.append(data)

该功能运行速度极慢。我想通过使用多处理来加快这个过程。这个SO question显示了如何在R中执行此操作。什么是python的等效代码？

这是运行后的数据样本：

df['difference'] = df.groupby('email_address')['ts_placed'].diff()



df
Out[6]: 
                           email_address           ts_placed       difference
0                  aaaaaaaaaaaaa@sky.com 2015-08-06 00:00:34              NaT
1              dfdfdfdfdfd@babcock.co.uk 2015-08-06 00:05:38              NaT
2        littlemifddreen85@hotmail.co.uk 2015-08-06 00:09:20              NaT
3                    smifdfddfms@aol.com 2015-08-06 00:10:01              NaT
4       terry.wfdfdfdfdfy-holdings.co.uk 2015-08-06 00:14:00              NaT
5              r.dfdfdfdfd16@hotmail.com 2015-08-06 00:14:00              NaT
6                 kdfdfdf979@outlook.com 2015-08-06 00:14:00              NaT
7           dd@ggggggggggg.eclipse.co.uk 2015-08-06 00:14:20              NaT
8                   gggz45@hotmail.co.uk 2015-08-06 00:14:43              NaT
9               gggggggggi@hotmail.co.uk 2015-08-06 00:17:03              NaT
10             mggggggggyke1@hotmail.com 2015-08-06 00:17:58              NaT
...
22                   ffdddfddd@yahoo.com 2015-08-06 00:46:12  0 days 00:04:15

Answer 1

IIUC然后您可以执行以下操作：

df['difference'] = df.groupby('email_address')['ts_placed'].diff()

df['repeat'] = df.groupby('email_address')['difference'].transform(lambda x: (x < 10).cumcount())

如何加速循环python

1 个答案: