将SFrame转换为输入数据集Sframe

时间:2016-06-15 14:39:43

标签: performance graphlab sframe

我将输入日志转换为输入数据集的方法非常糟糕。 我有一个SFrame sf,格式如下:

user_id     int
timestamp   datetime.datetime
action      int
reasoncode  str

操作列占用9个值,范围从1到9。

因此,每个user_id可以执行多次操作,不止一次。

我正在尝试从sf获取所有唯一的user_id,并按以下方式创建op_sf:

y = 225

def calc_class(a,x):
  diffd = a['timestamp'].apply(lambda x: (dte - x).days)
  g = 0
  b = 0
  for i in diffd:
    if i > y:
    g += 1
  else:
    b += 1
  if b>= x:
    return 4
  elif b!= 0:
    return 3
  elif g>= 0:
    return 2
  else:
    return 1

l1 = []
ids = z['user_id'].unique()

for idd in ids:
 temp = sf[sf['user_id']== idd]
 zero1 = temp[temp['action'] == 1]
 zero2 = temp[temp['action'] == 2]
 zero3 = temp[temp['action'] == 3]
 zero4 = temp[temp['action'] == 4]
 zero5 = temp[temp['action'] == 5]
 zero6 = temp[temp['action'] == 6]
 zero7 = temp[temp['action'] == 7]
 zeroh8 = temp[temp['reasoncode'] == 'xyz']
 zero9 = temp[temp['reasoncode'] == 'abc']
 /* I'm getting clas1 to clas9 from function calc_class for each action
    clas1 to clas9 are 4 integers ranging from 1 to 4
 */ 
 clas1 = calc_class(zero1,2)
 clas2 = calc_class(zero2,2)
 clas3 = calc_class(zero3,2)
 clas4 = calc_class(zero4,2)
 clas5 = calc_class(zero5,2)
 clas6 = calc_class(zero6,2)
 clas7 = calc_class(zero7,2)
 clas8 = calc_class(zero8,2)
 clas9 = calc_class(zero9,2)
 l1.append([idd,clas1,clas2,clas3,clas4,clas5*(-1),clas6*(-1),clas7*(-1),clas8*(-1),clas9])

我想知道这是否是最快的方法。具体来说,如果可以在不生成zero1到zero9 SFrame的情况下执行相同的操作。

示例sf:

user_id timestamp action reasoncode 
574 23/09/15 12:43  1   None
574 23/09/15 11:15  2   None
574 06/10/15 11:20  2   None
574 06/10/15 11:21  3   None
588 04/11/15 10:00  1   None
588 05/11/15 10:00  1   None
555 15/12/15 13:00  1   None
585 22/12/15 17:30  1   None
585 15/01/16 07:44  7   xyz
588 06/01/16 08:10  7   abc

l1对应上面的sf:

574 1   2   2   0   0   0   0   0   0
588 3   0   0   0   0   0   0   0   3
555 3   0   0   0   0   0   0   0   0
585 3   0   0   0   0   0   0   3   0

1 个答案:

答案 0 :(得分:1)

我认为您的逻辑相对复杂,但在整个数据集上使用逐列操作仍然更有效,而不是为每个用户提取行的子集。关键工具包括SFrame.groupbySFrame.applySFrame.unstackSFrame.unpack。 API文档在这里:

https://dato.com/products/create/docs/generated/graphlab.SFrame.html

这是一个使用比您的示例稍微简单的数据的解决方案,以及稍微简单的逻辑来编码旧操作和新操作。

# Set up and make the data
import graphlab as gl
import datetime as dt

sf = gl.SFrame({'user': [574, 574, 574, 588, 588, 588],
                'timestamp': [dt.datetime(2015, 9, 23), dt.datetime(2015, 9, 23),
                              dt.datetime(2015, 10, 6), dt.datetime(2015, 11, 4),
                              dt.datetime(2015, 11, 5), dt.datetime(2016, 1, 6)],
                'action': [1, 2, 3, 1, 1, 7]})

# Count old vs. new actions.
sf['days_elapsed'] = (dt.datetime.today() - sf['timestamp']) / (3600 * 24)
sf['old_threshold'] = sf['days_elapsed'] > 225

aggregator = {'total_count': gl.aggregate.COUNT('user'),
              'old_count': gl.aggregate.SUM('old_threshold')}
grp = sf.groupby(['user', 'action'], aggregator)

# Code the actions according to old vs. new. Use your own logic here.
grp['action_code'] = grp.apply(
                       lambda x: 2 if x['total_count'] > x['old_count'] else 1)
grp = grp[['user', 'action', 'action_code']]

# Reshape the results into columns.
sf_new = (grp.unstack(['action', 'action_code'], new_column_name='action_code')
             .unpack('action_code'))

# Fill in zeros for entries with no actions.
for c in sf_new.column_names():
    sf_new[c] = sf_new[c].fillna(0)

print sf_new
+------+---------------+---------------+---------------+---------------+
| user | action_code.1 | action_code.2 | action_code.3 | action_code.7 |
+------+---------------+---------------+---------------+---------------+
| 588  |       2       |       0       |       0       |       2       |
| 574  |       1       |       1       |       1       |       0       |
+------+---------------+---------------+---------------+---------------+
[2 rows x 5 columns]