根据超过 2 个组的 % 随机分配控制组和治疗组

Assign control vs. treatment groupings randomly based on % for more than 2 groups

小猪回避我之前的问题

感谢@maxU,我知道如何将随机 control/treatment 分组分配给 2 个组;但是如果我有 3 个或更多组怎么办?

例如:

df.head()

customer_id | Group | many other columns
ABC             1
CDE             3
BHF             2
NID             1
WKL             3
SDI             2
JSK             1
OSM             3
MPA             2
MAD             1

pd.pivot_table(df,index=['Group'],values=["customer_id"],aggfunc=lambda x: len(x.unique()))

Group 1  : 270
Group 2  : 180
Group 3  : 330

当我只有两组时,我有一个很好的答案:

df['Flag'] = df.groupby('Group')['customer_id']\
             .transform(lambda x: np.random.choice(['Control','Test'], len(x), 
                                                  p=[.5,.5] if x.name==1 else [.4,.6]))

但是如果我想这样拆分呢:

@MaxU 的回答很好,可惜分割的不准确

d = {1:[.5,.5], 2:[.4,.6], 3:[.2,.8]}

df['Flag'] = df.groupby('Group')['customer_id'] \
             .transform(lambda x: np.random.choice(['Control','Test'], len(x), p=d[x.name]))

当我测试它时,我没有得到精确的拆分。

pd.pivot_table(df,index=['Group'],values=["customer_id"],columns=['Flag'], aggfunc=lambda x: len(x.unique()))

           Control  Treatment
Group 1:    138       132
Group 2:    78        102
Group 3:    79        251

第 1 组应为 135/135。

In [13]: df
Out[13]:
  customer_id  Group
0         ABC      1
1         CDE      3
2         BHF      2
3         NID      1
4         WKL      3
5         SDI      2
6         JSK      1
7         OSM      3
8         MPA      2
9         MAD      1

In [14]: d = {1:[.5,.5], 2:[.4,.6], 3:[.2,.8]}

In [15]: df['Flag'] = \
    ...: df.groupby('Group')['customer_id'] \
    ...:   .transform(lambda x: np.random.choice(['Control','Test'], len(x), p=d[x.name]))
    ...:

In [16]: df
Out[16]:
  customer_id  Group     Flag
0         ABC      1  Control
1         CDE      3     Test
2         BHF      2     Test
3         NID      1  Control
4         WKL      3  Control
5         SDI      2     Test
6         JSK      1     Test
7         OSM      3     Test
8         MPA      2  Control
9         MAD      1     Test

听起来您正在寻找一种方法来将您的 customer_id 分成精确的比例,而不是依赖机会。这是使用 pandas.qcutnp.random.permutation.

的一种方法
In [228]: df = pd.DataFrame({'customer_id': np.random.normal(size=10000), 
                             'group': np.random.choice(['a', 'b', 'c'], size=10000)})

In [229]: proportions = {'a':[.5,.5], 'b':[.4,.6], 'c':[.2,.8]}

In [230]: df.head()
Out[230]:
   customer_id group
0       0.6547     c
1       1.4190     a
2       0.4205     a
3       2.3266     a
4      -0.5691     b

In [231]: def assigner(gp):
     ...:     group = gp['group'].iloc[0]
     ...:     cut = pd.qcut(
                  np.arange(gp.shape[0]), 
                  q=np.cumsum([0] + proportions[group]), 
                  labels=range(len(proportions[group]))
              ).get_values()
     ...:     return pd.Series(cut[np.random.permutation(gp.shape[0])], index=gp.index, name='assignment')
     ...:

In [232]: df['assignment'] = df.groupby('group', group_keys=False).apply(assigner)

In [233]: df.head()
Out[233]:
   customer_id group  assignment
0       0.6547     c           1
1       1.4190     a           1
2       0.4205     a           0
3       2.3266     a           1
4      -0.5691     b           0

In [234]: (df.groupby(['group', 'assignment'])
             .size()
             .unstack()
             .assign(proportion=lambda x: x[0] / (x[0] + x[1])))
Out[234]:
assignment     0     1  proportion
group
a           1659  1658      0.5002
b           1335  2003      0.3999
c            669  2676      0.2000

这是怎么回事?

  1. 在每个组中我们调用函数 assigner
  2. assigner从预定义字典中抓取组名和比例调用pd.qcut拆分为0(control) 1(treatment)
  3. np.random.permutation 然后随机分配作业
  4. 将其创建为原始数据框中的新列