Python 大熊猫多栏排名和观察结果之间的标记关系

我有一个类似于：

   group name    sum count      max_size
 1     1 aaa       3     2            4
 2     1 bbb       3     1            4
 3     1 ccc       2     2            4
 4     1 ddd       2     2            4
 5     1 eee       1     0            4
 6     2 aaa       3     2            3
 7     2 bbb       3     1            3
 8     2 ccc       2     3            3
 9     2 ddd       2     1            3
10     3 aaa       3     4            4
11     3 bbb       3     2            4
12     3 ccc       2     5            4
13     3 ddd       2     1            4
14     3 eee       2     1            4
15     3 fff       2     1            4

我想根据这一决策推理为每个观察结果贴上标签：

• 首先按组排列groupby（），然后按和排列名称，然后按计数降序排列名称
• 在

  max_size

  中选择前n个元素，这是组中要选择的最大元素数

在类似

组2

的情况下，有一个待选择元素的最大大小（3）和3个清除候选元素

  group name  decision       sum count     max_size
1     2 aaa   winner           3     2            3
2     2 bbb   winner           3     1            3
3     2 ccc   winner           2     3            3
4     2 ddd   loser            2     1            3

aaa

、

bbb

和

ccc

是前三位的排序方式，先是

sum

，然后是

count

，而

ddd

则不在列

对于第3组，尽管：

  group name  decision          sum count     max_size
1     3 aaa   winner              3     4            4
2     3 bbb   winner              3     2            4
3     3 ccc   winner              2     5            4
4     3 ddd   unclear             2     1            4
5     3 eee   unclear             2     1            4
6     3 fff   unclear             2     1            4

aaa

、

bbb

、

ccc

是前三名，但第四名（假设max_size=4）尚不清楚

ddd

、

eee

和

fff

具有相同的计数和总和

我希望得出一个最终的数据框，将观察结果标记为：

   name  decision   sum count max_size
 1 aaa   winner       3     2        4
 2 bbb   winner       3     1        4
 3 ccc   unclear      2     2        4
 4 ddd   unclear      2     2        4
 5 eee   winner       1     0        4
 6 aaa   winner       3     2        3
 7 bbb   winner       3     1        3
 8 ccc   winner       2     3        3
 9 ddd   loser        2     1        3
10 aaa   winner       3     4        4
11 bbb   winner       3     2        4
12 ccc   winner       2     5        4
13 ddd   unclear      2     1        4
14 eee   unclear      2     1        4
15 fff   unclear      2     1        4

可复制示例：

{'group': {0: 1, 1: 1, 2: 1, 3: 1, 4: 1, 5: 2, 6: 2, 7: 2, 8: 2, 9: 3, 10: 3, 11: 3, 12: 3, 13: 3, 14: 3}, 'name': {0: 'aaa', 1: 'bbb', 2: 'ccc', 3: 'ddd', 4: 'eee', 5: 'aaa', 6: 'bbb', 7: 'ccc', 8: 'ddd', 9: 'aaa', 10: 'bbb', 11: 'ccc', 12: 'ddd', 13: 'eee', 14: 'fff'}, 'decision': {0: 'winner', 1: 'winner', 2: 'unclear', 3: 'unclear', 4: 'winner', 5: 'winner', 6: 'winner', 7: 'winner', 8: 'loser', 9: 'winner', 10: 'winner', 11: 'winner', 12: 'unclear', 13: 'unclear', 14: 'unclear'}, 'sum': {0: 3, 1: 3, 2: 2, 3: 2, 4: 1, 5: 3, 6: 3, 7: 2, 8: 2, 9: 3, 10: 3, 11: 2, 12: 2, 13: 2, 14: 2}, 'count': {0: 2, 1: 1, 2: 2, 3: 2, 4: 0, 5: 2, 6: 1, 7: 3, 8: 1, 9: 4, 10: 2, 11: 5, 12: 1, 13: 1, 14: 1}, 'max_size': {0: 4, 1: 4, 2: 4, 3: 4, 4: 4, 5: 3, 6: 3, 7: 3, 8: 3, 9: 4, 10: 4, 11: 4, 12: 4, 13: 4, 14: 4}}

---

您可以缩短以下代码，但它应该可以工作：

# sort values
df = df.sort_values(['group', 'sum', 'count'], ascending=[True, False, False])

# duplicated performance columns are candidates for unclear
df['dup'] = df.duplicated(['group', 'sum', 'count'], False)

# set decision column
df['decision'] = 'winner'
# if dup, those are unclear
df.loc[df.dup == True, 'decision'] = 'unclear'

# allocate just a fraction of weight for unclear entries
df['alloc'] = df.loc[df.dup == True].groupby(['group']).decision.transform(lambda x: 1/np.size(x)+1e-6)
# if not dup, then allocate 1
df.loc[df.dup == False, 'alloc'] = 1

# cumulative allocation should add up to compare with max_size
df['cum_alloc'] = df.groupby('group').alloc.cumsum().astype(int)
# decide loser with clear logic
df.loc[df.cum_alloc > df.max_size, 'decision'] = 'loser'

# finally trim columns
df = df[['name', 'decision', 'sum', 'count', 'max_size']]

输出：

>>> df
   name decision  sum  count  max_size
1   aaa   winner    3      2         4
2   bbb   winner    3      1         4
3   ccc  unclear    2      2         4
4   ddd  unclear    2      2         4
5   eee   winner    1      0         4
6   aaa   winner    3      2         3
7   bbb   winner    3      1         3
8   ccc   winner    2      3         3
9   ddd    loser    2      1         3
10  aaa   winner    3      4         4
11  bbb   winner    3      2         4
12  ccc   winner    2      5         4
13  ddd  unclear    2      1         4
14  eee  unclear    2      1         4
15  fff  unclear    2      1         4

---

让我们看第1组，我有5个元素，我必须选择4个（最大元素值）。我首先用总和来排列它们，然后用优先级来排列。aaa和bbb排名前二。ccc和ddd的总和计数值相同-我只选择其中一个，但将它们标记为“pick_random”，然后eee是在它们之后最后一个被选择的值：组的最终大小将是4。我认为根据您的逻辑，您对第一组的输出是不正确的。有4个赢家aaa、bbb、ccc和ddd是明确的赢家，eee是明显的输家不？你只需要从不明确的赢家中选择一个@ALollzYou需要提供输入数据帧，作为可以复制并粘贴到Python中的东西（就像输出一样）。