我有一个数据库如下:
indexID matchID order userClean Probability
0 0 1 0 clean 35
1 0 2 1 clean 75
2 0 2 2 clean 25
5 3 4 5 clean 40
6 3 5 6 clean 85
9 4 5 9 clean 74
12 6 7 12 clean 23
13 6 8 13 clean 72
14 7 8 14 clean 85
15 9 10 15 clean 76
16 10 11 16 clean 91
19 13 14 19 clean 27
23 13 17 23 clean 10
28 13 18 28 clean 71
32 20 21 32 clean 97
33 20 22 33 clean 30
我想要做的是,对于每个重复的indexID,我想选择概率较高的条目并将其标记为干净而另一个标记为脏。
输出应该如下所示:
indexID matchID order userClean Probability
0 0 1 0 dirty 35
1 0 2 1 clean 75
2 0 2 2 dirty 25
5 3 4 5 dirty 40
6 3 5 6 clean 85
9 4 5 9 clean 74
12 6 7 12 dirty 23
13 6 8 13 clean 72
14 7 8 14 clean 85
15 9 10 15 clean 76
16 10 11 16 clean 91
19 13 14 19 dirty 27
23 13 17 23 dirty 10
28 13 18 28 clean 71
32 20 21 32 clean 97
33 20 22 33 dirty 30
如果需要pandas
解决方案通过将Probability
(Series.ne
)的!=
列与max
创建的每个组的transform
值进行比较来创建布尔掩码,因为需要与Series
大小相同的df
:
mask = df['Probability'].ne(df.groupby('indexID')['Probability'].transform('max'))
df.loc[mask, 'userClean'] = 'dirty'
print (df)
indexID matchID order userClean Probability
0 0 1 0 dirty 35
1 0 2 1 clean 75
2 0 2 2 dirty 25
5 3 4 5 dirty 40
6 3 5 6 clean 85
9 4 5 9 clean 74
12 6 7 12 dirty 23
13 6 8 13 clean 72
14 7 8 14 clean 85
15 9 10 15 clean 76
16 10 11 16 clean 91
19 13 14 19 dirty 27
23 13 17 23 dirty 10
28 13 18 28 clean 71
32 20 21 32 clean 97
33 20 22 33 dirty 30
详情:
print (df.groupby('indexID')['Probability'].transform('max'))
0 75
1 75
2 75
5 85
6 85
9 74
12 72
13 72
14 85
15 76
16 91
19 71
23 71
28 71
32 97
33 97
Name: Probability, dtype: int64
如果想比较mean
与gt
(>)
:
mask = df['Probability'].gt(df['Probability'].mean())
df.loc[mask, 'userClean'] = 'dirty'
print (df)
indexID matchID order userClean Probability
0 0 1 0 clean 35
1 0 2 1 dirty 75
2 0 2 2 clean 25
5 3 4 5 clean 40
6 3 5 6 dirty 85
9 4 5 9 dirty 74
12 6 7 12 clean 23
13 6 8 13 dirty 72
14 7 8 14 dirty 85
15 9 10 15 dirty 76
16 10 11 16 dirty 91
19 13 14 19 clean 27
23 13 17 23 clean 10
28 13 18 28 dirty 71
32 20 21 32 dirty 97
33 20 22 33 clean 30