[英]How to find column-index of top-n values within each row of huge dataframe
[英]Find top N values within each group
我有一個類似於以下示例的數據集:
| id | size | old_a | old_b | new_a | new_b |
|----|--------|-------|-------|-------|-------|
| 6 | small | 3 | 0 | 21 | 0 |
| 6 | small | 9 | 0 | 23 | 0 |
| 13 | medium | 3 | 0 | 12 | 0 |
| 13 | medium | 37 | 0 | 20 | 1 |
| 20 | medium | 30 | 0 | 5 | 6 |
| 20 | medium | 12 | 2 | 3 | 0 |
| 12 | small | 7 | 0 | 2 | 0 |
| 10 | small | 8 | 0 | 12 | 0 |
| 15 | small | 19 | 0 | 3 | 0 |
| 15 | small | 54 | 0 | 8 | 0 |
| 87 | medium | 6 | 0 | 9 | 0 |
| 90 | medium | 11 | 1 | 16 | 0 |
| 90 | medium | 25 | 0 | 4 | 0 |
| 90 | medium | 10 | 0 | 5 | 0 |
| 9 | large | 8 | 1 | 23 | 0 |
| 9 | large | 19 | 0 | 2 | 0 |
| 1 | large | 1 | 0 | 0 | 0 |
| 50 | large | 34 | 0 | 7 | 0 |
這是上表的輸入:
data=[[6,'small',3,0,21,0],[6,'small',9,0,23,0],[13,'medium',3,0,12,0],[13,'medium',37,0,20,1],[20,'medium',30,0,5,6],[20,'medium',12,2,3,0],[12,'small',7,0,2,0],[10,'small',8,0,12,0],[15,'small',19,0,3,0],[15,'small',54,0,8,0],[87,'medium',6,0,9,0],[90,'medium',11,1,16,0],[90,'medium',25,0,4,0],[90,'medium',10,0,5,0],[9,'large',8,1,23,0],[9,'large',19,0,2,0],[1,'large',1,0,0,0],[50,'large',34,0,7,0]]
data= pd.DataFrame(data,columns=['id','size','old_a','old_b','new_a','new_b'])
我想要一個輸出,它將根據大小對數據集進行分組,並根據每組大小中“new_a”列的值列出前 2 個 id。 由於某些 id 重復多次,我想對這些 id 的 new_a 值求和,然后找到前 2 個值。 我的決賽桌應該如下所示:
| size | id | new_a |
|--------|----|-------|
| large | 9 | 25 |
| large | 50 | 7 |
| medium | 13 | 32 |
| medium | 90 | 25 |
| small | 6 | 44 |
| small | 10 | 12 |
我已經嘗試了下面的代碼,但它沒有顯示“大小”列中每個組的 new_a 的前 2 個值。
nlargest = data.groupby(['size','id'])['new_a'].sum().nlargest(2).reset_index()
print(
df.groupby('size').apply(
lambda x: x.groupby('id').sum().nlargest(2, columns='new_a')
).reset_index()[['size', 'id', 'new_a']]
)
印刷:
size id new_a
0 large 9 25
1 large 50 7
2 medium 13 32
3 medium 90 25
4 small 6 44
5 small 10 12
您可以在這里設置size
、 id
作為索引以避免雙重 groupby,並使用Series.sum
利用level
參數。
df.set_index(["size", "id"]).groupby(level=0).apply(
lambda x: x.sum(level=1).nlargest(2)
).reset_index()
size id new_a
0 large 9 25
1 large 50 7
2 medium 13 32
3 medium 90 25
4 small 6 44
5 small 10 12
您可以鏈接兩個groupby
方法:
data.groupby(['id', 'size'])['new_a'].sum().groupby('size').nlargest(2)\
.droplevel(0).to_frame('new_a').reset_index()
輸出:
id size new_a
0 9 large 25
1 50 large 7
2 13 medium 32
3 90 medium 25
4 6 small 44
5 10 small 12
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