英文:
Select n rows of dataframe for each value in column A where n is the value in column C
问题
以下是要翻译的内容：
我有下面的数据框
从中，我想选择每个销售员的n行，其中n是每个销售员的相应销售限制
因此，从下面的示例创建的新数据框将不包含第3行（客户ID - ghi）和第六行（客户ID - pqr），因为A和B的相应销售限制都为2。
对于C，将选择所有行，因为限制（4）高于分配的行数。
| 销售员 | 客户ID | 销售限制 |
|---|---|---|
| A | abc | 2 |
| A | def | 2 |
| A | ghi | 2 |
| B | jkl | 2 |
| B | mno | 2 |
| B | pqr | 2 |
| C | stu | 4 |
| C | vwx | 4 |
| C | yz | 4 |
请注意：销售员列表是动态的，表格中的名称和限制将每周更改
我有一个想法，可以根据销售员名称进行分组，然后选择每个销售员的:n-1行，但由于表格行数可能会更改，所以我无法找到考虑到这一点的方法。
有人能否请提供Python代码来实现上述更改，先谢谢。
英文:
I have the below dataframe
From it, I want to select only n number of rows per salesman where n is the corresponding sales_limit for each salesperson
So, the new dataframe created from the below example, will not have the 3rd row (clint id - ghi) and the sixth rows (client id - pqr) as the corresponding sales limit for both A and B are 2.
For C, all will be selected as the limit (4) is higher than the number of rows assigned.
| salesman | client_id | sales_limit |
|---|---|---|
| A | abc | 2 |
| A | def | 2 |
| A | ghi | 2 |
| B | jkl | 2 |
| B | mno | 2 |
| B | pqr | 2 |
| C | stu | 4 |
| C | vwx | 4 |
| C | yz | 4 |
Please note: the list of salesman is dynamic, the names and their limits will change every week in the table
I had an idea of doing a groupby based on the salesman names and then select :n-1 rows per salesman but since the table row numbers might change, I wasnt able to figure a way to account for that.
Can someone please help with the code for above changes in python, thanks in advance
答案1
得分: 3
你可以使用 groupby 然后用 sales_limit 切片每个分组。
out = (df.groupby('salesman', as_index=False, group_keys=False).apply(lambda g: g.iloc[:g['sales_limit'].iloc[0]]))
print(out)salesman client_id sales_limit0 A abc 21 A def 23 B jkl 24 B mno 26 C stu 47 C vwx 48 C yz 4
英文:
You can do groupby then slice each group with sales_limit
out = (df.groupby('salesman', as_index=False, group_keys=False).apply(lambda g: g.iloc[:g['sales_limit'].iloc[0]]))
print(out)salesman client_id sales_limit0 A abc 21 A def 23 B jkl 24 B mno 26 C stu 47 C vwx 48 C yz 4
答案2
得分: 2
为了提高效率，使用 groupby.cumcount 和布尔索引 进行比较：
out = df[df.groupby('salesman').cumcount().lt(df['sales_limit'])]
输出：
salesman client_id sales_limit0 A abc 21 A def 23 B jkl 24 B mno 26 C stu 47 C vwx 48 C yz 4
中间结果：
salesman client_id sales_limit cumcount cumcount < sales_limit0 A abc 2 0 True1 A def 2 1 True2 A ghi 2 2 False3 B jkl 2 0 True4 B mno 2 1 True5 B pqr 2 2 False6 C stu 4 0 True7 C vwx 4 1 True8 C yz 4 2 True
英文:
For efficiency, use a comparison with groupby.cumcount and boolean indexing:
out = df[df.groupby('salesman').cumcount().lt(df['sales_limit'])]
Output:
salesman client_id sales_limit0 A abc 21 A def 23 B jkl 24 B mno 26 C stu 47 C vwx 48 C yz 4
Intermediates:
salesman client_id sales_limit cumcount cumcount < sales_limit0 A abc 2 0 True1 A def 2 1 True2 A ghi 2 2 False3 B jkl 2 0 True4 B mno 2 1 True5 B pqr 2 2 False6 C stu 4 0 True7 C vwx 4 1 True8 C yz 4 2 True
答案3
得分: 0
以下是翻译好的部分：
为此，我将利用 pandas groupby 和 apply。
导入 pandas 并生成 df：
import pandas as pddf = pd.DataFrame({'salesman':['A']*3+['B']*3+['C']*3,'client_id':list('ABCDEFGHI'),'sales_limit':[2]*6+[4]*3})
生成销售员限制的字典：
limits = df.groupby('salesman')['sales_limit'].max().to_dict()
使用 apply 和 iloc 打印前 x 行：
df.groupby('salesman').apply(lambda x:x.iloc[:limits[x.name]])
英文:
For this I would leverage pandas groupby, as well as apply.
import pandas and generate df:
import pandas as pddf = pd.DataFrame({'salesman':['A']*3+['B']*3+['C']*3,'client_id':list('ABCDEFGHI'),'sales_limit':[2]*6+[4]*3})
generate a dictionary with the limit per salesperson
limits = df.groupby('salesman')['sales_limit'].max().to_dict()
use apply and iloc to print the first x rows
df.groupby('salesman').apply(lambda x:x.iloc[:limits[x.name]])
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