问题

Sure, here is the translation of the code you provided without the code parts:

我尝试优化“Totpod”之和，约束条件是每个“Cat”只能被选择一次，但如果我改变我的代码，要么什么都得不到，要么得到所有的组件。

这是我的代码：

导入库：
```python
import numpy as np
import pandas as pd
import scipy.optimize as sc
from pulp import LpProblem, LpVariable, LpMaximize, lpSum, LpMinimize, value
prob = LpProblem("Maximize_TotPod", LpMaximize)

数据部分：

data={'a':['Amu',508], 'b':['Coif',508], 'c':['Amu', 253]}
df= pd.DataFrame.from_dict(data, orient='index', columns=['Cat', 'Totpod'])

创建二进制决策变量：

cat_vars = LpVariable.dicts("Cat", df['Cat'], cat="Binary")

计算'Totpod'值的总和：

totpod_sum = df['Totpod'].sum()

目标函数：最大化'Totpod'的总和：

prob += lpSum(cat_vars[cat] * df.loc[df['Cat'] == cat, 'Totpod'].sum() for cat in df['Cat'])

约束条件：每个'Cat'只能选择一次：

for cat in df['Cat']:
    prob += cat_vars[cat] <= 1

解决问题：

prob.solve()

选择的'Cat'和最大的Totpod值：

selected_cats = [cat for cat in df['Cat'] if cat_vars[cat].value() == 1]
total_sum = totpod_sum + value(prob.objective)

print("Selected index:", selected_cats)
print("Max Totpod:", total_sum)

它返回[‘Amu’, ‘Coif’, ‘Amu’]和3299.0，但我想要[‘a’, ‘b’]和(508+508) 1016。

你知道如何修改以使其正常工作吗？我不介意使用另一个优化器。

<details>
<summary>英文:</summary>

I tried to optimize the sum of &#39;Totpod&#39; under the constraint that each &#39;Cat&#39; could only be taken once but if I change my code I either got nothing or all the components.

Here is my code:

    

Cat	Fo	Pod	Pa2	Pa3	Pa4	Pa5	Pa6	Coif Amu An1 An2 Bot Ceint Arm	Totpod

a Amu 101 3 4 5 6 7 8 9 10 11 12 13 14 15 508
b Coif 101 3 4 5 6 7 8 9 10 11 12 13 14 15 508
c Amu 50 3 4 5 6 7 8 9 10 11 12 13 14 15 253

    import numpy as np
    import pandas as pd
    import scipy.optimize as sc
    from pulp import LpProblem, LpVariable, LpMaximize, lpSum, LpMinimize, value
    prob = LpProblem("Maximize_TotPod", LpMaximize)
    
    data={'a':['Amu',508], 'b':['Coif',508], 'c':['Amu', 253]}
    df= pd.DataFrame.from_dict(data, orient='index', columns=['Cat', 'Totpod'])

    # Creating binary decision variables for each 'Cat' value
    cat_vars = LpVariable.dicts("Cat", df['Cat'], cat="Binary")

    # Calculating the total sum of 'Totpod' values
    totpod_sum = df['Totpod'].sum()

    # Objective function: maximize the sum of 'Totpod'
    prob += lpSum(cat_vars[cat] * df.loc[df['Cat'] == cat, 'Totpod'].sum() for cat in df['Cat'])

    # Constraint: select each 'Cat' only once
    for cat in df['Cat']:
        prob += cat_vars[cat] <= 1

    # Solving the problem
    prob.solve()

    selected_cats = [cat for cat in df['Cat'] if cat_vars[cat].value() == 1]
    total_sum = totpod_sum + value(prob.objective)

    # Printing the selected index and the max of Totpod
    print("Selected index:", selected_cats)
    print("Max Totpod':", total_sum)

It returns me ['Amu', 'Coif', 'Amu'] and 3299.0 while I want ['a', 'b'] and (508+508) 1016

Do you know how I can modify to make it works, I don't mind using another optimizer 

</details>


# 答案1
**得分**: 1

以下是您要翻译的部分：

IIUC，您可以将`selected_cats`简单生成为唯一`Cat`值的列表，以及通过按`Cat`分组，取最大`Totpod`值并对它们求和来获得最优（我假设您指的是最大）`Totpod`总和：

```python
selected_cats = df['Cat'].unique().tolist()
total_sum = df.groupby('Cat')['Totpod'].max().sum()

print("Selected index:", selected_cats)
print("Max Totpod:", total_sum)

输出（对于您的示例数据）：

Selected index: ['Amu', 'Coif']
Max Totpod: 1016

selected_cats = df['Cat'].unique().tolist()
total_sum = df.groupby('Cat')['Totpod'].max().sum()

print("Selected index:", selected_cats)
print("Max Totpod':", total_sum)

Selected index: ['Amu', 'Coif']
Max Totpod': 1016