python中具有约束、边界和数据帧的Scipy或bayesian优化函数

Question

有了下面的数据帧，我想要优化总回报，同时满足某些界限。

d = {'Win':[0,0,1, 0, 0, 1, 0],'Men':[0,1,0, 1, 1, 0, 0], 'Women':[1,0,1, 0, 0, 1,1],'Matches' :[0,5,4, 7, 4, 10,13],
     'Odds':[1.58,3.8,1.95, 1.95, 1.62, 1.8, 2.1], 'investment':[0,0,6, 10, 5, 25,0],}

data = pd.DataFrame(d)

我想最大化下面的等式：

totalreturn = np.sum(data['Odds'] * data['investment'] * (data['Win'] == 1))

该函数应最大化，并满足以下界限：

for i in range(len(data)):
    
    investment = data['investment'][i]
    
    C = alpha0 + alpha1*data['Men'] + alpha2 * data['Women'] + alpha3 * data['Matches']
    
    if (lb < investment ) & (investment < ub) & (investment > C) == False:
        data['investment'][i] = 0

因此，lb和ub对于数据帧中的每一行都是常量。然而，阈值C对于每一行都是不同的。因此，有6个参数需要优化：lb, ub, alph0, alpha1, alpha2, alpha3。

谁能告诉我如何在python中做到这一点？到目前为止，我一直在进行scipy (Approach1)和贝叶斯(Approach2)优化，只尝试对lb和ub进行优化。Approach1：

import pandas as pd
from scipy.optimize import minimize

def objective(val, data):
    
    # Approach 1
    # Lowerbound and upperbound
    lb, ub = val
    
    # investments
    # These matches/bets are selected to put wager on
    tf1 = (data['investment'] > lb) & (data['investment'] < ub) 
    data.loc[~tf1, 'investment'] = 0
    
        
    # Total investment
    totalinvestment = sum(data['investment'])
    
    # Good placed bets 
    data['reward'] = data['Odds'] * data['investment'] * (data['Win'] == 1)
    totalreward = sum(data['reward'])

    # Return and cumalative return
    data['return'] = data['reward'] - data['investment']
    totalreturn = sum(data['return'])
    data['Cum return'] = data['return'].cumsum()
    
    # Return on investment
    print('\n',)
    print('lb, ub:', lb, ub)
    print('TotalReturn: ',totalreturn)
    print('TotalInvestment: ', totalinvestment)
    print('TotalReward: ', totalreward)
    print('# of bets', (data['investment'] != 0).sum())
          
    return totalreturn
          

# Bounds and contraints
b = (0,100)
bnds = (b,b,)
x0 = [0,100]

sol = minimize(objective, x0, args = (data,), method = 'Nelder-Mead', bounds = bnds)

和approach2：

import pandas as pd
import time
import pickle
from hyperopt import fmin, tpe, Trials
from hyperopt import STATUS_OK
from hyperopt import  hp

def objective(args):
    # Approach2

    # Lowerbound and upperbound
    lb, ub = args
    
    # investments
    # These matches/bets are selected to put wager on
    tf1 = (data['investment'] > lb) & (data['investment'] < ub) 
    data.loc[~tf1, 'investment'] = 0
    
        
    # Total investment
    totalinvestment = sum(data['investment'])
    
    # Good placed bets 
    data['reward'] = data['Odds'] * data['investment'] * (data['Win'] == 1)
    totalreward = sum(data['reward'])

    # Return and cumalative return
    data['return'] = data['reward'] - data['investment']
    totalreturn = sum(data['return'])
    data['Cum return'] = data['return'].cumsum()
    
    # store results
    d = {'loss': - totalreturn, 'status': STATUS_OK, 'eval time': time.time(),
    'other stuff': {'type': None, 'value': [0, 1, 2]},
    'attachments': {'time_module': pickle.dumps(time.time)}}
    
    return d

          

trials = Trials()

parameter_space  = [hp.uniform('lb', 0, 100), hp.uniform('ub', 0, 100)]

best = fmin(objective,
    space= parameter_space,
    algo=tpe.suggest,
    max_evals=500,
    trials = trials)


print('\n', trials.best_trial)

有人知道我应该怎么做吗？Scipy没有生成预期的结果。Hyperopt优化确实会产生所需的结果。在这两种方法中，我都不知道如何合并依赖于行的边界(C(i))。

任何事都可以帮上忙！(关于这种优化的任何相关文章、练习或有用的解释也非常受欢迎)

Answer 1

我在这里假设你不能遍历整个数据集，或者它是不完整的，或者你想要外推，所以你不能计算所有的组合。

在没有先验的情况下，如果你不确定平滑度，或者评估可能是昂贵的，我会使用贝叶斯优化。您可以控制勘探/开发，并最大限度地防止卡住。

我会使用scikit-optimize，它能更好地实现贝叶斯优化。它们有更好的初始化技术，比如Sobol'方法，这里正确实现了该方法。这将确保您的搜索空间将被正确采样。

from skopt import gp_minimize

res = gp_minimize(objective, bnds, initial_point_generator='sobol')

Answer 2

我认为你的公式还需要一个变量，它将是二进制的，并将定义投资是否应保存为0或是否应具有其初始值。假设这个变量将保存在另一个名为'new_binary‘的列中，您的目标函数可以按如下方式进行更改：

totalreturn = np.sum(data['Odds'] * data['investment'] * data['new_binary'] * data['Win'])

然后，唯一缺少的就是引入变量本身。

for i in range(len(data)):
    investment = data['investment'][i]
    C = alpha0 + alpha1*data['Men'] + alpha2 * data['Women'] + alpha3 * data['Matches']
    data['new_binary'] = (lb < data['investment'] ) & ( data['investment'] < ub) & (data['investment'] > C)
    # This should be enough to make the values in the columns binary, while in python it is easily replaced with 0 and 1. 

我现在看到的唯一问题是这个问题变成了整数，所以我不确定scipy.optimize.minimize是否可以。我不确定有什么替代方案，但根据this的说法，PuLP和Pyomo可以工作。