Scipy或贝叶斯优化函数,带有python中的约束、边界和数据帧



对于下面的数据帧,我希望在满足某些边界的情况下优化总回报。

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)

我想最大化以下等式:

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

函数应最大化,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,lbub对于数据帧中的每一行是恒定的。然而,阈值C对于每一行是不同的。因此有6个参数需要优化:lb, ub, alph0, alpha1, alpha2, alpha3

有人能告诉我如何在python中做到这一点吗?到目前为止,我的研究方向是scipy(方法1(和Bayesian(方法2(优化,只有lbub试图进行优化。方法1:

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)

和方法2:

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)

有人知道我该怎么做吗?Scipy没有产生想要的结果。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)(。

任何事情都会有帮助!(任何关于优化类型的相关文章、练习或有用的解释也非常受欢迎(

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

在没有先验知识的情况下,如果你不确定平滑度,或者评估可能代价高昂,我会使用贝叶斯优化。你可以控制探索/利用,并防止陷入最低限度。

我会使用scikit optimize,它可以更好地实现贝叶斯优化。他们有更好的初始化技术,比如Sobol'方法,它在这里得到了正确的实现。这样可以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')

我认为您的公式还需要一个变量,它将是二进制的,并定义投资是应该保存为0还是应该有其初始值。假设这个变量将保存在另一个名为"new_binary"的列中,您的目标函数可以更改如下:

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.deminime是否可行。我不确定有什么替代方案,但根据这一点,PuLP和Pyomo可以工作。

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