如何解决使用数据优化熊猫中的步进功能
我有两个数据框(df_1和df_2)和一些变量(A,B,C):
df_1 = pd.DataFrame({'O' : [1,2,3],'M' : [2,8,3]})
df_2 = pd.DataFrame({'O' : [1,1,3,'M' : [9,4,6,7,5,4],'X' : [2,9],'Y' : [3,7],'Z' : [2,1]})
下面我有一种算法,该算法使用A,B,C为df_2中的每一行计算得分(S)。它在df_2中找到得分最高(S)的行。它将df_2和df_1中得分最高的行进行比较,并计算出p_hat(衡量它们之间的相似性):
M_G = df_1.M
df_1 = df_1.set_index('O')
A = 1
B = 1
C = 1
# Score
df_2['S'] = df_2['X']*A + df_2['Y']*B + df_2['Z']*C
# Top score
df_Sort = df_2.sort_values(['S','X','M'],ascending=[False,True,True])
df_O = df_Sort.set_index('O')
M_Top = df_O[~df_O.index.duplicated(keep='first')].M
M_Top = M_Top.sort_index()
# Compare the top scoring row for each O to df_1
df_1_M = df_1.M
df_1_M = df_1_M.sort_index()
df_1_R = df_1_M.reindex(M_Top.index)
T_N_T = M_Top == df_1_R
# Record the results for the given values of A,B,C
df_Res = pd.DataFrame({'it_is':T_N_T}) # is this row of df_1 the same as this row of M_Top?
# p_hat = TP / (TP + FP)
p_hat = df_Res.sum() / len(df_Res.index)
对于示例中的A,B,C值,其p_hat = 0.333。我想找到给出p_hat的最大可能值的A,B,C值。我想使用优化算法来确保获得最大值。
该图适用于C = 2:
解决方法
我想您可以使用Optuna软件包,它非常易于使用。您定义一个目标函数,在该目标函数中以所需的方式计算要优化的变量,并为其创建一个study
对象,然后Optuna基本上完成其余的工作。
简单的2D功能的小示例:
import optuna
def optimize_me(trial):
x = trial.suggest_uniform('x',-10,10)
y = trial.suggest_uniform('y',10)
return ((y - 1) ** 2) + ((x + 2) ** 2)
study = optuna.create_study()
study.optimize(optimize_me,n_trials = 10)
您可以了解有关Optuna here
的更多信息我主要用它来优化RNN的超级参数,这是一个非常强大的软件包。
,我找到了一种使用全局蛮力优化的方法:
from scipy.optimize import brute
df_1 = pd.DataFrame({'O' : [1,2,3],'M' : [2,8,3]})
df_2 = pd.DataFrame({'O' : [1,1,3,'M' : [9,4,6,7,5,4],'X' : [2,9],'Y' : [3,7],'Z' : [2,1]})
# Range
min_ = -2
max_ = 2
step = .5
ran_ge = slice(min_,max_+step,step)
ranges = (ran_ge,ran_ge,ran_ge)
# Params
params = (df_1,df_2)
# Index
df_1 = df_1.set_index('O')
df_1_M = df_1.M
df_1_M = df_1_M.sort_index()
# Fun
def fun(z,*params):
A,B,C = z
# Score
df_2['S'] = df_2['X']*A + df_2['Y']*B + df_2['Z']*C
# Top score
df_Sort = df_2.sort_values(['S','X','M'],ascending=[False,True,True])
df_O = df_Sort.set_index('O')
M_Top = df_O[~df_O.index.duplicated(keep='first')].M
M_Top = M_Top.sort_index()
# Compare the top scoring row for each O to df_1
df_1_R = df_1_M.reindex(M_Top.index) # Nan
T_N_T = M_Top == df_1_R
# Record the results for the given values of A,C
df_Res = pd.DataFrame({'it_is':T_N_T}) # is this row of df_1 the same as this row of M_Top?
# p_hat = TP / (TP + FP)
p_hat = df_Res.sum() / len(df_Res.index)
return -p_hat
# Brute
resbrute = brute(fun,ranges,args=params,full_output=True,finish=None)
print('Global maximum ',resbrute[0])
print('Function value at global maximum ',-resbrute[1])
它给出:
Global maximum [-2. 0.5 1.5]
Function value at global maximum 0.6666666666666666
,
或使用全局,渐进式优化:(代码由@Aviv Yaniv固定)
from scipy.optimize import differential_evolution
df_1 = pd.DataFrame({'O' : [1,1]})
# Index
df_1 = df_1.set_index('O')
df_1_M = df_1.M
df_1_M = df_1_M.sort_index()
# Fun
def fun(z,C
df_Res = pd.DataFrame({'it_is':T_N_T}) # is this row of df_1 the same as this row of M_Top?
# p_hat = TP / (TP + FP)
p_hat = df_Res.sum() / len(df_Res.index)
print(z)
return -p_hat[0]
# Bounds
min_ = -2
max_ = 2
ran_ge = (min_,max_)
bounds = [ran_ge,ran_ge]
# Params
params = (df_1,df_2)
# DE
DE = differential_evolution(fun,bounds,args=params)
print('Function value at global maximum ',-DE.fun)
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