问题描述
我有两个数据框（df_1，df_2），一些变量（A，B，C），一个函数（fun）和一个全局遗传优化器，可以找到给定范围A，B的最大乐趣， C。
from scipy.optimize import differential_evolution
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]})
# 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)
print(z)
return -p_hat[0]
# Bounds
min_ = 0
max_ = 1
ran_ge = (min_,max_)
bounds = [ran_ge,ran_ge,ran_ge]
# Params
params = (df_1,df_2)
# DE
DE = differential_evolution(fun,bounds,args=params)
它在每次迭代中打印出[A B C]，例如最后三行是：
[0.04003901 0.50504249 0.56332845]
[0.040039 0.5050425 0.56332845]
[0.040039 0.50504249 0.56332846]
要查看其收敛情况，请问我如何绘制 A，B，C以防止迭代？
我试图将A，B，C存储在：
df_P = pd.DataFrame({0})
在增添乐趣的同时：
df_P.append(z)
但是我得到了：
RuntimeError: The map-like callable must be of the form f(func,iterable),returning a sequence of numbers the same length as 'iterable'
解决方法
因此，我不确定是否找到了最好的方法，但我找到了一种。 它使用列表通过引用传递的事实。这意味着如果将列表传递给函数并对其进行修改，即使该函数未返回该列表，它也会在程序的其余部分被修改。
# Params
results = [] # this list will hold our restuts
params = (df_1,df_2,results) # add it to the params of the functions
# now in the function add the output to the list,Instead of the mean here I used the distance to the origin (as if you 3 value were a 3d vector)
p_hat = df_Res.sum() / len(df_Res.index)
distance_to_zeros = sum([e**2 for e in z]) ** 1/2
results.append(distance_to_zeros)
# Indeed you can also append z directly.
# Then after DE call
DE = differential_evolution(fun,bounds,args=params)
x = range(0,len(results))
plt.scatter(x,results,alpha=0.5)
plt.show()
使用RomainL的代码绘制A，B，C：
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import differential_evolution
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]})
# 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)
results.append(z)
return -p_hat[0]
# Bounds
min_ = 0
max_ = 1
ran_ge = (min_,max_)
bounds = [ran_ge,ran_ge,ran_ge]
# Params
results = []
params = (df_1,results)
# DE
DE = differential_evolution(fun,args=params)
# Plot
df_R = pd.DataFrame(list(map(np.ravel,results)),columns=('A','B','C'))
plt.figure()
ax = df_R.plot()
ax.set_ylim(min_,max_)
