我正在尝试使用scipy.optimize优化预测参数。我遵循了教程，在stackoverflow上也找到了一些不错的例子，但我面临着一个无法解决的问题。我开始怀疑，使用熊猫是否是一个糟糕的选择与scipy？
我已经设置了我的代码如下：
import simpy
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm
import pandas as pd
import statistics as stat
import math as m
#from sklearn.grid_search import ParameterGrid
from scipy.optimize import minimize
###dataframe for the simulation
df = pd.read_csv('simulation_df_data_2018_2.csv')
with pd.option_context("max_rows", None,"max_columns", None):
print(df.head())
for i in df.index:
alpha = 0.2
beta = 0.3
x = np.array([alpha, beta])
def holts(x):
LO = np.int(df['average_demand'].loc[i])
print(type(LO))
TO = ((df['m2'].loc[i] - df['m3'].loc[i]) + (df['m1'].loc[i] - df['m2'].loc[i])) / 2
L1 = round(x[0] * df['m3'].loc[i] + (1 - x[0]) * (
LO + TO))
T1 = x[1] * (L1 - LO) + (1 - x[1]) * TO
L2 = round(x[0] * df['m2'].loc[i] + (1 - x[0]) * (
L1 + T1))
T2 = x[1] * (L2 - L1) + (1 - x[1]) * T1
L3 = round(x[0] * df['m1'].loc[i] + (1 - x[0]) * (
L2 + T2))
T3 = beta * (L3 - L2) + (1 - beta) * T2
LT1 = round(L3 + 1 * T3)
MSE = ((df['m3'].loc[i] - L1) + (df['m2'].loc[i] - L2) + (
df['m2'].loc[i] - L3)) ** 2 / 3
return MSE
#print(holts(x))
x0 = [0.1,0.1]
result = minimize(holts, x0, bounds=[(0,1),(0,1)], method="SLSQP")
print(result)
print(x)
df看起来是这样的：
m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m11
0 0.0 8.0 2.0 0.0 14.0 0.0 5.0 2.0 4.0 4.0 10.0
1 4.0 55.0 2.0 72.0 38.0 87.0 113.0 2.0 0.0 165.0 2.0
2 18.0 34.0 6.0 63.0 14.0 18.0 33.0 35.0 51.0 0.0 24.0
3 0.0 21.0 3.0 10.0 15.0 0.0 32.0 1.0 3.0 17.0 0.0
4 96.0 106.0 237.0 136.0 138.0 116.0 167.0 158.0 110.0 115.0 161.0
m12 m13 m14 m15 m16 m17 m18 m19 m20 m21 m22
0 0.0 6.0 10.0 0.0 2.0 2.0 17.0 0.0 0.0 0.0 0.0
1 35.0 7.0 88.0 6.0 3.0 103.0 18.0 59.0 6.0 20.0 152.0
2 6.0 5.0 99.0 7.0 17.0 15.0 8.0 3.0 21.0 6.0 4.0
3 30.0 5.0 88.0 1.0 6.0 10.0 9.0 17.0 9.0 0.0 1.0
4 116.0 77.0 48.0 96.0 69.0 77.0 96.0 74.0 94.0 101.0 115.0
m23 m24 average_demand low_demand high_demand
0 0.0 0.0 3.583333 0.0 17.0
1 6.0 0.0 43.458333 0.0 165.0
2 14.0 12.0 21.375000 0.0 99.0
3 0.0 0.0 11.583333 0.0 88.0
4 158.0 167.0 117.833333 48.0 237.0
我对不断出现的错误感到非常困惑，这是回溯
Traceback (most recent call last):
File "/Users/pierre/Desktop/simul/forecast_holts_alpha.py", line 121, in <module>
result = minimize(holts, x0, args= coef_list,bounds=[(0,1),(0,1)], method="SLSQP")
File "/Users/pierre/Desktop/Django-app/lib/python3.7/site-packages/scipy/optimize/_minimize.py", line 618, in minimize
constraints, callback=callback, **options)
File "/Users/pierre/Desktop/Django-app/lib/python3.7/site-packages/scipy/optimize/slsqp.py", line 399, in _minimize_slsqp
fx = func(x)
File "/Users/pierre/Desktop/Django-app/lib/python3.7/site-packages/scipy/optimize/optimize.py", line 327, in function_wrapper
return function(*(wrapper_args + args))
File "/Users/pierre/Desktop/simul/forecast_holts_alpha.py", line 63, in holts
LO = np.int(df['average_demand'].loc[i])
IndexError: only integers, slices (`:`), ellipsis (`...`), numpy.newaxis (`None`) and integer or boolean arrays are valid indices
我不明白为什么会出现这个错误，特别是因为如果我搜索LO的类型，我会得到这个：
print(type(LO))
<class 'int'>
我不是一个经验丰富的程序员，所以我很难弄清楚发生了什么，任何帮助都将不胜感激！
更新：
fun: 56.333333333333336
jac: array([0., 0.])
message: 'Optimization terminated successfully.'
nfev: 4
nit: 1
njev: 1
status: 0
success: True
x: array([0.1, 0.1])
[0.2,0.3]
OUTPUT看起来像这样，但似乎没有优化任何
由于我不知道您的代码到底在做什么，所以我无法判断以下代码是否正确。但为了指导如何使用scipy.optimize.minimize函数，它应该如下所示：
import numpy as np
import pandas as pd
from scipy.optimize import minimize
df = pd.read_csv('simulation_df_data_2018_2.csv')
for i in df.index:
# Since you said "alpha" and "beta" are the variable of "holts" function
# I changed all "alpha" and "beta" to "x[0]" and "x[1]" respectively.
def holts(x):
LO = np.int(df['average_demand'].loc[i])
TO = ((df['m2'].loc[i] - df['m3'].loc[i]) + (df['m1'].loc[i] - df['m2'].loc[i])) / 2
L1 = x[0] * df['m3'].loc[i] + (1 - x[0]) * (LO + TO)
T1 = x[1] * (L1 - LO) + (1 - x[1]) * TO
L2 = x[0] * df['m2'].loc[i] + (1 - x[0]) * (L1 + T1)
T2 = x[1] * (L2 - L1) + (1 - x[1]) * T1
L3 = x[0] * df['m1'].loc[i] + (1 - x[0]) * (L2 + T2)
MSE = ((df['m3'].loc[i] - L1) + (df['m2'].loc[i] - L2) + (df['m2'].loc[i] - L3)) ** 2 / 3
return MSE
x0 = np.array([0.1, 0.1]) # initial guess for the points "x" which "holts(x)" attains its minimum.
result = minimize(holts, x0, bounds=[(0, 1), (0, 1)], method="SLSQP", options={'disp': True})
print(result)
print()
它打印
Optimization terminated successfully (Exit mode 0)
Current function value: 16.921875002444228
Iterations: 5
Function evaluations: 15
Gradient evaluations: 5
fun: 16.921875002444228
jac: array([ 8.06808472e-04, -6.23427415e+00])
message: 'Optimization terminated successfully'
nfev: 15
nit: 5
njev: 5
status: 0
success: True
x: array([0.75000606, 1. ])
Optimization terminated successfully (Exit mode 0)
Current function value: 3.0792324286388828e-09
Iterations: 7
Function evaluations: 24
Gradient evaluations: 7
fun: 3.0792324286388828e-09
jac: array([-0.00833083, -0.00081578])
message: 'Optimization terminated successfully'
nfev: 24
nit: 7
njev: 7
status: 0
success: True
x: array([0.1495232 , 0.25665903])
Optimization terminated successfully (Exit mode 0)
Current function value: 1.9951658230345873e-10
Iterations: 9
Function evaluations: 32
Gradient evaluations: 9
fun: 1.9951658230345873e-10
jac: array([0.03791677, 0.00858221])
message: 'Optimization terminated successfully'
nfev: 32
nit: 9
njev: 9
status: 0
success: True
x: array([0.37155192, 0.19218713])
Optimization terminated successfully (Exit mode 0)
Current function value: 114.53090716252409
Iterations: 11
Function evaluations: 37
Gradient evaluations: 11
fun: 114.53090716252409
jac: array([ 3.06129456e-04, -3.04457455e+01])
message: 'Optimization terminated successfully'
nfev: 37
nit: 11
njev: 11
status: 0
success: True
x: array([0.7171298, 1. ])
Optimization terminated successfully (Exit mode 0)
Current function value: 33.333333333333336
Iterations: 2
Function evaluations: 6
Gradient evaluations: 2
fun: 33.333333333333336
jac: array([-403.33331966, -0. ])
message: 'Optimization terminated successfully'
nfev: 6
nit: 2
njev: 2
status: 0
success: True
x: array([1., 1.])
更新：round函数似乎是minimize函数不起作用的原因，因为round函数会在holts函数的图上创建许多平台。我删除了round函数，并更新了打印的结果。