Scipy Objective 函数

Question

我正在尝试将我的 gurobi 代码转换为 scipy，但我在定义 objective 函数时遇到了问题。当测试我是否正确定义函数时，我收到错误：

TypeError: 'float' object is not iterable

代码在这里：

import pandas as pd
import numpy as np
import scipy as sp
from scipy.optimize import minimize
import matplotlib.pyplot as plt
%matplotlib inline
step=80
f1load=[44,48,53,28,32,36,41,48,38,32,38,34,44,36,41,48,38,32,44,48,53,28,32,36,41,48,38,32,38,34,44,36,41,48,38,32,44,48,53,28,32,36,41,48,38,32,38,34,44,36,41,48,38,32,44,48,53,28,32,36,41,48,38,32,38,34,44,36,41,48,38,32,44,48,53,28,32,36,41,48,38,32,38,34,44,36,41,48]
fload=f1load[0:step+1]
i1load=[40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40,40]
iload=i1load[0:step+1]
#Following command converts fload&iload to arrays and subtracts them from each other 
load1=np.array(iload)-np.array(fload)
load1
#This command converts array to list so we can use it as a list in the rest of the code
load2=load1.tolist()
load=load2
x = np.zeros(80)
x = x.tolist()
def objective(x,load):
# this line is from my gurobi code    obj1=sum(((load[i+1]-(6*x[i]))*(load[i+1]-(6*x[i])) for i in range (n)))
    for i in range(step):
        obj1 = sum((load[i+1]-(6*x[i]))*(load[i+1]-(6*x[i])))
        obj2 = obj2 + obj1
    return obj2

objective(x,load)

错误的完整堆栈跟踪：

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-26-cef8470baf0d> in <module>
----> 1 objective(x,load)

<ipython-input-25-a9de3ab9ff2b> in objective(x, load)
      2 #     obj1=sum(((load[i+1]-(6*x[i]))*(load[i+1]-(6*x[i])) for i in range (n)))
      3     for i in range(step):
----> 4         obj1 = sum((load[i+1]-(6*x[i]))*(load[i+1]-(6*x[i])))
      5         obj2 = obj2 + obj1
      6     return obj2

TypeError: 'float' object is not iterable

Answer 1

这一行：

obj1 = sum((load[i+1]-(6*x[i]))*(load[i+1]-(6*x[i])))

括号内的表达式 (load[i+1]-(6*x[i]))*(load[i+1]-(6*x[i])) 的计算结果为 float，并且对该表达式调用 sum 会给出错误，因为 sum 函数需要一个可迭代的参数。

Scipy Objective 函数

Scipy Objective function

python

numpy

scipy

pandas

scipy-optimize-minimize