有什么方法可以在 PuLP 优化中从字符串调用 LpVariables 吗？

Question

在 PuLP 优化程序中，我想从 str 列表中调用定义的 LpVariable 值。

我试图通过某种方式转换 'x2'，但我做不到。
除了使用 many IF/ELIF list if test_str=='x1': ... ?

之外，有什么办法可以做到这一点吗？

from pulp import *


def pulp_test():
    # define the problem
    prob = LpProblem("The_Problem", LpMinimize)
    x1 = LpVariable('x1', 0, None, LpContinuous)
    x2 = LpVariable('x2', 0, None, LpContinuous)
    x3 = LpVariable('x3', 0, None, LpContinuous)
    prob += 3 * x1 + 11 * x2 + 2 * x3
    prob += -1 * x1 + 3 * x2 <= 5
    prob += 3 * x1 + 3 * x2 <= 4
    prob += 3 * x2 + 2 * x3 <= 6
    prob += 3 * x1 + 5 * x3 >= 4
    status = prob.solve()

    test_str = 'x2'

    print("type(x2): ", type(x2))
    print("type('x2'): ", type(test_str))

    print("value(x2): ", value(x2))
    # want to call LpVariable from str list
    print("value(convert from str 'x2'): ", value(pulp.LpVariable(test_str)))

pulp_test()

结果是，

type(x2):  <class 'pulp.pulp.LpVariable'>
type('x2'):  <class 'str'>
value(x2):  0.0
value(convert from str 'x2'):  None

Answer 1

检查LpVariable.dicts方法。它将创建一个变量字典。这是一个使用它的例子：https://coin-or.github.io/pulp/CaseStudies/a_blending_problem.html

来自那个例子：

from pulp import *
Ingredients = ['CHICKEN', 'BEEF', 'MUTTON', 'RICE', 'WHEAT', 'GEL']
ingredient_vars = LpVariable.dicts("Ingr",Ingredients,0)

那么如果你这样做：

ingredient_vars['CHICKEN'] # you get the variable for chicken

有什么方法可以在 PuLP 优化中从字符串调用 LpVariables 吗？

Is there any way to call LpVariables from string in PuLP optimization?

python

pulp