本文整理匯總了Python中pulp.LpProblem.sortedVariables方法的典型用法代碼示例。如果您正苦於以下問題：Python LpProblem.sortedVariables方法的具體用法？Python LpProblem.sortedVariables怎麽用？Python LpProblem.sortedVariables使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類pulp.LpProblem的用法示例。
在下文中一共展示了LpProblem.sortedVariables方法的2個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的Python代碼示例。
示例1: pulp_solve
# 需要導入模塊: from pulp import LpProblem [as 別名]
# 或者: from pulp.LpProblem import sortedVariables [as 別名]
def pulp_solve(A, b, c):
problem = LpProblem(sense = pulp.constants.LpMaximize)
x = [LpVariable('x' + str(i + 1),
0,
None,
LpInteger)
for i in xrange(len(c))]
problem += lpSum(ci * xi for ci, xi in zip(c, x))
for ai, bi in zip(A, b):
problem += lpSum(aij * xj for aij, xj in zip(ai, x)) == bi
status = problem.solve()
return (pulp.constants.LpStatus[status],
[variable.varValue for variable in problem.sortedVariables()],
problem.objective.value())示例2: pulp_solve
# 需要導入模塊: from pulp import LpProblem [as 別名]
# 或者: from pulp.LpProblem import sortedVariables [as 別名]
def pulp_solve(A, b, c, bounds):
problem = LpProblem(sense = pulp.constants.LpMaximize)
x = [LpVariable('x' + str(i + 1),
bound['low_bound'],
bound['up_bound'],
LpInteger)
for i, bound in enumerate(bounds)]
problem += lpSum(ci * xi for ci, xi in zip(c, x))
for ai, bi in zip(A, b):
problem += lpSum(aij * xj for aij, xj in zip(ai, x)) == bi
status = problem.solve()
return (pulp.constants.LpStatus[status],
[variable.varValue for variable in problem.sortedVariables()],
problem.objective.value())