CPLEX API for Python

GAMS 文档： 很好的说明书

cplex.SparsePair

cplex.SparseTriple

l = cplex.SparsePair(ind = ['x'], val = [1.0])
q = cplex.SparseTriple(ind1 = ['x'], ind2 = ['y'], val = [1.0])
 c.quadratic_constraints.add(name = "my_quad",
							    	lin_expr = l,
                                    quad_expr = q,
                                    rhs = 1.0,
                                    sense = "G")

这里一定要注意：SparsePair 和 SparseTriple 的入参顺序不能调换，必须参数 ind 在前，系数值 val 在后。否则会报错：

TypeError: non-float value in input sequence

cplex.objective.set_linear

给定 $N$ 个变量，set_linear 的作用就是设置每个变量的系数（相当于为一个数组赋值），从而构建出一个线性目标函数。set_linear 的入参是一个列表，列表中每个元素 $(i,V_i)$ 的含义就是设置第 $i+1$ 个变量的系数为 $V_i$。get_linear() 的出参也是一个列表，它实际上是构建出的线性目标函数的系数向量。

#   实例化
cpx = cplex.Cplex()
cpx.set_results_stream(None)
cpx.set_log_stream(None)
cpx.parameters.preprocessing.reduce.set(0)
cpx.parameters.lpmethod.set(cpx.parameters.lpmethod.values.primal)
cpx.objective.set_sense(cpx.objective.sense.maximize)
#  添加变量
cpx.variables.add(obj=[0.0],lb=[0.0],ub=[cplex.infinity],names=['X1'])
cpx.variables.add(obj=[0.0],lb=[0.0],ub=[cplex.infinity],names=['X2'])
cpx.variables.add(obj=[0.0],lb=[0.0],ub=[cplex.infinity],names=['X3'])
cpx.variables.add(obj=[0.0],lb=[0.0],ub=[cplex.infinity],names=['X4'])
cpx.variables.add(obj=[0.0],lb=[0.0],ub=[cplex.infinity],names=['X5'])
#  设置线性变量
cpx.objective.set_linear([(0,21),(2,22)])
cpx.objective.set_linear([(1,23),(3,24),(4,25)])
cpx.objective.set_linear([(2,11)])
print(cpx.objective.get_linear())

运行结果如下：

[21.0, 23.0, 11.0, 24.0, 25.0]

备注：在 set_linear 之前必须添加变量，否则会报错

cplex.exceptions.errors.CplexSolverError: CPLEX Error  1201: Column index 0 out of range.

objective	set_linear
linear_constraints	add
solution	get_status()
solution	get_values(*)
solution	status	unbounded
solution	advanced	get_ray()
SparsePair
parameters	preprocessing	reduce.set(0)
parameters	lpmethod	set(*)
parameters	lpmethod	values	primal	单纯形法

Python 用户的 CPLEX

python教程

模块 cplex 中的主类是类 Cplex。 它将优化问题的数学方程与用户所指定的有关 CPLEX 应如何对问题求解的信息封装在一起。 类 Cplex 提供了方法以修改问题、对问题求解以及查询问题本身及其解法。 类 Cplex 的方法分组为相关功能的类别。 例如，用于添加、修改和查询与变量相关的数据的方法包含在类 Cplex 的成员 variables 中。

可以在 CPLEX 的各组件中对标准方程中的此线性规划模型求解。
要求解的问题是：
最大化 $$x_1+2x_2+3x_3$$ 约束 $$-x_1+x_2+x_3\le 20$$ $$x_1-3x_2+x_3\le 30$$
使用这些界限 $$0\le x_1\le 40,x_2\ge 0,x_3\ge 0$$

以下是使用 CPLEX 对示例求解的 Python 应用程序:

execfile("cplexpypath.py")

import cplex
from cplex.exceptions import CplexError
import sys

# data common to all populateby functions
my_obj      = [1.0, 2.0, 3.0]
my_ub       = [40.0, cplex.infinity, cplex.infinity]
my_colnames = ["x1", "x2", "x3"]
my_rhs      = [20.0, 30.0]
my_rownames = ["c1", "c2"]
my_sense    = "LL"


def populatebyrow(prob):
    prob.objective.set_sense(prob.objective.sense.maximize)

    # since lower bounds are all 0.0 (the default), lb is omitted here
    prob.variables.add(obj = my_obj, ub = my_ub, names = my_colnames)

    # can query variables like the following bounds and names:

    # lbs is a list of all the lower bounds
    lbs = prob.variables.get_lower_bounds()

    # ub1 is just the first lower bound
    ub1 = prob.variables.get_upper_bounds(0)

    # names is ["x1", "x3"]
    names = prob.variables.get_names([0, 2])

    rows = [[[0,"x2","x3"],[-1.0, 1.0,1.0]],
            [["x1",1,2],[ 1.0,-3.0,1.0]]]

    prob.linear_constraints.add(lin_expr = rows, senses = my_sense,
                                rhs = my_rhs, names = my_rownames)

    # because there are two arguments, they are taken to specify a range
    # thus, cols is the entire constraint matrix as a list of column vectors
    cols = prob.variables.get_cols("x1", "x3")


def populatebycolumn(prob):
    prob.objective.set_sense(prob.objective.sense.maximize)

    prob.linear_constraints.add(rhs = my_rhs, senses = my_sense,
                                names = my_rownames)

    c = [[[0,1],[-1.0, 1.0]],
         [["c1",1],[ 1.0,-3.0]],
         [[0,"c2"],[ 1.0, 1.0]]]

    prob.variables.add(obj = my_obj, ub = my_ub, names = my_colnames,
                       columns = c)

def populatebynonzero(prob):
    prob.objective.set_sense(prob.objective.sense.maximize)

    prob.linear_constraints.add(rhs = my_rhs, senses = my_sense,
                                names = my_rownames)
    prob.variables.add(obj = my_obj, ub = my_ub, names = my_colnames)

    rows = [0,0,0,1,1,1]
    cols = [0,1,2,0,1,2]
    vals = [-1.0,1.0,1.0,1.0,-3.0,1.0]

    prob.linear_constraints.set_coefficients(zip(rows, cols, vals))
    # can also change one coefficient at a time

    # prob.linear_constraints.set_coefficients(1,1,-3.0)
    # or pass in a list of triples
    # prob.linear_constraints.set_coefficients([(0,1,1.0), (1,1,-3.0)])


def lpex1(pop_method):
    try:
        my_prob = cplex.Cplex()

        if pop_method == "r":
            handle = populatebyrow(my_prob)
        if pop_method == "c":
            handle = populatebycolumn(my_prob)
        if pop_method == "n":
            handle = populatebynonzero(my_prob)

        my_prob.solve()
    except CplexError, exc:
        print exc
        return

    numrows = my_prob.linear_constraints.get_num()
    numcols = my_prob.variables.get_num()

    print
    # solution.get_status() returns an integer code
    print "Solution status = " , my_prob.solution.get_status(), ":",
    # the following line prints the corresponding string
    print my_prob.solution.status[my_prob.solution.get_status()]
    print "Solution value  = ", my_prob.solution.get_objective_value()
    slack = my_prob.solution.get_linear_slacks()
    pi    = my_prob.solution.get_dual_values()
    x     = my_prob.solution.get_values()
    dj    = my_prob.solution.get_reduced_costs()
    for i in range(numrows):
        print "Row %d:  Slack = %10f  Pi = %10f" % (i, slack[i], pi[i])
    for j in range(numcols):
        print "Column %d:  Value = %10f Reduced cost = %10f" % (j, x[j], dj[j])

    my_prob.write("lpex1.lp")

if __name__ == "__main__":
    if len(sys.argv) != 2 or sys.argv[1] not in  ["-r", "-c", "-n"]:
        print "Usage: lpex1.py -X"
        print "   where X is one of the following options:"
        print "      r          generate problem by row"
        print "      c          generate problem by column"
        print "      n          generate problem by nonzero"
        print " Exiting..."
        sys.exit(-1)
    lpex1(sys.argv[1][1])
else:
    prompt = """Enter the letter indicating how the problem data should be populated:
    r : populate by rows
    c : populate by columns
    n : populate by nonzeros\n ? > """
    r = 'r'
    c = 'c'
    n = 'n'
    lpex1(input(prompt))