优化变量的命名索引
创建命名索引
优化变量可以使用名称来索引元素。您可以在创建变量时或之后指定名称。例如，在创建变量时给出名称。
x = optimvar("x",["United","Lufthansa","Virgin Air"])
x =
1×3 OptimizationVariable array with properties:
Array-wide properties:
Name: 'x'
Type: 'continuous'
IndexNames: {{} {1×3 cell}}
Elementwise properties:
LowerBound: [-Inf -Inf -Inf]
UpperBound: [Inf Inf Inf]
See variables with show.
See bounds with showbounds.
optimvar 会自动将您指定的名称按照变量的顺序映射到索引号。例如，"United" 对应索引 1，"Lufthansa" 对应索引 2，"Virgin Air" 对应索引 3。显示最后一个变量以供确认。
show(x(3))
[ x('Virgin Air') ]
索引名称使您能够通过索引名称来寻址 x 的元素。例如：
route = 2*x("United") + 3*x("Virgin Air")
route =
Linear OptimizationExpression
2*x('United') + 3*x('Virgin Air')
创建变量后，您可以创建或更改索引名称。但是，构造之后就无法更改优化变量的大小。因此，您只能通过设置与原始变量索引大小相同的新名称来更改索引名称。例如：
x = optimvar("x",3,2); x.IndexNames = { ["row1","row2","row3"], ["col1","col2"] };
您可以为每个维度单独设置索引名称：
x.IndexNames{1} = ["row1", "row2", "row3"];
x.IndexNames{2} = ["col1", "col2"];您还可以为特定元素设置索引名称：
x.IndexNames{1}{2} = 'importantRow';检查变量的索引名称。
x.IndexNames{1}ans = 1×3 cell
{'row1'} {'importantRow'} {'row3'}
x.IndexNames{2}ans = 1×2 cell
{'col1'} {'col2'}
使用命名索引
您可以使用命名索引变量轻松创建和调试一些问题。例如，考虑由 x 中的名称索引的变量 vars：
vars = {'P1','P2','I1','I2','C','LE1','LE2','HE1','HE2',...
'HPS','MPS','LPS','BF1','BF2','EP','PP'};
x = optimvar('x',vars,'LowerBound',0);使用命名索引为 x 创建边界、目标函数和线性约束。
x('P1').LowerBound = 2500; x('I2').UpperBound = 244000; linprob = optimproblem; linprob.Objective = 0.002614*x('HPS') + 0.0239*x('PP') + 0.009825*x('EP'); linprob.Constraints.cons1 = x('I1') - x('HE1') <= 132000;
您可以在索引变量中随意使用字符串（" "）或字符向量（' '）。例如：
x("P2").LowerBound = 3000; x('MPS').LowerBound = 271536; showbounds(x)
2500 <= x('P1')
3000 <= x('P2')
0 <= x('I1')
0 <= x('I2') <= 244000
0 <= x('C')
0 <= x('LE1')
0 <= x('LE2')
0 <= x('HE1')
0 <= x('HE2')
0 <= x('HPS')
271536 <= x('MPS')
0 <= x('LPS')
0 <= x('BF1')
0 <= x('BF2')
0 <= x('EP')
0 <= x('PP')
使用字符串指定的变量（例如 x("P2")）与使用字符向量指定的变量（例如 x('MPS')）之间没有区别。
由于命名索引变量具有数值等效项，因此即使您有命名索引变量，也可以使用普通求和和冒号运算符。例如，您可以有以下形式的约束：
constr = sum(x) <= 100; show(constr)
x('P1') + x('P2') + x('I1') + x('I2') + x('C') + x('LE1') + x('LE2') + x('HE1') + x('HE2') + x('HPS') + x('MPS') + x('LPS') + x('BF1') + x('BF2') + x('EP') + x('PP') <= 100
y = optimvar('y',{'red','green','blue'},{'plastic','wood','metal'},... 'Type','integer','LowerBound',0); constr2 = y("red",:) == [5,7,3]; show(constr2)
(1, 1)
y('red', 'plastic') == 5
(1, 2)
y('red', 'wood') == 7
(1, 3)
y('red', 'metal') == 3
用索引变量查看解
使用指定的索引变量创建和求解优化问题。问题描述：将水果运往多个机场，使利润加权运量最大化，同时确保加权运量满足约束。
rng(0) % For reproducibility p = optimproblem(ObjectiveSense="maximize"); flow = optimvar("flow", ... ["apples", "oranges", "bananas", "berries"], ["NYC", "BOS", "LAX"], ... LowerBound=0,Type="integer"); p.Objective = sum(sum(rand(4,3).*flow)); p.Constraints.NYC = rand(1,4)*flow(:,"NYC") <= 10; p.Constraints.BOS = rand(1,4)*flow(:,"BOS") <= 12; p.Constraints.LAX = rand(1,4)*flow(:,"LAX") <= 35; sol = solve(p);
Solving problem using intlinprog.
Running HiGHS 1.11.0: Copyright (c) 2025 HiGHS under MIT licence terms
MIP has 3 rows; 12 cols; 12 nonzeros; 12 integer variables (0 binary)
Coefficient ranges:
Matrix [4e-02, 1e+00]
Cost [1e-01, 1e+00]
Bound [0e+00, 0e+00]
RHS [1e+01, 4e+01]
Presolving model
3 rows, 12 cols, 12 nonzeros 0s
3 rows, 12 cols, 12 nonzeros 0s
Solving MIP model with:
3 rows
12 cols (0 binary, 12 integer, 0 implied int., 0 continuous, 0 domain fixed)
12 nonzeros
Src: B => Branching; C => Central rounding; F => Feasibility pump; J => Feasibility jump;
H => Heuristic; L => Sub-MIP; P => Empty MIP; R => Randomized rounding; Z => ZI Round;
I => Shifting; S => Solve LP; T => Evaluate node; U => Unbounded; X => User solution;
z => Trivial zero; l => Trivial lower; u => Trivial upper; p => Trivial point
Nodes | B&B Tree | Objective Bounds | Dynamic Constraints | Work
Src Proc. InQueue | Leaves Expl. | BestBound BestSol Gap | Cuts InLp Confl. | LpIters Time
J 0 0 0 0.00% inf 945.5907645 Large 0 0 0 0 0.0s
S 0 0 0 0.00% 1079.371705 1027.233133 5.08% 0 0 0 0 0.0s
1 0 1 100.00% 1027.233133 1027.233133 0.00% 0 0 0 3 0.0s
Solving report
Status Optimal
Primal bound 1027.23313332
Dual bound 1027.23313332
Gap 0% (tolerance: 0.01%)
P-D integral 3.719551084e-05
Solution status feasible
1027.23313332 (objective)
0 (bound viol.)
0 (int. viol.)
0 (row viol.)
Timing 0.01 (total)
0.00 (presolve)
0.00 (solve)
0.00 (postsolve)
Max sub-MIP depth 0
Nodes 1
Repair LPs 0 (0 feasible; 0 iterations)
LP iterations 3 (total)
0 (strong br.)
0 (separation)
0 (heuristics)
Optimal solution found.
Intlinprog stopped at the root node because the objective value is within a gap tolerance of the optimal value, options.AbsoluteGapTolerance = 1e-06. The intcon variables are integer within tolerance, options.ConstraintTolerance = 1e-06.
找出运送至纽约和洛杉矶的橙子和浆果的最佳运量。
[idxFruit,idxAirports] = findindex(flow, ["oranges","berries"], ["NYC", "LAX"])
idxFruit = 1×2
2 4
idxAirports = 1×2
1 3
orangeBerries = sol.flow(idxFruit, idxAirports)
orangeBerries = 2×2
0 980
70 0
此结果表示不向 NYC 运送橙子，只将 70 份浆果运至 NYC，同时将 980 份橙子运至 LAX，而不向 LAX 运送浆果。
列出以下最佳运量：
Fruit Airports
----- --------
Berries NYC
Apples BOS
Oranges LAX
idx = findindex(flow, ["berries", "apples", "oranges"], ["NYC", "BOS", "LAX"])
idx = 1×3
4 5 10
optimalFlow = sol.flow(idx)
optimalFlow = 1×3
70 28 980
此结果表示将 70 份浆果运送至 NYC，将 28 份苹果运送至 BOS，将 980 份橙子运送至 LAX。
