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findindex

查找命名索引变量的等效数值索引

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语法

[numindex1,numindex2,...,numindexk] = findindex(var,strindex1,strindex2,...,strindexk)
numindex = findindex(var,strindex1,strindex2,...,strindexk)

说明

[numindex1,numindex2,...,numindexk] = findindex(var,strindex1,strindex2,...,strindexk) 在优化变量 var 中查找命名索引变量的等效数值索引。

示例

numindex = findindex(var,strindex1,strindex2,...,strindexk) 查找命名索引变量的等效线性索引。

示例

示例

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打开实时脚本

创建一个名为 colors 的优化变量，该变量按主要加色名称和主要减色名称进行索引。在加色名称中包括 "black" 和 "white"，在减色名称中包括 "black"。

colors = optimvar("colors",["black","white","red","green","blue"],["cyan","magenta","yellow","black"]);

查找加色 "red" 和 "black" 以及减色 "black" 的索引编号。

[idxadd,idxsub] = findindex(colors,["red","black"],["black"])
idxadd = 1×2

     3     1


idxsub = 
4
打开实时脚本

创建一个名为 colors 的优化变量，该变量按主要加色名称和主要减色名称进行索引。在加色名称中包括 "black" 和 "white"，在减色名称中包括 "black"。

colors = optimvar("colors",["black","white","red","green","blue"],["cyan","magenta","yellow","black"]);

查找 ["white","black"]、["red","cyan"]、["green","magenta"] 和 ["blue","yellow"] 组合的等效线性索引。

idx = findindex(colors,["white","red","green","blue"],["black","cyan","magenta","yellow"])
idx = 1×4

    17     3     9    15
打开实时脚本

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。

打开实时脚本

为一个问题创建命名索引变量，该问题涉及不同土地类型、可能种植的农作物和耕作方法。

land = ["irr-good","irr-poor","dry-good","dry-poor"];
crops = ["wheat-lentil","wheat-corn","barley-chickpea","barley-lentil","wheat-onion","barley-onion"];
plow = ["tradition","mechanized"];
xcrop = optimvar("xcrop",land,crops,plow,LowerBound=0);

将初始点设置为正确大小的由零组成的数组。

x0.xcrop = zeros(size(xcrop));

对于可在任何干旱条件下种植的传统耕作农作物 "wheat-onion" 和 "wheat-lentil"，将初始值设置为 3000。

[idxLand, idxCrop, idxPlough] = findindex(xcrop, ["dry-good","dry-poor"], ...
             ["wheat-onion","wheat-lentil"],"tradition");
x0.xcrop(idxLand,idxCrop,idxPlough) = 3000;

为以下三个点设置初始值。

Land      Crops           Method      Value
dry-good  wheat-corn      mechanized  2000
irr-poor  barley-onion    tradition   5000
irr-good  barley-chickpea mechanized  3500

idx = findindex(xcrop,...
    ["dry-good","irr-poor","irr-good"],...
    ["wheat-corn","barley-onion","barley-chickpea"],...
    ["mechanized","tradition","mechanized"]);
x0.xcrop(idx) = [2000,5000,3500];

输入参数

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优化变量，指定为 OptimizationVariable 对象。使用 optimvar 创建 var。

示例: var = optimvar("var",4,6)

命名索引，指定为字符向量元胞数组、字符向量、字符串向量或整数向量。strindex 参量的数目必须等于 var 中的维数。

示例: ["small","medium","large"]

数据类型: double | char | string | cell

输出参量

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等效数值索引，以整数向量形式返回。输出参量的数目必须为下列值之一：

  • var 中的维数。每个输出向量 numindexj 是对应输入参量 strindexj 的等效数值索引。

  • 1。在这种情况下，对于所有 j，每个输入 strindexj 的大小必须相同，并且输出满足线性索引条件

    var(numindex(j)) = var(strindex1(j),...,strindexk(j))（对于全部 j）。

版本历史记录

在 R2018a 中推出

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