findindex - Find numeric index equivalents of named index variables - MATLAB (original) (raw)

Find numeric index equivalents of named index variables

Syntax

Description

[[numindex](#d126e101200)1,[numindex](#d126e101200)2,...,[numindex](#d126e101200)k] = findindex([var](#d126e101133),[strindex](#d126e101154)1,[strindex](#d126e101154)2,...,[strindex](#d126e101154)k) finds the numeric index equivalents of the named index variables in the optimization variable var.

example

[numindex](#d126e101200) = findindex([var](#d126e101133),[strindex](#d126e101154)1,[strindex](#d126e101154)2,...,[strindex](#d126e101154)k) finds the linear index equivalents of the named index variables.

example

Examples

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Create an optimization variable named colors that is indexed by the primary additive color names and the primary subtractive color names. Include 'black' and 'white' as additive color names and 'black' as a subtractive color name.

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

Find the index numbers for the additive colors 'red' and 'black' and for the subtractive color 'black'.

[idxadd,idxsub] = findindex(colors,{'red','black'},{'black'})

Create an optimization variable named colors that is indexed by the primary additive color names and the primary subtractive color names. Include 'black' and 'white' as additive color names and 'black' as a subtractive color name.

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

Find the linear index equivalents to the combinations ["white","black"], ["red","cyan"], ["green","magenta"], and ["blue","yellow"].

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

Create and solve an optimization problem using named index variables. The problem is to maximize the profit-weighted flow of fruit to various airports, subject to constraints on the weighted flows.

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 1.7.1: Copyright (c) 2024 HiGHS under MIT licence terms 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) 12 nonzeros

    Nodes      |    B&B Tree     |            Objective Bounds              |  Dynamic Constraints |       Work      
 Proc. InQueue |  Leaves   Expl. | BestBound       BestSol              Gap |   Cuts   InLp Confl. | LpIters     Time

     0       0         0   0.00%   1160.150059     -inf                 inf        0      0      0         0     0.0s

S 0 0 0 0.00% 1160.150059 1027.233133 12.94% 0 0 0 0 0.0s

Solving report Status Optimal Primal bound 1027.23313332 Dual bound 1027.23313332 Gap 0% (tolerance: 0.01%) Solution status feasible 1027.23313332 (objective) 0 (bound viol.) 0 (int. viol.) 0 (row viol.) Timing 0.00 (total) 0.00 (presolve) 0.00 (postsolve) Nodes 1 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.

Find the optimal flow of oranges and berries to New York and Los Angeles.

[idxFruit,idxAirports] = findindex(flow, {'oranges','berries'}, {'NYC', 'LAX'})

orangeBerries = sol.flow(idxFruit, idxAirports)

orangeBerries = 2×2

 0   980
70     0

This display means that no oranges are going to NYC, 70 berries are going to NYC, 980 oranges are going to LAX, and no berries are going to LAX.

List the optimal flow of the following:

Fruit Airports

----- --------

Berries NYC

Apples BOS

Oranges LAX

idx = findindex(flow, {'berries', 'apples', 'oranges'}, {'NYC', 'BOS', 'LAX'})

optimalFlow = sol.flow(idx)

optimalFlow = 1×3

70    28   980

This display means that 70 berries are going to NYC, 28 apples are going to BOS, and 980 oranges are going to LAX.

Create named index variables for a problem with various land types, potential crops, and plowing methods.

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);

Set the initial point to a zero array of the correct size.

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

Set the initial value to 3000 for the "wheat-onion" and "wheat-lentil" crops that are planted in any dry condition and are plowed traditionally.

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

Set the initial values for the following three points.

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];

Input Arguments

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Named index, specified as a cell array of character vectors, character vector, string vector, or integer vector. The number ofstrindex arguments must be the number of dimensions in var.

Example: ["small","medium","large"]

Data Types: double | char | string | cell

Output Arguments

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Numeric index equivalent, returned as an integer vector. The number of output arguments must be one of the following:

Version History

Introduced in R2018a