findindex

Find numeric index equivalents of named index variables

Syntax

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

numindex = findindex(var,strindex1,strindex2,...,strindexk)

Description

[numindex1,numindex2,...,numindexk] = findindex(var,strindex1,strindex2,...,strindexk) finds the numeric index equivalents of the named index variables in the optimization variable var.

numindex = findindex(var,strindex1,strindex2,...,strindexk) finds the linear index equivalents of the named index variables.

Examples

Find Numeric Equivalents of Named Index Variables

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'})

idxadd = 1×2

     3     1

idxsub = 
4

Find Linear Index Equivalents of Named Index Variables

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"])

idx = 1×4

    17     3     9    15

View Solution with Index Variables

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 1.7.0: 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'})

idxFruit = 1×2

     2     4

idxAirports = 1×2

     1     3

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'})

idx = 1×3

     4     5    10

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 Initial Point with Named Index Variables

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

var — Optimization variable

OptimizationVariable object

Optimization variable, specified as an OptimizationVariable object. Create var using optimvar.

Example: var = optimvar('var',4,6)

strindex — Named index

cell array of character vectors | character vector | string vector | integer vector

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

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

Data Types: double | char | string | cell

Output Arguments

numindex — Numeric index equivalent

integer vector

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

• The number of dimensions in var. Each output vector numindexj is the numeric equivalent of the corresponding input argument strindexj.

• One. In this case, the size of each input strindexj must be the same for all j, and the output satisfies the linear indexing criterion

  var(numindex(j)) = var(strindex1(j),...,strindexk(j)) for all j.

Version History

Introduced in R2018a