Find position of cell array within another cell array or table or in a structure

79 vues (au cours des 30 derniers jours)
Timothy
Timothy le 17 Déc 2025 à 15:28
Commenté : Fangjun Jiang le 19 Déc 2025 à 19:11
I have a number of nodes on a graph and a number of edges that connect them. Attached to each edge is a structure wear the first field is the EndNodes data from the edge table which defines the edges. I want to search the struct array to find the structure that matches the selected edge EndNodes but I can't figure out the syntax. The closest I've come is something like this where EdgeData is the struct array and data is the 2x1 cell containing the desired EndNodes:
selectedEdge=EdgeData(ismember(data,array2table({app.EdgeData.EndNodes}).Var1))
This fails because the inputs of ismember must be character arrays rather than cell arrays containing character arrays. Concatenating them could work but I would then simply be detecting if the two names of the edges are present and not whether they are contained together within one field. I've tried ==, ismember, strcmp, isequal and I've run out of things to try and ideas of documentation to comb through. It's possible that arrayfun might be the way forward but I don't really understand it honestly
PS yes I'm using app designer hence the referencing of EdgeData as a property
  2 commentaires
Fangjun Jiang
Fangjun Jiang le 17 Déc 2025 à 15:53
Providing an example of the data would be most helpful to understand the need. You can construct a simplifed data using code.
Timothy
Timothy le 17 Déc 2025 à 16:35
Of course. As mentioned the data is all EndNode data which, in case you're not familiar with the required format of edges on graphs, are 1x2 cells (sorry not 2x1) where each cell in the array is the identifier which references a specific node - in this case each cell array contains two character arrays which is the name of a train junction.
For example each cell would look something like this:
data={{'Littleport'}{'West River'}}

Connectez-vous pour commenter.

Réponse acceptée

Fangjun Jiang
Fangjun Jiang le 17 Déc 2025 à 17:05
Something like this?
data={{'Littleport'},{'West River'}};
EndNodes={{'Bigport'},{'West River'};
{'Littleport'},{'West River'};
{'Littleport'},{'East River'}};
Index=ismember(string(EndNodes),string(data),'row')
Index = 3×1 logical array
0 1 0
  5 commentaires
Timothy
Timothy le 19 Déc 2025 à 17:04
This is perfect. Been knocking my head against the table for 3 days on this. I knew it would be something to do with cellfun but I still don't know how to use it. Only note for anyone who looks at this in the future is that for a list derived from a struct array, curly braces {} are needed to create a cell otherwise the transpose function sees each value separately and considers it "too many arguments"
Fangjun Jiang
Fangjun Jiang le 19 Déc 2025 à 19:11
The lesson learned here is to provide an "EXACT" example inputs and the expected outputs for the most efficient Q&A. Imaging if you had provided the correct "data" and "EndNodes" initially, there would be no need for this lengthy discussion. You don't even need to mention your specific field like "graph" or "edge".
You can use the "Insert a line of code" button and then "RUN" it to provide a valid example data. Note that the example you previously provided in your comment text is not valid.
data={{'Littleport'}{'West River'}}
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

Connectez-vous pour commenter.

Plus de réponses (1)

Steven Lord
Steven Lord le 18 Déc 2025 à 15:53
Are you using a graph or digraph object? Do you have the additional struct information stored as a separate variable or have you added that information as custom attributes in the Edges and/or Nodes properties of the graph/digraph object? I recommend the latter if you're using the object.
If you have that information as custom attributes of a graph or digraph object, to locate the data for a specific edge given the end nodes use the findedge function. Then use that edge index to index into the Edges property.
s = [1 1 2 2 3];
t = [2 4 3 4 4];
G = digraph(s,t);
G.Edges.Weight = [10 20 30 40 50]';
disp(G.Edges)
EndNodes Weight ________ ______ 1 2 10 1 4 20 2 3 30 2 4 40 3 4 50
whichEdge = findedge(G, 2, 3)
whichEdge = 3
theWeight = G.Edges{whichEdge, 'Weight'}
theWeight = 30
Note that if you've constructed the digraph using a possibly unordered set of source and target indices, and the additional edge information is stored in a separate variable, the order may be different. The EndNodes list order is not guaranteed to be the same as the order in which the edges were specified when the graph/digraph object was created.
s2 = [1 1 3 2 2];
t2 = [2 4 4 3 4];
w2 = [10 20 50 30 40];
D = digraph(s2, t2, w2);
D.Edges
ans = 5×2 table
EndNodes Weight ________ ______ 1 2 10 1 4 20 2 3 30 2 4 40 3 4 50
The third edge in the Edges property is not the one specified by the third elements in s2, t2, and w2. So if you were to findedge and use that to index into w2, you'd get the wrong answer. That's one reason I recommend storing both the edge information and the additional attributes inside the graph/digraph object.
E = findedge(D, 2, 3) % Correct edge located
E = 3
weight = w2(E) % but w2 is in a different order than Weight so wrongly returns 50
weight = 50
weight = D.Edges{E, "Weight"} % Correctly returns weight of 30
weight = 30
  5 commentaires
Steven Lord
Steven Lord le 19 Déc 2025 à 13:20
What's in the cell? Is it text data? If so there are lots of functions for searching for data in cell arrays containing text.
C = {'apple', 'banana', 'cherry', 'durian'}
C = 1×4 cell array
{'apple'} {'banana'} {'cherry'} {'durian'}
namesContainingTheLetterA = contains(C, 'a')
namesContainingTheLetterA = 1×4 logical array
1 1 0 1
C(namesContainingTheLetterA)
ans = 1×3 cell array
{'apple'} {'banana'} {'durian'}
namesWith6Letters = strlength(C) == 6
namesWith6Letters = 1×4 logical array
0 1 1 1
C(namesWith6Letters)
ans = 1×3 cell array
{'banana'} {'cherry'} {'durian'}
startsWithB = startsWith(C, 'b')
startsWithB = 1×4 logical array
0 1 0 0
C(startsWithB)
ans = 1×1 cell array
{'banana'}
The set functions like ismember may also be of use.
Timothy
Timothy le 19 Déc 2025 à 13:32
Each cell is a cell array of two strings as I said. Using the fruit example it could be {{'apple', 'banana'} {'cherry', 'durian'}}. This why ismember and contains havent been working for me

Connectez-vous pour commenter.

Catégories

En savoir plus sur Matrices and Arrays dans Help Center et File Exchange

Produits


Version

R2024b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by