OBTW, in this case A,B are floating point values and am actually using ismembertol because exact comparison fails for some elements...
ISMEMBER() modified result desired
17 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
[lia,locb]=ismember(A,B);
returns the first/last matching location in B of the elements in A that occur in B depending on which optional parameters one chooses. That's fine for many cases, but...
If it is desired to do a 1:1 matching of elements in A to their corresponding mates in B, a modified behavior would be a desirable option if there are repeated elements -- locB could return the next match after the previous for those elements that may not be unique in either array.
This can be done in a loop, of course, by making a search for each element in A in turn, either keeping an indexing variable as to last location for each element previously located or, conversely, marking the location in B with a missing value or somesuch.
That's the problem, the question is --
Is there some function which already does this that I'm not aware of/can't find or a "more cleverer" way to make the match?
5 comentarios
Jan
el 21 de En. de 2019
What about a small example?
A = [1, 2, 3, 1, 2, 3];
B = [2, 2, 2, 3, 3];
Is the wanted output this:
lia = [0, 1, 1, 0, 1, 1] % as logical
locb = [0, 1, 4, 0, 2, 5]
% ^ ^ would be 1 and 4 for standard ISMEMBER
Respuestas (2)
Jan
el 21 de En. de 2019
Editada: Jan
el 21 de En. de 2019
Do you mean something like this:
function [lia, locb] = ismemberNext(A, B)
lia = false(size(A));
locb = zeros(size(A));
for iA = 1:numel(A)
match = (A(iA) == B);
if any(match)
lia(iA) = true;
index = find(match, 1);
locb(iA) = index;
B(index) = NaN; % Mask first match for following calls
end
end
end
2 comentarios
Guillaume
el 21 de En. de 2019
Maybe I misunderstood the problem, but ismembertol already has the option to return all matching indices in B:
[found, where] = ismembertol(A, B, 'OutputAllIndices', true)
3 comentarios
Guillaume
el 21 de En. de 2019
Are the A that match the same set of B exactly identical? If so, then it's easily solved. If there's a small tolerance on the As as well, then I'll have to think about it.
Ver también
Categorías
Más información sobre Creating and Concatenating Matrices en Help Center y File Exchange.
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!