Fastest way to cycle through a structure for a number

given: a structure and a double

return: the index of a structure that has that number

data:

structure(1).val = 50

structure(2).val = 47

structure(3).val = 22

given: structure, 50

return: 1

given: structure, 22

return: 3

% my attempt
function y = return_struct_index(structure,inputval)
for i = 1:length(structure)
    if structure(i).val==inputval
        y = i;
        return
    end
end
y = [];

This is rather slow and comes in at about 2.5 seconds for 1000 iterations for my code. I know for a fact there is a faster way to do this but haven't been able to track it down. Obviously I haven't vectored my function. I have also tried converting the structure to a cell array and then vectoring the solution, but the conversion makes it even slower.

Help greatly appreciated!!

Will

[Stephen23]

If there are non-scalar arrays in the field val:

S(1).val = 50;
S(2).val = 47;
S(3).val = [23,5];
fun = @(n)find(arrayfun(@(s)any(s.val(:)==n),S));

>> fun(22)
ans =
   Empty matrix: 1-by-0
>> fun(23)
ans =
     3

Answer 1

[James Tursa]

result = find([structure.val]==value);

E.g.,

>> structure(1).val = 50;
>> structure(2).val = 47;
>> structure(3).val = 22;
>> find([structure.val]==50)
ans =
     1
>> find([structure.val]==47)
ans =
     2
>> find([structure.val]==22)
ans =
     3
>> find([structure.val]==99)
ans =
   Empty matrix: 1-by-0

[Will Kinsman]

thanks so much. massively improved my programs performance by about 10 minutes. run time per 1000 iterations came in at about .5 seconds

Answer 2

[Azzi Abdelmalek]

s(1).val = 50
s(2).val = 47
s(3).val = 22
find([s(:).val]==50)