Both of those cell arrays contain scalar structures in each cell, where all of the structures have the same fields. This is an inefficient way to store that data: it would be much simpler to access if these were non-scalar structures, so the first thing that we will do is convert from these superfluous cell arrays and combine into simpler structure arrays:
>> SN1 = cat(1,StateName1{:});
>> SN2 = cat(1,StateName2{:});
At this point we can see that each non-scalar structure has the same fields:
>> fieldnames(SN1)
ans =
MCOS
string
>> fieldnames(SN2)
ans =
MCOS
string
The string field is always empty:
>> any(~cellfun('isempty',{SN1.string}))
ans = 0
>> any(~cellfun('isempty',{SN2.string}))
ans = 0
Whereas the MCOS field is always a six element column vector:
>> find(~cellfun(@iscolumn,{SN1.MCOS}))
ans = []
>> find(~cellfun(@iscolumn,{SN2.MCOS}))
ans = []
It is not clear from your question how you expect to compare these arrays, but at this point you have several choices, such as using isequal in a loop, or perhaps concatenating into numeric matrices, transposing, and then comparing the rows:
>> mat1 = cat(2,SN1.MCOS).';
>> mat2 = cat(2,SN2.MCOS).';
>> find(ismember(mat2,mat1,'rows'))
ans = []
>> find(ismember(mat1,mat2,'rows'))
ans = []
None of the MCOS field values (i.e. column vectors) exist in the other structure array, so it is not clear what you expect to obtain from this comparison.
