# Check if any elements of cell array are equal ?

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Hirak Basumatary on 13 Jun 2019
Commented: VK Bhardwaj on 14 Jun 2019
Suppose if I have cell array a{1}=[1 1 0]; a{2}=[0 0 0]; a{3}=[0 0 0]. I want to check if any elements of the cell array are equal. If i use " isequal(a{:}) " then it returns "Logical 0". However, we can see that a{2} == a{3}. So, i need the answer to be "Logical 1" as some of the elements of the cell array are equal. Is there any built-in function to check that directly in MATLAB.

KALYAN ACHARJYA on 13 Jun 2019
Yes, but it can be done using multiple steps, not in single function.
Hirak Basumatary on 13 Jun 2019
Can you suggest me an algo?

Stephen Cobeldick on 14 Jun 2019
Edited: Stephen Cobeldick on 14 Jun 2019
Matching:
>> a{1}=[1,1,0];
>> a{2}=[9,9,9,9];
>> a{3}=[1,1,0];
>> a{4}=[];
>> N = numel(a);
>> [X,Y] = ndgrid(1:N);
>> Z = tril(true(N),-1);
>> any(arrayfun(@(x,y)isequal(a{x},a{y}),X(Z),Y(Z)))
ans = 1
vs. non-matching:
>> a{1}=[1,1,1];
>> any(arrayfun(@(x,y)isequal(a{x},a{y}),X(Z),Y(Z)))
ans = 0

madhan ravi on 14 Jun 2019
+1, dope solution!
Stephen Cobeldick on 14 Jun 2019
Hirak Basumatary on 14 Jun 2019
@stephen cobeldick: thank you very much. Wil remember this technique for my future problems.

madhan ravi on 13 Jun 2019
Edited: madhan ravi on 14 Jun 2019
This method works for cells with contents of any sizes:
a{1}=[1 1 0]; % example array
a{2}=[0 10];
a{3}=[1 1 0];
Result = false;
for k = 1:numel(a)
for l = 1:numel(a)
if k~=l
if isequal(a{k},a{l})
Result = true;
break
end
end
end
end
Note: The below two methods assume each cell has the same number of elements.
a{1}=[1 1 0];
a{2}=[0 0 0];
a{3}=[0 0 0];
A = vertcat(a{:});
Result = false;
for k = 1:numel(a)
if nnz(ismember(A,a{k},'rows'))>1
Result = true;
break
end
end
% or
idx=all(A'==permute(reshape(A,[],1,size(A,1)),[3,2,1]));
Result=any(squeeze(sum(idx,2))>1)
Choose which method you like the best , some mould can be given to the above but I’m off for the day perhaps will look into it tomorrow.

KALYAN ACHARJYA on 13 Jun 2019
+1
Stephen Cobeldick on 14 Jun 2019
+1 the nested loops (first algorithm) is probably the most efficient approach to this. Note that break only exits the inner loop while the outer loop keeps running (but this does not affect the result). An improvement to avoid testing pairs of arrays twice is to use the outer-loop's iteration variable to set the range of the inner loop (here the short-circuit || makes the code quite efficient without break), e.g.:
>> a{1}=[1,1,0];
>> a{2}=[9,9,9,9];
>> a{3}=[1,1,0];
>> a{4}=[];
>> N=numel(a);
>> Z=false;
>> for x=1:N, for y=x+1:N, Z=Z||isequal(a{x},a{y}); end, end
>> Z
Z = 1
madhan ravi on 14 Jun 2019
@Stephen Cobeldick: Thank you very much!

VK Bhardwaj on 13 Jun 2019
Edited: VK Bhardwaj on 14 Jun 2019
function y = checkequal(x)
% Input 'x' should be cell array
% Output 'y' logical value true. If any input cell array index is equal to
% another else false
% Example1:
% a{1}=[1 1 0]; a{2}=[0 0 0]; a{3}=[0 0 0];
% y = checkequal(a);
% Output is y = logical(1)
% Example2:
% a{1}=[1 1 0]; a{2}=[0 1 0]; a{3}=[0 0 0];
% y = checkequal(a);
% Output is y = logical(0)
y = false;
num = numel(x);
for i = 1:num
for j = 1:num
if i~=j
if isequal(x{i},x{j})
y = true;
return;
end
end
end
end
end

madhan ravi on 14 Jun 2019
@Stephen Cobeldick: Thank you!
Hirak Basumatary on 14 Jun 2019
@madhan ravi: thank you very much. Didn't spot this.
VK Bhardwaj on 14 Jun 2019
@madhan ravi: Thanks for spotting the issue. I have updated the code.