## Check if any elements of cell array are equal ?

### Hirak Basumatary (view profile)

on 13 Jun 2019
Latest activity Commented on by VK Bhardwaj

on 14 Jun 2019

### Stephen Cobeldick (view profile)

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

### KALYAN ACHARJYA (view profile)

on 13 Jun 2019
Yes, but it can be done using multiple steps, not in single function.
Hirak Basumatary

### Hirak Basumatary (view profile)

on 13 Jun 2019
Can you suggest me an algo?

### Products ### Stephen Cobeldick (view profile)

on 14 Jun 2019
Edited by Stephen Cobeldick

### Stephen Cobeldick (view profile)

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

on 14 Jun 2019
+1, dope solution!
Stephen Cobeldick

on 14 Jun 2019
Hirak Basumatary

### Hirak Basumatary (view profile)

on 14 Jun 2019
@stephen cobeldick: thank you very much. Wil remember this technique for my future problems.

on 13 Jun 2019

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

### KALYAN ACHARJYA (view profile)

on 13 Jun 2019
+1
Stephen Cobeldick

### Stephen Cobeldick (view profile)

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

on 14 Jun 2019
@Stephen Cobeldick: Thank you very much!

### VK Bhardwaj (view profile)

on 13 Jun 2019
Edited by VK Bhardwaj

### VK Bhardwaj (view profile)

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

on 14 Jun 2019
@Stephen Cobeldick: Thank you!
Hirak Basumatary

### Hirak Basumatary (view profile)

on 14 Jun 2019
@madhan ravi: thank you very much. Didn't spot this.
VK Bhardwaj

### VK Bhardwaj (view profile)

on 14 Jun 2019
@madhan ravi: Thanks for spotting the issue. I have updated the code.