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Check if any elements of cell array are equal ?

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Hirak Basumatary
Hirak Basumatary am 13 Jun. 2019
Kommentiert: VK Bhardwaj am 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.
  2 Kommentare
KALYAN ACHARJYA
KALYAN ACHARJYA am 13 Jun. 2019
Bearbeitet: KALYAN ACHARJYA am 13 Jun. 2019
Yes, but it can be done using multiple steps, not in single function.
Hirak Basumatary
Hirak Basumatary am 13 Jun. 2019
Can you suggest me an algo?

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Akzeptierte Antwort

Stephen23
Stephen23 am 14 Jun. 2019
Bearbeitet: Stephen23 am 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
  3 Kommentare
Stephen23
Stephen23 am 14 Jun. 2019
Bearbeitet: Stephen23 am 14 Jun. 2019
@madhan ravi: thank you :)
Hirak Basumatary
Hirak Basumatary am 14 Jun. 2019
@stephen cobeldick: thank you very much. Wil remember this technique for my future problems.

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Weitere Antworten (2)

madhan ravi
madhan ravi am 13 Jun. 2019
Bearbeitet: madhan ravi am 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.
  3 Kommentare
Stephen23
Stephen23 am 14 Jun. 2019
Bearbeitet: Stephen23 am 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
madhan ravi am 14 Jun. 2019
@Stephen Cobeldick: Thank you very much!

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VK Bhardwaj
VK Bhardwaj am 13 Jun. 2019
Bearbeitet: VK Bhardwaj am 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
  5 Kommentare
Hirak Basumatary
Hirak Basumatary am 14 Jun. 2019
@madhan ravi: thank you very much. Didn't spot this.
VK Bhardwaj
VK Bhardwaj am 14 Jun. 2019
@madhan ravi: Thanks for spotting the issue. I have updated the code.

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