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Check if any elements of cell array are equal ?

조회 수: 86 (최근 30일)
Hirak Basumatary
Hirak Basumatary 2019년 6월 13일
댓글: VK Bhardwaj 2019년 6월 14일
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.
  댓글 수: 2
KALYAN ACHARJYA
KALYAN ACHARJYA 2019년 6월 13일
편집: KALYAN ACHARJYA 2019년 6월 13일
Yes, but it can be done using multiple steps, not in single function.
Hirak Basumatary
Hirak Basumatary 2019년 6월 13일
Can you suggest me an algo?

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채택된 답변

Stephen23
Stephen23 2019년 6월 14일
편집: Stephen23 2019년 6월 14일
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
  댓글 수: 3
Stephen23
Stephen23 2019년 6월 14일
편집: Stephen23 2019년 6월 14일
@madhan ravi: thank you :)
Hirak Basumatary
Hirak Basumatary 2019년 6월 14일
@stephen cobeldick: thank you very much. Wil remember this technique for my future problems.

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추가 답변 (2개)

madhan ravi
madhan ravi 2019년 6월 13일
편집: madhan ravi 2019년 6월 14일
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.
  댓글 수: 3
Stephen23
Stephen23 2019년 6월 14일
편집: Stephen23 2019년 6월 14일
+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 2019년 6월 14일
@Stephen Cobeldick: Thank you very much!

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VK Bhardwaj
VK Bhardwaj 2019년 6월 13일
편집: VK Bhardwaj 2019년 6월 14일
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
  댓글 수: 5
Hirak Basumatary
Hirak Basumatary 2019년 6월 14일
@madhan ravi: thank you very much. Didn't spot this.
VK Bhardwaj
VK Bhardwaj 2019년 6월 14일
@madhan ravi: Thanks for spotting the issue. I have updated the code.

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