Find double repetitions in a (sorted) array.
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bbb_bbb
el 20 de Oct. de 2017
Comentada: Andrei Bobrov
el 23 de Oct. de 2017
Given an array submitted in a form of struct field, containing integer numbers. For convenience, let's assume that the numbers are already sorted in ascending order:
>> s.x
ans =
2
ans =
2
ans =
5
ans =
5
ans =
5
ans =
8
ans =
8
Find indexes of elements, which occur exact 2 times:
ind =
1 2 6 7
4 comentarios
Jan
el 21 de Oct. de 2017
Editada: Jan
el 21 de Oct. de 2017
The iterative growing of arrays is a standard mistake from the view point of efficiency. Simply pre-allocate:
d = diff(x);
j = 0;
ind = zeros(1, numel(d));
indi = 1;
for i=1:numel(d)
if d(i)==0
j=j+1;
else
if j==1
ind(indi) = i-1;
ind(indi+1) = i;
indi = indi + 2;
end
j=0;
end
end
ind = ind(1:indi-1);
This does not catch the case, if the last two elements are equal.
Respuesta aceptada
Andrei Bobrov
el 20 de Oct. de 2017
Editada: Andrei Bobrov
el 23 de Oct. de 2017
x= [2; 2; 5; 5; 5; 8; 8; 13; 13; 13; 13];
[~,~,g] = unique(x); % OR for last versions of MATLAB: g = findgroups(x)
c = accumarray(g,1:numel(x),[],@(x){x});
out = cell2mat(c(cellfun(@numel,c) == 2));
or
[a,~,g] = unique(x);
out = find(ismember(x,a(accumarray(g,1) == 2)));
or (FIXED)
out = reshape(strfind([1,diff(x(:)')~=0,1],[1 0 1]) + [0;1],[],1);
out = reshape(bsxfun(@plus,strfind([1,diff(x(:)')~=0,1],[1 0 1]),[0;1]),[],1); % for old MATLAB
13 comentarios
Más respuestas (4)
Rik
el 20 de Oct. de 2017
Editada: Rik
el 20 de Oct. de 2017
It always pays off to get rid of loops and/or pre-allocating your output.
x= [2; 2; 5; 5; 5; 8; 8; 13; 13; 13; 13];
s = struct('x', num2cell(x));
x=[s.x];
%only newer releases: 0.000778 seconds
tic
count=histcounts(x,0.5 : max(x)+0.5);
ind=find(sum(x==find(count==2)'));
toc
%should work on most releases: 0.000628 seconds
tic
count=histcounts(x,0.5 : max(x)+0.5);
count=find(count==2);
ind=find(sum(repmat(x,length(count),1)==repmat(count',1,length(x))));
toc
%your loop: 0.001100 seconds
tic
d=diff(x); j=0; ind=[];
for i=1:numel(d)
if d(i)==0
j=j+1;
else
if j==1
ind(end+1)=i-1;
ind(end+1)=i;
end
j=0;
end
end
toc
8 comentarios
Rik
el 21 de Oct. de 2017
That's because x has a different shape:
x1= [2; 2; 5; 5; 5; 8; 8; 13; 13; 13; 13];
s = struct('x', num2cell(x1));
x2=[s.x];
x1 is 11x1 and x2 is 1x11
Image Analyst
el 20 de Oct. de 2017
You didn't tag it as homework. Is it? This will do it:
% Assignment of a struct with a field containing integer numbers
x= [2; 2; 5; 5; 5; 8; 8; 13; 13; 13; 13];
s = struct('x', num2cell(x));
numbers = [s.x]
[groupNumber, groupValue] = findgroups(numbers)
counts = histcounts(groupNumber)
ofGroupSize2 = find(counts == 2) % Find those only if they have a length of 2.
values = groupValue(ofGroupSize2)
indexes = find(ismember(numbers, values))
2 comentarios
Image Analyst
el 20 de Oct. de 2017
Editada: Image Analyst
el 21 de Oct. de 2017
You can use regionprops() instead of findgroups() if you have an old version and have the Image Processing Toolbox. See my separate answer with demo code.
Jan
el 21 de Oct. de 2017
Your code looks like the input is sorted. The other approaches do not have this limitation. If it is really sorted:
d = [true; diff(x) ~= 0]; % TRUE if values change
b = x(d); % Elements without repetitions
k = find([d', true]); % Indices of changes
n = diff(k);
is2 = find(n==2);
ind4 = reshape([k(is2); k(is2)+1], 1, []);
Code taken from FEX: RunLength.
0 comentarios
Image Analyst
el 21 de Oct. de 2017
Editada: Image Analyst
el 21 de Oct. de 2017
If you have the Image Processing Toolbox, you can use regionprops():
% Assignment of a struct with a field containing integer numbers
x= [2; 2; 5; 5; 5; 8; 8; 13; 13; 13; 13];
s = struct('x', num2cell(x));
numbers = [s.x] % A labeled "image"
% Find lengths of each run of numbers plus the indexes where they occur.
props = regionprops(numbers, 'Area', 'PixelIdxList')
% Extract from structure into one vector.
allLengths = [props.Area]
% Find those only if they have a length of 2.
ofGroupSize2 = find(allLengths == 2)
% Find indexes of those runs with length 2.
indexes = [props(ofGroupSize2).PixelIdxList]
% Shape into row vector
indexes = reshape(sort(indexes(:)), 1, [])
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