清空筛选条件
清空筛选条件

Find double repetitions in a (sorted) array.

1 次查看（过去 30 天）
bbb_bbb
bbb_bbb 2017-10-20
评论： Andrei Bobrov 2017-10-23
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 个评论
显示 2更早的评论
Jan
Jan 2017-10-21
编辑：Jan 2017-10-21
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.
bbb_bbb
bbb_bbb 2017-10-21
This does not catch the case, if the last two elements are equal.
Adding this line repairs this:
if d(end)==0, d(end+1)=1; end
This variant seems to be the fastest.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Andrei Bobrov
Andrei Bobrov 2017-10-20
编辑：Andrei Bobrov 2017-10-23
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 个评论
显示 11更早的评论
bbb_bbb
bbb_bbb 2017-10-23
This is ok. I sligtly modified it for speed. The fastest and concisiest algorithm of all suggested!
out = strfind([true,diff(x')~=0,true],[1 0 1]);
out = reshape([out;out+1],1,[]);

请先登录，再进行评论。

更多回答（4 个）

Rik
Rik 2017-10-20
编辑：Rik 2017-10-20
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 个评论
显示 6更早的评论
Rik
Rik 2017-10-21
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
bbb_bbb
bbb_bbb 2017-10-21
This variant isn't working at big arrays:
x=randi([1,1e6],1e5,1); x=sort(x)';
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))));
Error using repmat
Maximum variable size allowed by the program is exceeded.

请先登录，再进行评论。

Image Analyst
Image Analyst 2017-10-20
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 个评论
bbb_bbb
bbb_bbb 2017-10-20
编辑：bbb_bbb 2017-10-21
No, its no homework - so called "just-for-fun project".
[groupNumber, groupValue] = findgroups(numbers)
Undefined function or variable 'findgroups'.
Matlab 2015a
Image Analyst
Image Analyst 2017-10-20
编辑：Image Analyst 2017-10-21
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.

请先登录，再进行评论。

Jan
Jan 2017-10-21
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.
Image Analyst
Image Analyst 2017-10-21
编辑：Image Analyst 2017-10-21
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, [])
  1 个评论
bbb_bbb
bbb_bbb 2017-10-21
This works, but the variant is the longest (6.2 sec on 1e6 elements vector). The fastest is 0.03 sec.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by