MATLAB - search and number of duplicates in an array
I have an array of values, some of which have duplicates, for example:
a = [5;5;4;7;7;3;3;9;5;7]
and I would like to find duplicates and then list each one sequentially, making the null duplicates zeros. For example:
b = [1;1;0;2;2;3;3;0;1;2]
I currently have a very inefficient and incomplete approach using function unique
and various loops for
and if
, but feel like there should be a simple answer.
What's the most efficient way to get this answer?
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Here are two liners that will also work for non-sequential duplicates
[c, ia, ic] = unique(a, 'stable');
[~, b] = ismember(a, a(ia(accumarray(ic,1)>1)));
I used some ideas from @ excaza's answer with modifications.
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You can use a combination unique
, accumarray
and ismember
make the necessary adjustments:
a = [5;5;4;7;7;3;3;9]; % Identify unique values and their counts [uniquevals, ~, ia] = unique(a, 'stable'); % Stable keeps it in the same order bincounts = accumarray(ia, 1); % Count the frequency of each index in ia % Zero out singles singles = uniquevals(bincounts <= 1); [~, singleidx] = intersect(a, singles); a(singleidx) = 0; % Overwrite repeats repeats = uniquevals(bincounts > 1); [~, a] = ismember(a, repeats);
Which returns a new one a
from:
a = 1 1 0 2 2 3 3 0
Step by step guide
We use unique
here to find all the unique values in our input array a
. We also store an additional third output, which is a mapping of values a
to their index in an array of unique values. Note that we are using a parameter stable
to retrieve the unique values in the order they were first encountered in a
; results unique
are sorted by default.
We then use accumarray
to accumulate the indices obtained from unique
, which gives us the score of each index. When using boolean indexing , we first use these counts to zero out single instances. After clearing them, we can abuse the use of the second pin ismember
to return the final answer.
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Here is a solution based on indexing, boolean operators and cumsum :
x = [false; a(2:end)==a(1:end-1)]; %logical indexes of repeated elements except the first element of each block
y = [x(2:end)|x(1:end-1) ;x(end)]; %logical indexes of repeated elements
result = cumsum(~x&y).*y %cumsum(...):number all elements sequentially and (... .* y): making non-duplicates zero
Edit:
As an edited question, to manipulate inconsistent duplicates, you can do this:
[s ii] = sort(a);
x = [false ;s(2:end)==s(1:end-1)];
y = [x(2:end)|x(1:end-1) ;x(end)];
first = ~x&y;
[~,ix]=sort(ii(first));
un(ix,1)=1:numel(ix);
result(ii,1)=un(cumsum(first)).*y;
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Here's a different approach:
a = [5;5;4;7;7;3;3;9;5;7]; [u, ~, w] = unique(a, 'stable'); s = find(sum(bsxfun(@eq, a, u.'), 1) > 1); b = sum(bsxfun(@times, bsxfun(@eq, w, s), 1:numel(s)), 2);
In R2016b onwards, you can simplify the syntax:
a = [5;5;4;7;7;3;3;9;5;7]; [u, ~, w] = unique(a, 'stable'); s = find(sum(a==u.', 1) > 1); b = sum((w==s).*(1:numel(s)), 2);
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