Can we do something like this similar to union()?

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Kevin
Kevin 2018-2-10
编辑： Stephen23 2018-2-13
Currently union() returns non-repeated result. But I want repeat result like the following,
>> myunion([2 3 3 5], [1 3 5])
ans =
1 2 3 3 5
I can write for-loop and if-statement to do this. But is there a better or easy way?
Thanks Kevin
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Stephen23
Stephen23 2018-2-10
Kevin's "Answer" moved here:
Sorry, my fault. I should be more specific in my first post.
>> myunion([2 3 3 5], [1 3 5])
ans =
1 2 3 3 5
  • The first element (in the output vector) is 1 because the second input vector has a 1.
  • The second element (in the output vector) is 2 because the first input vector has a 2.
  • The third and forth elements (in the output vector) are 3 because the first input vector has [3 3]. Now if first input vector has [3 3 3], then output vector vector should have three 3's, i.e. I want the output vector should have the repeated elements (number of repeated elements = maximum of the number of elements in either input vector).
So MATLAB's union() is almost perfect except that it does not returned the repeated elements.
>> union([2 3 3 5], [1 3 5])
ans =
1 2 3 5
union() would be perfect if the output vector has two 3's since the first input vector has 2 3's.
Hope this is clear.
Kevin
John D'Errico
John D'Errico 2018-2-11
编辑：John D'Errico 2018-2-11
I'm a bit confused by the logic in what you want. As you wrote it, you want
myunion([2 3 3 5], [1 3 5])
to return the vector [1 2 3 3 5]. But by extension, if you simply swapped the arguments, calling it as
myunion([1 3 5], [2 3 3 5])
then it would return [1 2 3 5].
SHIVER. This is just asking to create the code from hell, i.e., code that will create all sorts of strange bugs, arising from that non-symmetrical behavior. Any code that claims to be a union should also be symmetrical in the arguments.
My personal prediction is that this code will soon be the source of a new, frenzied post, "Why do I have this strange bug in my code? Forewarned is forearmed. :)

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Stephen23
Stephen23 2018-2-11
编辑：Stephen23 2018-2-13
Method one: unique, arrayfun, ones:
>> A = [2,3,3,5];
>> B = [1,3,5];
>> fun = @(n)n*ones(1,max(nnz(A==n),nnz(B==n)));
>> C = arrayfun(fun,unique([A(:);B(:)]),'uni',0);
>> U = [C{:}]
U =
1 2 3 3 5
Note: the arrayfun call is just for convenience, not for speed!
Method two: unique, histc, repelem:
>> A = [2,3,3,5];
>> B = [1,3,5];
>> Q = unique([A(:);B(:)].');
>> repelem(Q,max([histc(A,Q);histc(B,Q)]))
ans =
1 2 3 3 5
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Kevin
Kevin 2018-2-12
Hi Stephen,
You are brilliant. Thank you for your idea on using HIST. Your method 2 is very elegant and much faster.
Found a corner case that breaks method 2,
A = [3 3]; B = [3 3 3];
This causes problem into the call of HIST because Q is a scalar and treated as number of bins. So use HISTC instead.
Fix:
Q = unique([A(:); B(:)].');
U = repelem(Q, max([histc(A, Q); histc(B, Q)]));
Thank you very much for your idea. Kevin
Stephen23
Stephen23 2018-2-13
编辑：Stephen23 2018-2-13
@Kevin: thank you for the feedback. I will include your suggested fix into my answer (for future readers).

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更多回答（1 个）

John D'Errico
John D'Errico 2018-2-10
编辑：John D'Errico 2018-2-10
My guess is that your ambiguous example was not even representative of your problem, given your words. So let me guess and try to answer your words. :)
Your words said that you want the union, but one that allows repeats. In that case it should have produced:
myunion([2 3 3 5], [1 3 5])
ans =
1 2 3 3 3 5 5
since there were 3 copies of the number 3, and 2 copies of the number 5.
If so, then the answer is trivial.
myunion = @(X,Y) sort([X,Y]);
You might need to worry about the orientation of the vectors, in case they might be row OR column vectors or both. But that is an easy change.

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