Problem 36. Find relatively common elements in matrix rows

Created by Cody Team

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You want to find all elements that exist in greater than 50% of the rows in the matrix.

For example, given

the answer would be the row vector

 [1  2  3  5]

since each of these appears three or more times in the rows of matrix A. Elements should be returned sorted ascending.

Note there is no guarantee that there are any such elements; if there are none, the routine should return the empty set. Also, the matrix might contain repeats, very large or very small numbers, or NaNs (though NaN should not be treated as a number to be returned in the output).

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Solution Stats

2328 Solutions
909 Solvers

Last Solution submitted on Jul 02, 2020

Last 200 Solutions

Problem Comments

6 Comments

6 Comments

Show 3 older comments

Jean-Marie Sainthillier on 2 Aug 2012

Tests aren't specifically adapted to this problem. The code for problem 67 (Find common elements in matrix works here).

Sergey on 7 Aug 2012

Absolutely agree with Jean-Marie.
Tests is not good enough.

Marlies on 4 Dec 2012

Not sure if the previous commenters have the same issue that I have, but I dislike that this test-suite makes a difference in how the empty-matrix is defined (See also comments for solution 167825).
I would suggest checking isempty' for the relevant solutions.

Heather on 5 Sep 2013

Ironically, simply including the example from the problem statement in the test suite would reject the current smallest ("best") solutions.

Peter on 18 Nov 2014

Indeed, the current leading solutions with size 29 yield only elements that exist in ALL of the rows in the matrix!

Simon Thomas on 20 Nov 2017

This needs another test case.
Many would fail this:

%%
x = [1 1 1 1;2 3 4 5;3 6 7 8];
y_correct = [3];
assert(isequal(common_by_row(x),y_correct))

Solution Comments

Solution 2251477

1 Comment

1 Comment

Haoyu Dai on 3 May 2020

function y = common_by_row(x)
tmp=[];
y=[];
z=[];
k=1;
[r,c]=size(x);
for i=1:r
for j=1:c
if x(i,j)~=NaN
if ismember(x(i,j),z)==0
z(k)=x(i,j);
k=k+1;
end
end
end
end
cnt=0;
for k=1:length(z)
for i=1:r
for j=1:c
if x(i,j)==z(k)
cnt=cnt+1;
end
end
end
tmp(k)=cnt;
cnt=0;
end
m=1;
for i=1:length(z)
if tmp(i)>0.5*r
y(m)=z(i);
m=m+1;
end
end
y=sort(y);
end

Solution 2047384

1 Comment

1 Comment

Paresh yeole on 5 Dec 2019

I know I have adapted code according to test cases. But was stuck in this one.

Solution 781377

2 Comments

2 Comments

Daniel Pereira on 20 Nov 2015

Don't know why an empty matrix 1-by-0 is not equal to the empty matrix [].

Jan Orwat on 20 Nov 2015

[] is 0-by-0. mtimes(ones(1,0), ones(0,1)) equals 0, mtimes([], ones(0,1)) is empty 0-by-1. This is one of the reasons.

Solution 435844

1 Comment

1 Comment

Pascal Geschwill on 28 Apr 2014

I used exactly the same code as for problem 67.

Solution 340331

1 Comment

1 Comment

Ze Zhang on 17 Mar 2015

Nice! But I think you made a mistake in if length(m)==size(x,1), according to the instruction, a number that appears more than half of total rows would be okay.

Solution 167825

2 Comments

2 Comments

Marlies on 25 Nov 2012

My solution fails on the magic, the wilkonson and the [1;2] inputs.
However, on my own computer each input works as expected.
Anyone has a hint on where I am going wrong?

Marlies on 4 Dec 2012

Found it.
Due to the way the test-suite is set-up (using 'isequal'), MATLAB makes a difference between the values "[]" (size [0 0]) and "Empty matrix: 1-by-0" (size [1 0]), even though they are both considered empty.

Problem Recent Solvers909

Problem Tags

matlab