Getting rows that correspond to second occurrence of a value

I have a matrix x:

x = [1 11;
     1 12;
     1 13;
     2 26;
     2 27;
     2 28;
     3 34;
     4 41;
     4 42]

x is guaranteed to be sorted by the first column and then the second column.

From the second column, I want the second occurrence (if any) for each unique value in the first column. So, in my example, I want my output to be

 y = [12; 27; 42];

My first thought was to do this with accumarray:

y = accumarray(x(:,1),x(:,2),[],@(x)x(2));

but this errors because there is no second row when x(:,1)==3.

Thoughts? I don't need this to be fast, or handle huge arrays; I'd like it to be elegant!

Answer 1

[Jos (10584)]

x(strfind([1 diff(x(:,1)).']==0,[0 1])+1,2)

[Jos (10584)]

oh, and add a transpose at the end ;-)

[Jan]

If you speak Matlab fluently, this looks elegant. For a beginner, this looks crude. +1

Answer 2

[Jan]

With FEX: RunLength:

[dummy, n, index] = RunLength(x(:, 1));
y = x(index(n >= 2) + 1, 2);

Answer 3

[Cedric]

Here is another one liner:

    >>   regexp('','(?@ buffer=urlread(''http://en.wikipedia.org/wiki/Sea_urchin''); eval(''x(strfind([1 diff(x(:,1)).'''']==0,[0 1])+1,str2num(buffer(regexp(buffer, ''''\(?<=size from 6 to 1\)\\d'''', ''''start''''))))''))') ;

Isn't that elegant? ;-)

C.

[Jan]

If you speak Matlab fluently, this looks crude. For a beginner, this looks crude. +1

Answer 4

[Andrei Bobrov]

[a,b] = unique(x(:,1),'first');
out = x(b(histc(x(:,1),a)>=2)+1,2);

or

[~,i1] = unique(x(:,1),'first');
[~,i2] = unique(x(:,1),'last');
out= x(i1((i2 - i1) > 0) + 1,2);