Getting rows that correspond to second occurrence of a value

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the cyclist
the cyclist 2013-7-16
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!

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回答（4 个）

Jos (10584)
Jos (10584) 2013-7-16
x(strfind([1 diff(x(:,1)).']==0,[0 1])+1,2)
  2 个评论
Jan
Jan 2013-7-16
编辑：Jan 2013-7-16
If you speak Matlab fluently, this looks elegant. For a beginner, this looks crude. +1

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Jan
Jan 2013-7-16
With FEX: RunLength:
[dummy, n, index] = RunLength(x(:, 1));
y = x(index(n >= 2) + 1, 2);
Cedric
Cedric 2013-7-16
编辑：Cedric 2013-7-16
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.
Andrei Bobrov
Andrei Bobrov 2013-7-31
编辑：Andrei Bobrov 2013-8-1
[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);

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