looking condition in matrix

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Nejc
Nejc 2013 年 4 月 10 日
Hey,
I got matrix
[ 2 9841 5 4 2 1;
3 79463 1 3 1 0;
3 79463 4 3 1 0;
3 79463 5 3 1 0;
3 79463 1 3 2 0;
3 79463 4 3 2 0;
3 79463 5 3 2 0;
3 79463 1 4 1 0;
3 79463 4 4 1 0;
3 79463 5 4 1 0;
3 79463 1 4 2 0;
3 79463 4 4 2 0;
3 79463 5 4 2 0;
3 79463 1 4 1 1;
3 79463 4 4 1 1;
3 79463 5 4 1 1;
3 79463 4 4 1 1;
3 79463 5 4 1 1;
3 79212 1 3 1 0;]
I have to look for every row in column 1,2,4,5, if they match, then i check last column 6th, where is information if row data is correct or wrong.
If there are both correct and wrong data for rows that all include same values in column 1,2,4,5 than i must keep only correct data and erase wrong. If there is only wrong data for row than i keeep it.
The result should erase those two rows...
3 79463 4 4 1 0;
3 79463 5 4 1 0;
Thank you for your suggestions and solutions, take care.
Nejc
  4 件のコメント
Yao Li
Yao Li 2013 年 4 月 10 日
Sorry, I haven't got it yet. Maybe you can take the first 5 rows for example. Pls. give the results for the first 5 rows.
Nejc
Nejc 2013 年 4 月 10 日
No problem, first 5 rows: 2 9841 5 4 2 1; this row is only one which got same 1st,2nd,4th,5th parameters, we look for 6th and it's 1, ok we keep that row...
now next 3 rows have same 1st,2nd,4th,5th parameters, again we check for 6th column and each row have value 0...becasue we dont have 1 in that 6th, we keep that 3 rows.
The crucial here are rows which consist of 1st-3 2nd- 79463 4th-4 5th-1 parameters....We have 8 rows that meet those parameters. So we look 6 column now and there are both 0-wrong and 1- correct, because there are both options i want to keep just correct rows and erase wrong.

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回答 (1 件)

Andrei Bobrov
Andrei Bobrov 2013 年 4 月 10 日
編集済み: Andrei Bobrov 2013 年 4 月 10 日
a = [ 2 9841 5 4 2 1;3 79463 1 3 1 0;3 79463 4 3 1 0;3 79463 5 3 1 0;3 79463 1 3 2 0;3 79463 4 3 2 0;3 79463 5 3 2 0;3 79463 1 4 1 0;3 79463 4 4 1 0;3 79463 5 4 1 0;3 79463 1 4 2 0;3 79463 4 4 2 0;3 79463 5 4 2 0;3 79463 1 4 1 1;3 79463 4 4 1 1;3 79463 5 4 1 1;3 79463 4 4 1 1;3 79463 5 4 1 1;3 79212 1 3 1 0;];
[~,~,ii] = unique(a(:,[1 2 4 5]),'rows');
i1 = histc(ii,1:max(ii));
out = a(ismember(ii,find(i1 > 1)) & a(:,end) == 1,:);
  1 件のコメント
Nejc
Nejc 2013 年 4 月 10 日
[ 2 9841 5 4 2 1;
3 79463 1 3 1 0;
3 79463 4 3 1 0;
3 79463 5 3 1 0;
3 79463 1 3 2 0;
3 79463 4 3 2 0;
3 79463 5 3 2 0;
3 79463 1 4 2 0;
3 79463 4 4 2 0;
3 79463 5 4 2 0;
3 79463 1 4 1 1;
3 79463 4 4 1 1;
3 79463 5 4 1 1;
3 79463 4 4 1 1;
3 79463 5 4 1 1;
3 79212 1 3 1 0;]
without
3 79463 1 4 1 0;
3 79463 4 4 1 0;
3 79463 5 4 1 0;
but i appreciate your effort.

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