What would be the way to remove duplicates for defined numbers in array?

1 次查看（过去 30 天）
Mantas Vaitonis
Mantas Vaitonis 2018-7-7
评论： Mantas Vaitonis 2018-7-8
Hello to All,
I am trying to solve the following situation. There is array IN (NxM), size my vary. My input is
IN=
9 -1 9 -1 -1
9 9 9 9 9
9 10 9 -1 -1
9 9 9 10 9
9 9 10 10 -1
9 -1 9 9 9
-1 9 -1 9 9
-1 -1 9 9 9
Output supposes to look like this:
9 -1 9 -1 -1
9 9 9 9 9
9 10 9 -1 -1
9 9 9 10 9
9 9 10 0 -1
9 -1 9 9 9
-1 9 -1 9 9
0 -1 9 9 9
Only numbers 10 and -1 can not have repeated values, if they are repeated, they should be changed to zero. How could it be achieved?
  3 个评论
显示 1更早的评论
Image Analyst
Image Analyst 2018-7-7
编辑：Image Analyst 2018-7-7
And for the 10 you zero out the second 10 but for the bottom row you zero out the first -1. Have you thought this through? Please read this link to make it easy for us to help you. Please explain the "use case", i.e. why you want to do this so we have some context.
Mantas Vaitonis
Mantas Vaitonis 2018-7-7
I forgot to mention that repeated values can not be column-vise, however I went through my algorithm and the logic of this part should be changed. The idea was to detect when signals in two arrays are not connect, one array A is filled randomly with ones and other array B with minus ones, thus i though i would do A*10+B, this is were are those numbers from. Suppose I will close this question. Thanks to all for their time trying to answer this question.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Paolo
Paolo 2018-7-7
You can use strfind for matching patterns in arrays. Read more about it here. (link). The code below defines two patterns, two consecutive -1s and two consecutive 10s.
strfind is used in a loop to search for the pattern in every column. When the pattern is found, the substitution is carried out.
IN= [
9 -1 9 -1 -1
9 9 9 9 9
9 10 9 -1 -1
9 9 9 10 9
9 9 10 10 -1
9 -1 9 9 9
-1 9 -1 9 9
-1 -1 9 9 9];
first_pattern = [-1 -1];
second_pattern = [10 10];
[row,col] = size(IN);
for i = 1:col
IN(strfind(IN(:,i)',first_pattern)+1,i) = 0;
IN(strfind(IN(:,i)',second_pattern)+1,i) = 0;
end
Resulting IN:
IN =
9 -1 9 -1 -1
9 9 9 9 9
9 10 9 -1 -1
9 9 9 10 9
9 9 10 0 -1
9 -1 9 9 9
-1 9 -1 9 9
0 -1 9 9 9

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by