find similar element in a matrix

2 Ansichten (letzte 30 Tage)
Niki
Niki am 2 Mai 2014
Kommentiert: Niki am 3 Mai 2014
I have several questions. First, I have several vectors with different dimensions. I want to put them all in one matrix. I would fill it with zero to make them the same size as follows.
Do you know how to do it?
X= [1 2 3 9 20 67 81 43 101 24;
43 42 88 20 43 67 101 0 0 0;
22 44 0 0 0 0 0 0 0 0;
10 20 67 43 101 0 0 0 0 0;
]
I have a matrix, I want to generate a vector of all elements that are similar in all raws of that matrix. Some raws might have nothing similar as other raws. I count those element similar as if they are similar (e.g. 22 is similar to 22) or they are + - 2 different (e.g. 22 is similar to 24 or 20) the dimension of each raw is different. for example The similarity should be at least between three raws that we select that element.
Then I want to get a vector of all similar elements in all raws as follows
Y=[20 43 67];

Akzeptierte Antwort

Sara
Sara am 2 Mai 2014
First question: create
X = zeros(10,4);
then plug in you vectors, e.g.,
X(1,1:numel(v1)) = v1;
Second point (I'm sure it can be written in a more compact way but it works):
X= [1 2 3 9 20 67 81 43 101 24;...
43 42 88 20 43 67 101 0 0 0;...
22 44 0 0 0 0 0 0 0 0;...
10 20 67 43 101 0 0 0 0 0;...
] ;
n = unique(X(1,:));
X(X==0)=NaN; % So you don't consider the zeros
nrow = size(X,1)-1;
Y = zeros(numel(X),1);cont = 0;
for i = 1:numel(n)
x = X(2:end,:);
[row,~] = find(x >= n(i)-2 & x <= n(i)+2);
if(numel(unique(row)) == nrow)
cont = cont + 1;
Y(cont) = n(i);
end
end
Y = Y(1:cont)
  7 Kommentare
Sara
Sara am 3 Mai 2014
Can you post the vectors you are using and how X looks like after you build it?
Niki
Niki am 3 Mai 2014
I think i found where the problem was. If you check the first line, you are building a matrix of 10*4. however, I succeeded to solve the problem. Thanks. I will accept your code.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Jos (10584)
Jos (10584) am 2 Mai 2014
As for your first question, take a look at my PADCAT function on the File Exchange as it does exactly what you want. http://www.mathworks.com/matlabcentral/fileexchange/22909-padcat
As for your second question, if you consider 22 to be the same as 20 and 24, are 20 and 24 then also the same?
  3 Kommentare
Jos (10584)
Jos (10584) am 3 Mai 2014
So, 1 is the same as 3 which is the same as 4 which is … the same as … is the same as 123456789… !
Are you looking for some kind of cluster analysis? I think you need to be more precise in your goal.
Sara
Sara am 3 Mai 2014
I think you take one element per time from a row and look if there are elements within +-2 in all the other lines. Then you start over with a new element.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by