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Asked by Liber-T on 8 Aug 2011

Hye, I have a matrix 1 by (a big number, like 1000), and I would like to extract the element of the matrix that appears most of the time.

Can you help me?

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Answer by Friedrich on 8 Aug 2011

Accepted answer

Hi,

i think the hist function can help here:

%create some random numbers between 1 and 20 a = ceil(rand(100,1)*20); %put them in as much bin as different numbers exists n = hist(a,numel(unique(a))); %show the histogram hist(a,numel(unique(a))); %number appears most element = find(n == max(n)) %numel(find(a==element)) should return the same as max(n)

Liber-T on 8 Aug 2011

Let say we have this vector: Y= 0.0084 0.0084 0.0084 0.0084 0.0084 0.0084 0.0084 0.0084 0.0084 0.0084 5.9139 5.9432 5.9140 5.9432 5.9434 5.9139

The element that appear the most is 0.0084, and that is the answer I want to extract in this case. I want the program to do it by itself. I used the line x=find(Y==max(Y)), but it give me 15. What have I done wrong?

Answer by Walter Roberson on 8 Aug 2011

Much more simple:

x = mode(Y);

Show 2 older comments

Paulo Silva on 8 Aug 2011

the mode function was introduced in the Service Pack 3 of Release 14 back in Setember 2005, I think it's the first time I see it, thanks for bring it up Walter :)

Liber-T on 9 Aug 2011

function X=modi(Y)

%Y is a vector, it cannot be a matrix, reshape it before entering it in

%modi.

%First, FOR MY NEED, I adjust the value of Y

Y=round(10000*Y)/10000;

%Parameter used in the boucle.

n=length(Y);

%Preallocation of the memories for the matrix T

T(2,n)=0;

%For the first iteration, the first row of T is the different value of Y,

%the second is number of time the value appears.

T(1,1)=Y(1);

T(2,1)=1;

%This is a parameter to know where will go the next new value in the matrix

%T.

s=1;

for j=2:n %First boucle to pass every element of Y.

for i=1:s %Second boucle to compare the value of Y(j) with every value of T.

k=1; %Parameter of control to know if the value exist in T or not

if Y(j)==T(1,i) %Condition to adjust the count of the value Y(j), if not already in T, the value will be added after.

T(2,i)=T(2,i)+1;

k=0;

break

end

end

if k==1 %If the value didn't exist in T, it is now added.

s=s+1;

T(1,s)=Y(j);

T(2,s)=1;

end

end

%Now looking for the most recurrent value by comparing the value of the

%second row.

H=T(2,1);

d=1;

for j=2:s

if H<T(2,j)

H=T(2,j);

d=j;

end

end

%Time to give the answer.

X=T(1,d);

Answer by Paulo Silva on 8 Aug 2011

Another possible way

a=randi([1 20],1,1000); u=unique(a); [C,I]=max(arrayfun(@(x)sum(a==u(x)),1:numel(u))); disp('The value that appears most times is:') u(I) disp('Number of times it appears:') C

In case of having two values that appear the same number of times it will choose just one of them.

Answer by huda nawaf on 10 Aug 2011

hi, see it please,

X=[1 1 1 2 3 4 4 4 0]; >> v=mode(X) ??? No appropriate methods for function mode. . I need the mode which compute frequencies of appearing of number. thanks

Liber-T on 10 Aug 2011

function T=modi(Y)

%Y is a vector, it cannot be a matrix, reshape it before entering it in

%modi.

%Parameter used in the boucle.

n=length(Y);

%Preallocation of the memories for the matrix T

T(2,n)=0;

%For the first iteration, the first row of T is the different value of Y,

%the second is number of time the value appears.

T(1,1)=Y(1);

T(2,1)=1;

%This is a parameter to know where will go the next new value in the matrix

%T.

s=1;

for j=2:n %First boucle to pass every element of Y.

for i=1:s %Second boucle to compare the value of Y(j) with every value of T.

k=1; %Parameter of control to know if the value exist in T or not

if Y(j)==T(1,i) %Condition to adjust the count of the value Y(j), if not already in T, the value will be added after.

T(2,i)=T(2,i)+1;

k=0;

break

end

end

if k==1 %If the value didn't exist in T, it is now added.

s=s+1;

T(1,s)=Y(j);

T(2,s)=1;

end

end

for j=n:-1:1

if T(2,j)==0

T(:,j)=[];

else

break

end

end

I change my code, and this one should return you a matrix of two row. The first row gives you the different number, and the second row return you the frequencies of their appearence. I didn't test it.

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